If $3x^2 -2x+7=0$ then $(x-frac{1}{3})^2 =$?












15












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If $ 3x^{2}-2x+7=0$ then $$left(x-frac{1}{3}right)^2 =text{?} $$



I am so confused. It is a self taught algebra book. The answer is: $ -frac{20}{9}$ but I don't know how it was derived.



Please explain.



Thanks for everyone who commented! I understand it now.






algebra-precalculus quadratics completing-the-square







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  • 4




    $begingroup$
    Hint: Multiply out $(x-1/3)^2$ and compare it to the original equality.
    $endgroup$
    – Michael Burr
    Aug 16 '15 at 17:52






  • 1




    $begingroup$
    You can learn from the two different kinds of solutions below. One strategy is to complete the square and hope to be lucky. That would occur to you if you had some experience knowing when and how to complete the square. The other is to start by multiplying out the sought for square, hoping to find something useful. That's a good strategy if nothing else comes to mind.
    $endgroup$
    – Ethan Bolker
    Aug 16 '15 at 19:50










  • $begingroup$
    Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    $endgroup$
    – Martin Sleziak
    Aug 17 '15 at 9:11
















15












$begingroup$


If $ 3x^{2}-2x+7=0$ then $$left(x-frac{1}{3}right)^2 =text{?} $$



I am so confused. It is a self taught algebra book. The answer is: $ -frac{20}{9}$ but I don't know how it was derived.



Please explain.



Thanks for everyone who commented! I understand it now.






algebra-precalculus quadratics completing-the-square







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Hint: Multiply out $(x-1/3)^2$ and compare it to the original equality.
    $endgroup$
    – Michael Burr
    Aug 16 '15 at 17:52






  • 1




    $begingroup$
    You can learn from the two different kinds of solutions below. One strategy is to complete the square and hope to be lucky. That would occur to you if you had some experience knowing when and how to complete the square. The other is to start by multiplying out the sought for square, hoping to find something useful. That's a good strategy if nothing else comes to mind.
    $endgroup$
    – Ethan Bolker
    Aug 16 '15 at 19:50










  • $begingroup$
    Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    $endgroup$
    – Martin Sleziak
    Aug 17 '15 at 9:11














15












15








15


9



$begingroup$


If $ 3x^{2}-2x+7=0$ then $$left(x-frac{1}{3}right)^2 =text{?} $$



I am so confused. It is a self taught algebra book. The answer is: $ -frac{20}{9}$ but I don't know how it was derived.



Please explain.



Thanks for everyone who commented! I understand it now.






algebra-precalculus quadratics completing-the-square







share|cite|improve this question











$endgroup$




If $ 3x^{2}-2x+7=0$ then $$left(x-frac{1}{3}right)^2 =text{?} $$



I am so confused. It is a self taught algebra book. The answer is: $ -frac{20}{9}$ but I don't know how it was derived.



Please explain.



Thanks for everyone who commented! I understand it now.






algebra-precalculus quadratics completing-the-square




algebra-precalculus quadratics completing-the-square






share|cite|improve this question















share|cite|improve this question













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edited Oct 19 '16 at 12:17









Bhaskara-III

1,1372827




1,1372827










asked Aug 16 '15 at 17:49









NickiNicki

96114




96114








  • 4




    $begingroup$
    Hint: Multiply out $(x-1/3)^2$ and compare it to the original equality.
    $endgroup$
    – Michael Burr
    Aug 16 '15 at 17:52






  • 1




    $begingroup$
    You can learn from the two different kinds of solutions below. One strategy is to complete the square and hope to be lucky. That would occur to you if you had some experience knowing when and how to complete the square. The other is to start by multiplying out the sought for square, hoping to find something useful. That's a good strategy if nothing else comes to mind.
    $endgroup$
    – Ethan Bolker
    Aug 16 '15 at 19:50










  • $begingroup$
    Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    $endgroup$
    – Martin Sleziak
    Aug 17 '15 at 9:11














  • 4




    $begingroup$
    Hint: Multiply out $(x-1/3)^2$ and compare it to the original equality.
    $endgroup$
    – Michael Burr
    Aug 16 '15 at 17:52






  • 1




    $begingroup$
    You can learn from the two different kinds of solutions below. One strategy is to complete the square and hope to be lucky. That would occur to you if you had some experience knowing when and how to complete the square. The other is to start by multiplying out the sought for square, hoping to find something useful. That's a good strategy if nothing else comes to mind.
    $endgroup$
    – Ethan Bolker
    Aug 16 '15 at 19:50










  • $begingroup$
    Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    $endgroup$
    – Martin Sleziak
    Aug 17 '15 at 9:11








4




4




$begingroup$
Hint: Multiply out $(x-1/3)^2$ and compare it to the original equality.
$endgroup$
– Michael Burr
Aug 16 '15 at 17:52




$begingroup$
Hint: Multiply out $(x-1/3)^2$ and compare it to the original equality.
$endgroup$
– Michael Burr
Aug 16 '15 at 17:52




1




1




$begingroup$
You can learn from the two different kinds of solutions below. One strategy is to complete the square and hope to be lucky. That would occur to you if you had some experience knowing when and how to complete the square. The other is to start by multiplying out the sought for square, hoping to find something useful. That's a good strategy if nothing else comes to mind.
$endgroup$
– Ethan Bolker
Aug 16 '15 at 19:50




$begingroup$
You can learn from the two different kinds of solutions below. One strategy is to complete the square and hope to be lucky. That would occur to you if you had some experience knowing when and how to complete the square. The other is to start by multiplying out the sought for square, hoping to find something useful. That's a good strategy if nothing else comes to mind.
$endgroup$
– Ethan Bolker
Aug 16 '15 at 19:50












$begingroup$
Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
$endgroup$
– Martin Sleziak
Aug 17 '15 at 9:11




$begingroup$
Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
$endgroup$
– Martin Sleziak
Aug 17 '15 at 9:11










8 Answers
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Notice, $$3x^2-2x+7=0$$ $$3x^2-2x+frac{1}{3}+7-frac{1}{3}=0$$ $$3left(x^2-frac{2x}{3}+frac{1}{9}right)+7-frac{1}{3}=0$$ $$3left(x-frac{1}{3}right)^2+frac{21-1}{3}=0$$
$$3left(x-frac{1}{3}right)^2=-frac{20}{3}$$
$$left(x-frac{1}{3}right)^2=-frac{20}{9}$$
Hence, we get
$$bbox[5px, border:2px solid #C0A000]{color{red}{left(x-frac{1}{3}right)^2=color{blue}{-frac{20}{9}}}}$$






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  • 2




    $begingroup$
    As is, the first step is not math, it is "magic". There is a "why" missing there.
    $endgroup$
    – Rolazaro Azeveires
    Aug 16 '15 at 22:40






  • 1




    $begingroup$
    @Rol looks like my edit went through that added the missing step so it's not as magical now
    $endgroup$
    – Kevin Brown
    Aug 17 '15 at 2:05






  • 2




    $begingroup$
    Yes, you are right
    $endgroup$
    – Harish Chandra Rajpoot
    Aug 17 '15 at 2:14






  • 1




    $begingroup$
    @Harish yes. But I still think Burr's answer is clearer because it it is clearer that we are aiming to get to (x-1/3)^2
    $endgroup$
    – Rolazaro Azeveires
    Aug 18 '15 at 2:08






  • 4




    $begingroup$
    (+1) But if roots were assumed real, this problem made no sense.
    $endgroup$
    – Amad27
    Oct 20 '15 at 12:52



















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Observe $$left(x-frac{1}{3}right)^2=x^2-frac{2}{3}x+frac{1}{9}$$$$=frac{1}{3}left(3x^2-2xright)+frac{1}{9}.$$
This is almost the original expression, we're just missing a $7$. Then,
$$left(x-frac{1}{3}right)^2=frac{1}{3}left(3x^2-2x+7-7right)+frac{1}{9}.$$
Now, use the original equality to simplify.
Then, we get
$$
left(x-frac{1}{3}right)^2=frac{1}{3}left(3x^2-2x+7-7right)+frac{1}{9}$$ $$=-frac{7}{3}+frac{1}{9}
$$ $$=-frac{20}{9}
$$






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$endgroup$





















    8












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    $$3x^2-2x+7=0Longleftrightarrow$$
    $$x=frac{-(-2)pmsqrt{(-2)^2-4cdot 3 cdot 7}}{2cdot 3}Longleftrightarrow$$
    $$x=frac{2pmsqrt{4-4cdot 3 cdot 7}}{6}Longleftrightarrow$$
    $$x=frac{2pmsqrt{4-84}}{6}Longleftrightarrow$$
    $$x=frac{2pmsqrt{-80}}{6}Longleftrightarrow$$
    $$x=frac{2pm isqrt{80}}{6}Longleftrightarrow$$
    $$x=frac{2pm 4isqrt{5}}{6}Longleftrightarrow$$
    $$x=frac{2 + 4isqrt{5}}{6} vee x=frac{2 - 4isqrt{5}}{6}Longleftrightarrow$$
    $$x=frac{2 + 4isqrt{5}}{6} vee x=frac{2 - 4isqrt{5}}{6}$$




    $$left(left(frac{2 + 4isqrt{5}}{6}right)-frac{1}{3}right)^2 =left(frac{2isqrt{5}}{3}right)^2 =frac{4i^2cdot 5}{9}=-frac{20}{9}$$



    $$left(left(frac{2 - 4isqrt{5}}{6}right)-frac{1}{3}right)^2 =left(frac{-2isqrt{5}}{3}right)^2=frac{4i^2cdot 5}{9}=-frac{20}{9}$$



    So as we see the answer is $color{red}{-frac{20}{9}}$






    share|cite|improve this answer









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    • 1




      $begingroup$
      I actually like the brute force approach for this one. Sometimes we need to get our hands "dirty"
      $endgroup$
      – Rolazaro Azeveires
      Aug 16 '15 at 22:43






    • 3




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      A pretty inefficient approach, and what if the OP doesn't know about complex numbers ?
      $endgroup$
      – Yves Daoust
      Sep 21 '16 at 13:07






    • 1




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      I also like the brute force. Of course, not as elegant as other solutions, but very very straightforward. And you can only find elegant solutions with experience.
      $endgroup$
      – Adrian
      Oct 21 '16 at 10:10










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      @YvesDaoust, if $x$ is supposed to be real, then $42$ is another valid answer.
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      – Carsten S
      Jan 10 '17 at 18:44










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      @CarstenS: of course. This exercise is about "completing the square" and can be solved without caring about $x$.
      $endgroup$
      – Yves Daoust
      Jan 10 '17 at 20:12





















    4












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    HINT:
    complete the square in $$3x^2-2x+7$$






    share|cite|improve this answer











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    • $begingroup$
      barak manos has given good hint
      $endgroup$
      – Bhaskara-III
      Sep 11 '16 at 12:10










    • $begingroup$
      i have edited a little
      $endgroup$
      – Bhaskara-III
      Sep 21 '16 at 12:53










    • $begingroup$
      Obviously, the OP is precisely stuck because he is hasn't learnt how to complete the square.
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      – Yves Daoust
      Sep 21 '16 at 12:55










    • $begingroup$
      @YvesDaoust why is that so obvious? If this is from a book, why couldn't the chapter this question is taken from be about completing the square? It might be that the OP just didn't know how to proceed while still being able to complete the square; he could have just not seen the advantage of this manipulation.
      $endgroup$
      – Hirshy
      Sep 21 '16 at 13:11



















    2












    $begingroup$

    Starting from



    $$3x^2-2x+7=0\$$
    $$x^2-frac{2}{3}x+frac{7}{3}=0\$$
    $$x^2-2cdotfrac{1}{3}cdot x+left(frac{1}{3}right)^2-left(frac{1}{3}right)^2+frac{7}{3}=0\$$
    $$left(x-frac{1}{3}right)^2+frac{21-1}{9}=0\$$
    $$ left(x-frac{1}{3}right)^2=frac{-20}{9}$$






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      HINT:



      $$3x^2-2x+7=3left(x-frac13right)^2+6+frac23$$






      share|cite|improve this answer











      $endgroup$









      • 3




        $begingroup$
        In general, I find the notation $6frac{2}{3}$ to be a source of confusion because it is unclear if one means $6+frac{2}{3}$ or $6cdotfrac{2}{3}$. I strongly suggest either improper fractions or explicitly writing out the operation instead of mixed fractions.
        $endgroup$
        – Michael Burr
        Aug 16 '15 at 20:11










      • $begingroup$
        @MichaelBurr: I thought about it while writing the answer, but figured it could hardly be seen as $6cdotfrac23$ without explicitly using the dot. In any case, I followed your suggestion and added a plus in order to avoid any possible confusion. Thanks.
        $endgroup$
        – barak manos
        Aug 16 '15 at 20:25



















      2












      $begingroup$

      Starting from, $$3x^2-2x+7=0$$ Dividing by 3:$$x^2-frac{2x}{3}+frac{7}{3}=0$$ Completing the square: $$left(x-frac 13right)^2-frac{1}{9}+frac{7}{3}=0$$ $$left(x-frac 13right)^2=frac{1}{9}-frac{7}{3}$$ $$left(x-frac 13right)^2=frac{1-21}{9}$$
      $$left(x-frac 13right)^2=-frac{20}{9}$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        I don't think further editing here is needed.
        $endgroup$
        – Daniel Fischer♦
        Jan 17 '17 at 10:40



















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      Expand the binomial and compare it to the trinomial.



      $$3x^2-2x+7iffleft(x-frac13right)^2=x^2-frac23x+frac19.$$



      If you divide the polynomial by $3$, you get closer, with two identical terms



      $$frac{3x^2-2x+7}3=x^2-frac23x+frac73.$$



      To get a perfect identity, it now suffices to add a well-chosen constant



      $$frac{3x^2-2x+7}3-frac{20}9=x^2-frac23x+frac19.$$



      Now as the polynomial is known to evaluate to $0$, you know the value of the RHS.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        As you removed my edit: $3x^2-2x+7$ can't be equivalent to anything, you are missing the $ldots =0$.
        $endgroup$
        – Hirshy
        Sep 21 '16 at 13:28










      • $begingroup$
        @Hirshy: no, you missed the intent. The double arrow relates the two expressions that are to be compared. Appending $=0$ turns the left one in an equation, while the right one is an identity, and this would result in a false proposition.
        $endgroup$
        – Yves Daoust
        Sep 21 '16 at 13:32












      • $begingroup$
        I did indeed miss that, sorry. You might want to use another arrow for your purpose?
        $endgroup$
        – Hirshy
        Sep 21 '16 at 13:35










      • $begingroup$
        @Hirshy: I didn't find one that pleased me.
        $endgroup$
        – Yves Daoust
        Sep 21 '16 at 13:54










      protected by Alex M. Nov 19 '16 at 19:43



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      8 Answers
      8






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      8 Answers
      8






      active

      oldest

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      active

      oldest

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      active

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      26












      $begingroup$

      Notice, $$3x^2-2x+7=0$$ $$3x^2-2x+frac{1}{3}+7-frac{1}{3}=0$$ $$3left(x^2-frac{2x}{3}+frac{1}{9}right)+7-frac{1}{3}=0$$ $$3left(x-frac{1}{3}right)^2+frac{21-1}{3}=0$$
      $$3left(x-frac{1}{3}right)^2=-frac{20}{3}$$
      $$left(x-frac{1}{3}right)^2=-frac{20}{9}$$
      Hence, we get
      $$bbox[5px, border:2px solid #C0A000]{color{red}{left(x-frac{1}{3}right)^2=color{blue}{-frac{20}{9}}}}$$






      share|cite|improve this answer











      $endgroup$









      • 2




        $begingroup$
        As is, the first step is not math, it is "magic". There is a "why" missing there.
        $endgroup$
        – Rolazaro Azeveires
        Aug 16 '15 at 22:40






      • 1




        $begingroup$
        @Rol looks like my edit went through that added the missing step so it's not as magical now
        $endgroup$
        – Kevin Brown
        Aug 17 '15 at 2:05






      • 2




        $begingroup$
        Yes, you are right
        $endgroup$
        – Harish Chandra Rajpoot
        Aug 17 '15 at 2:14






      • 1




        $begingroup$
        @Harish yes. But I still think Burr's answer is clearer because it it is clearer that we are aiming to get to (x-1/3)^2
        $endgroup$
        – Rolazaro Azeveires
        Aug 18 '15 at 2:08






      • 4




        $begingroup$
        (+1) But if roots were assumed real, this problem made no sense.
        $endgroup$
        – Amad27
        Oct 20 '15 at 12:52
















      26












      $begingroup$

      Notice, $$3x^2-2x+7=0$$ $$3x^2-2x+frac{1}{3}+7-frac{1}{3}=0$$ $$3left(x^2-frac{2x}{3}+frac{1}{9}right)+7-frac{1}{3}=0$$ $$3left(x-frac{1}{3}right)^2+frac{21-1}{3}=0$$
      $$3left(x-frac{1}{3}right)^2=-frac{20}{3}$$
      $$left(x-frac{1}{3}right)^2=-frac{20}{9}$$
      Hence, we get
      $$bbox[5px, border:2px solid #C0A000]{color{red}{left(x-frac{1}{3}right)^2=color{blue}{-frac{20}{9}}}}$$






      share|cite|improve this answer











      $endgroup$









      • 2




        $begingroup$
        As is, the first step is not math, it is "magic". There is a "why" missing there.
        $endgroup$
        – Rolazaro Azeveires
        Aug 16 '15 at 22:40






      • 1




        $begingroup$
        @Rol looks like my edit went through that added the missing step so it's not as magical now
        $endgroup$
        – Kevin Brown
        Aug 17 '15 at 2:05






      • 2




        $begingroup$
        Yes, you are right
        $endgroup$
        – Harish Chandra Rajpoot
        Aug 17 '15 at 2:14






      • 1




        $begingroup$
        @Harish yes. But I still think Burr's answer is clearer because it it is clearer that we are aiming to get to (x-1/3)^2
        $endgroup$
        – Rolazaro Azeveires
        Aug 18 '15 at 2:08






      • 4




        $begingroup$
        (+1) But if roots were assumed real, this problem made no sense.
        $endgroup$
        – Amad27
        Oct 20 '15 at 12:52














      26












      26








      26





      $begingroup$

      Notice, $$3x^2-2x+7=0$$ $$3x^2-2x+frac{1}{3}+7-frac{1}{3}=0$$ $$3left(x^2-frac{2x}{3}+frac{1}{9}right)+7-frac{1}{3}=0$$ $$3left(x-frac{1}{3}right)^2+frac{21-1}{3}=0$$
      $$3left(x-frac{1}{3}right)^2=-frac{20}{3}$$
      $$left(x-frac{1}{3}right)^2=-frac{20}{9}$$
      Hence, we get
      $$bbox[5px, border:2px solid #C0A000]{color{red}{left(x-frac{1}{3}right)^2=color{blue}{-frac{20}{9}}}}$$






      share|cite|improve this answer











      $endgroup$



      Notice, $$3x^2-2x+7=0$$ $$3x^2-2x+frac{1}{3}+7-frac{1}{3}=0$$ $$3left(x^2-frac{2x}{3}+frac{1}{9}right)+7-frac{1}{3}=0$$ $$3left(x-frac{1}{3}right)^2+frac{21-1}{3}=0$$
      $$3left(x-frac{1}{3}right)^2=-frac{20}{3}$$
      $$left(x-frac{1}{3}right)^2=-frac{20}{9}$$
      Hence, we get
      $$bbox[5px, border:2px solid #C0A000]{color{red}{left(x-frac{1}{3}right)^2=color{blue}{-frac{20}{9}}}}$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Aug 17 '15 at 0:38









      Kevin Brown

      1034




      1034










      answered Aug 16 '15 at 17:56









      Harish Chandra RajpootHarish Chandra Rajpoot

      29.6k103671




      29.6k103671








      • 2




        $begingroup$
        As is, the first step is not math, it is "magic". There is a "why" missing there.
        $endgroup$
        – Rolazaro Azeveires
        Aug 16 '15 at 22:40






      • 1




        $begingroup$
        @Rol looks like my edit went through that added the missing step so it's not as magical now
        $endgroup$
        – Kevin Brown
        Aug 17 '15 at 2:05






      • 2




        $begingroup$
        Yes, you are right
        $endgroup$
        – Harish Chandra Rajpoot
        Aug 17 '15 at 2:14






      • 1




        $begingroup$
        @Harish yes. But I still think Burr's answer is clearer because it it is clearer that we are aiming to get to (x-1/3)^2
        $endgroup$
        – Rolazaro Azeveires
        Aug 18 '15 at 2:08






      • 4




        $begingroup$
        (+1) But if roots were assumed real, this problem made no sense.
        $endgroup$
        – Amad27
        Oct 20 '15 at 12:52














      • 2




        $begingroup$
        As is, the first step is not math, it is "magic". There is a "why" missing there.
        $endgroup$
        – Rolazaro Azeveires
        Aug 16 '15 at 22:40






      • 1




        $begingroup$
        @Rol looks like my edit went through that added the missing step so it's not as magical now
        $endgroup$
        – Kevin Brown
        Aug 17 '15 at 2:05






      • 2




        $begingroup$
        Yes, you are right
        $endgroup$
        – Harish Chandra Rajpoot
        Aug 17 '15 at 2:14






      • 1




        $begingroup$
        @Harish yes. But I still think Burr's answer is clearer because it it is clearer that we are aiming to get to (x-1/3)^2
        $endgroup$
        – Rolazaro Azeveires
        Aug 18 '15 at 2:08






      • 4




        $begingroup$
        (+1) But if roots were assumed real, this problem made no sense.
        $endgroup$
        – Amad27
        Oct 20 '15 at 12:52








      2




      2




      $begingroup$
      As is, the first step is not math, it is "magic". There is a "why" missing there.
      $endgroup$
      – Rolazaro Azeveires
      Aug 16 '15 at 22:40




      $begingroup$
      As is, the first step is not math, it is "magic". There is a "why" missing there.
      $endgroup$
      – Rolazaro Azeveires
      Aug 16 '15 at 22:40




      1




      1




      $begingroup$
      @Rol looks like my edit went through that added the missing step so it's not as magical now
      $endgroup$
      – Kevin Brown
      Aug 17 '15 at 2:05




      $begingroup$
      @Rol looks like my edit went through that added the missing step so it's not as magical now
      $endgroup$
      – Kevin Brown
      Aug 17 '15 at 2:05




      2




      2




      $begingroup$
      Yes, you are right
      $endgroup$
      – Harish Chandra Rajpoot
      Aug 17 '15 at 2:14




      $begingroup$
      Yes, you are right
      $endgroup$
      – Harish Chandra Rajpoot
      Aug 17 '15 at 2:14




      1




      1




      $begingroup$
      @Harish yes. But I still think Burr's answer is clearer because it it is clearer that we are aiming to get to (x-1/3)^2
      $endgroup$
      – Rolazaro Azeveires
      Aug 18 '15 at 2:08




      $begingroup$
      @Harish yes. But I still think Burr's answer is clearer because it it is clearer that we are aiming to get to (x-1/3)^2
      $endgroup$
      – Rolazaro Azeveires
      Aug 18 '15 at 2:08




      4




      4




      $begingroup$
      (+1) But if roots were assumed real, this problem made no sense.
      $endgroup$
      – Amad27
      Oct 20 '15 at 12:52




      $begingroup$
      (+1) But if roots were assumed real, this problem made no sense.
      $endgroup$
      – Amad27
      Oct 20 '15 at 12:52











      21












      $begingroup$

      Observe $$left(x-frac{1}{3}right)^2=x^2-frac{2}{3}x+frac{1}{9}$$$$=frac{1}{3}left(3x^2-2xright)+frac{1}{9}.$$
      This is almost the original expression, we're just missing a $7$. Then,
      $$left(x-frac{1}{3}right)^2=frac{1}{3}left(3x^2-2x+7-7right)+frac{1}{9}.$$
      Now, use the original equality to simplify.
      Then, we get
      $$
      left(x-frac{1}{3}right)^2=frac{1}{3}left(3x^2-2x+7-7right)+frac{1}{9}$$ $$=-frac{7}{3}+frac{1}{9}
      $$ $$=-frac{20}{9}
      $$






      share|cite|improve this answer











      $endgroup$


















        21












        $begingroup$

        Observe $$left(x-frac{1}{3}right)^2=x^2-frac{2}{3}x+frac{1}{9}$$$$=frac{1}{3}left(3x^2-2xright)+frac{1}{9}.$$
        This is almost the original expression, we're just missing a $7$. Then,
        $$left(x-frac{1}{3}right)^2=frac{1}{3}left(3x^2-2x+7-7right)+frac{1}{9}.$$
        Now, use the original equality to simplify.
        Then, we get
        $$
        left(x-frac{1}{3}right)^2=frac{1}{3}left(3x^2-2x+7-7right)+frac{1}{9}$$ $$=-frac{7}{3}+frac{1}{9}
        $$ $$=-frac{20}{9}
        $$






        share|cite|improve this answer











        $endgroup$
















          21












          21








          21





          $begingroup$

          Observe $$left(x-frac{1}{3}right)^2=x^2-frac{2}{3}x+frac{1}{9}$$$$=frac{1}{3}left(3x^2-2xright)+frac{1}{9}.$$
          This is almost the original expression, we're just missing a $7$. Then,
          $$left(x-frac{1}{3}right)^2=frac{1}{3}left(3x^2-2x+7-7right)+frac{1}{9}.$$
          Now, use the original equality to simplify.
          Then, we get
          $$
          left(x-frac{1}{3}right)^2=frac{1}{3}left(3x^2-2x+7-7right)+frac{1}{9}$$ $$=-frac{7}{3}+frac{1}{9}
          $$ $$=-frac{20}{9}
          $$






          share|cite|improve this answer











          $endgroup$



          Observe $$left(x-frac{1}{3}right)^2=x^2-frac{2}{3}x+frac{1}{9}$$$$=frac{1}{3}left(3x^2-2xright)+frac{1}{9}.$$
          This is almost the original expression, we're just missing a $7$. Then,
          $$left(x-frac{1}{3}right)^2=frac{1}{3}left(3x^2-2x+7-7right)+frac{1}{9}.$$
          Now, use the original equality to simplify.
          Then, we get
          $$
          left(x-frac{1}{3}right)^2=frac{1}{3}left(3x^2-2x+7-7right)+frac{1}{9}$$ $$=-frac{7}{3}+frac{1}{9}
          $$ $$=-frac{20}{9}
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Sep 22 '16 at 9:40









          Bhaskara-III

          1,1372827




          1,1372827










          answered Aug 16 '15 at 17:55









          Michael BurrMichael Burr

          26.7k23262




          26.7k23262























              8












              $begingroup$

              $$3x^2-2x+7=0Longleftrightarrow$$
              $$x=frac{-(-2)pmsqrt{(-2)^2-4cdot 3 cdot 7}}{2cdot 3}Longleftrightarrow$$
              $$x=frac{2pmsqrt{4-4cdot 3 cdot 7}}{6}Longleftrightarrow$$
              $$x=frac{2pmsqrt{4-84}}{6}Longleftrightarrow$$
              $$x=frac{2pmsqrt{-80}}{6}Longleftrightarrow$$
              $$x=frac{2pm isqrt{80}}{6}Longleftrightarrow$$
              $$x=frac{2pm 4isqrt{5}}{6}Longleftrightarrow$$
              $$x=frac{2 + 4isqrt{5}}{6} vee x=frac{2 - 4isqrt{5}}{6}Longleftrightarrow$$
              $$x=frac{2 + 4isqrt{5}}{6} vee x=frac{2 - 4isqrt{5}}{6}$$




              $$left(left(frac{2 + 4isqrt{5}}{6}right)-frac{1}{3}right)^2 =left(frac{2isqrt{5}}{3}right)^2 =frac{4i^2cdot 5}{9}=-frac{20}{9}$$



              $$left(left(frac{2 - 4isqrt{5}}{6}right)-frac{1}{3}right)^2 =left(frac{-2isqrt{5}}{3}right)^2=frac{4i^2cdot 5}{9}=-frac{20}{9}$$



              So as we see the answer is $color{red}{-frac{20}{9}}$






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                I actually like the brute force approach for this one. Sometimes we need to get our hands "dirty"
                $endgroup$
                – Rolazaro Azeveires
                Aug 16 '15 at 22:43






              • 3




                $begingroup$
                A pretty inefficient approach, and what if the OP doesn't know about complex numbers ?
                $endgroup$
                – Yves Daoust
                Sep 21 '16 at 13:07






              • 1




                $begingroup$
                I also like the brute force. Of course, not as elegant as other solutions, but very very straightforward. And you can only find elegant solutions with experience.
                $endgroup$
                – Adrian
                Oct 21 '16 at 10:10










              • $begingroup$
                @YvesDaoust, if $x$ is supposed to be real, then $42$ is another valid answer.
                $endgroup$
                – Carsten S
                Jan 10 '17 at 18:44










              • $begingroup$
                @CarstenS: of course. This exercise is about "completing the square" and can be solved without caring about $x$.
                $endgroup$
                – Yves Daoust
                Jan 10 '17 at 20:12


















              8












              $begingroup$

              $$3x^2-2x+7=0Longleftrightarrow$$
              $$x=frac{-(-2)pmsqrt{(-2)^2-4cdot 3 cdot 7}}{2cdot 3}Longleftrightarrow$$
              $$x=frac{2pmsqrt{4-4cdot 3 cdot 7}}{6}Longleftrightarrow$$
              $$x=frac{2pmsqrt{4-84}}{6}Longleftrightarrow$$
              $$x=frac{2pmsqrt{-80}}{6}Longleftrightarrow$$
              $$x=frac{2pm isqrt{80}}{6}Longleftrightarrow$$
              $$x=frac{2pm 4isqrt{5}}{6}Longleftrightarrow$$
              $$x=frac{2 + 4isqrt{5}}{6} vee x=frac{2 - 4isqrt{5}}{6}Longleftrightarrow$$
              $$x=frac{2 + 4isqrt{5}}{6} vee x=frac{2 - 4isqrt{5}}{6}$$




              $$left(left(frac{2 + 4isqrt{5}}{6}right)-frac{1}{3}right)^2 =left(frac{2isqrt{5}}{3}right)^2 =frac{4i^2cdot 5}{9}=-frac{20}{9}$$



              $$left(left(frac{2 - 4isqrt{5}}{6}right)-frac{1}{3}right)^2 =left(frac{-2isqrt{5}}{3}right)^2=frac{4i^2cdot 5}{9}=-frac{20}{9}$$



              So as we see the answer is $color{red}{-frac{20}{9}}$






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                I actually like the brute force approach for this one. Sometimes we need to get our hands "dirty"
                $endgroup$
                – Rolazaro Azeveires
                Aug 16 '15 at 22:43






              • 3




                $begingroup$
                A pretty inefficient approach, and what if the OP doesn't know about complex numbers ?
                $endgroup$
                – Yves Daoust
                Sep 21 '16 at 13:07






              • 1




                $begingroup$
                I also like the brute force. Of course, not as elegant as other solutions, but very very straightforward. And you can only find elegant solutions with experience.
                $endgroup$
                – Adrian
                Oct 21 '16 at 10:10










              • $begingroup$
                @YvesDaoust, if $x$ is supposed to be real, then $42$ is another valid answer.
                $endgroup$
                – Carsten S
                Jan 10 '17 at 18:44










              • $begingroup$
                @CarstenS: of course. This exercise is about "completing the square" and can be solved without caring about $x$.
                $endgroup$
                – Yves Daoust
                Jan 10 '17 at 20:12
















              8












              8








              8





              $begingroup$

              $$3x^2-2x+7=0Longleftrightarrow$$
              $$x=frac{-(-2)pmsqrt{(-2)^2-4cdot 3 cdot 7}}{2cdot 3}Longleftrightarrow$$
              $$x=frac{2pmsqrt{4-4cdot 3 cdot 7}}{6}Longleftrightarrow$$
              $$x=frac{2pmsqrt{4-84}}{6}Longleftrightarrow$$
              $$x=frac{2pmsqrt{-80}}{6}Longleftrightarrow$$
              $$x=frac{2pm isqrt{80}}{6}Longleftrightarrow$$
              $$x=frac{2pm 4isqrt{5}}{6}Longleftrightarrow$$
              $$x=frac{2 + 4isqrt{5}}{6} vee x=frac{2 - 4isqrt{5}}{6}Longleftrightarrow$$
              $$x=frac{2 + 4isqrt{5}}{6} vee x=frac{2 - 4isqrt{5}}{6}$$




              $$left(left(frac{2 + 4isqrt{5}}{6}right)-frac{1}{3}right)^2 =left(frac{2isqrt{5}}{3}right)^2 =frac{4i^2cdot 5}{9}=-frac{20}{9}$$



              $$left(left(frac{2 - 4isqrt{5}}{6}right)-frac{1}{3}right)^2 =left(frac{-2isqrt{5}}{3}right)^2=frac{4i^2cdot 5}{9}=-frac{20}{9}$$



              So as we see the answer is $color{red}{-frac{20}{9}}$






              share|cite|improve this answer









              $endgroup$



              $$3x^2-2x+7=0Longleftrightarrow$$
              $$x=frac{-(-2)pmsqrt{(-2)^2-4cdot 3 cdot 7}}{2cdot 3}Longleftrightarrow$$
              $$x=frac{2pmsqrt{4-4cdot 3 cdot 7}}{6}Longleftrightarrow$$
              $$x=frac{2pmsqrt{4-84}}{6}Longleftrightarrow$$
              $$x=frac{2pmsqrt{-80}}{6}Longleftrightarrow$$
              $$x=frac{2pm isqrt{80}}{6}Longleftrightarrow$$
              $$x=frac{2pm 4isqrt{5}}{6}Longleftrightarrow$$
              $$x=frac{2 + 4isqrt{5}}{6} vee x=frac{2 - 4isqrt{5}}{6}Longleftrightarrow$$
              $$x=frac{2 + 4isqrt{5}}{6} vee x=frac{2 - 4isqrt{5}}{6}$$




              $$left(left(frac{2 + 4isqrt{5}}{6}right)-frac{1}{3}right)^2 =left(frac{2isqrt{5}}{3}right)^2 =frac{4i^2cdot 5}{9}=-frac{20}{9}$$



              $$left(left(frac{2 - 4isqrt{5}}{6}right)-frac{1}{3}right)^2 =left(frac{-2isqrt{5}}{3}right)^2=frac{4i^2cdot 5}{9}=-frac{20}{9}$$



              So as we see the answer is $color{red}{-frac{20}{9}}$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Aug 16 '15 at 20:57









              JanJan

              21.8k31240




              21.8k31240








              • 1




                $begingroup$
                I actually like the brute force approach for this one. Sometimes we need to get our hands "dirty"
                $endgroup$
                – Rolazaro Azeveires
                Aug 16 '15 at 22:43






              • 3




                $begingroup$
                A pretty inefficient approach, and what if the OP doesn't know about complex numbers ?
                $endgroup$
                – Yves Daoust
                Sep 21 '16 at 13:07






              • 1




                $begingroup$
                I also like the brute force. Of course, not as elegant as other solutions, but very very straightforward. And you can only find elegant solutions with experience.
                $endgroup$
                – Adrian
                Oct 21 '16 at 10:10










              • $begingroup$
                @YvesDaoust, if $x$ is supposed to be real, then $42$ is another valid answer.
                $endgroup$
                – Carsten S
                Jan 10 '17 at 18:44










              • $begingroup$
                @CarstenS: of course. This exercise is about "completing the square" and can be solved without caring about $x$.
                $endgroup$
                – Yves Daoust
                Jan 10 '17 at 20:12
















              • 1




                $begingroup$
                I actually like the brute force approach for this one. Sometimes we need to get our hands "dirty"
                $endgroup$
                – Rolazaro Azeveires
                Aug 16 '15 at 22:43






              • 3




                $begingroup$
                A pretty inefficient approach, and what if the OP doesn't know about complex numbers ?
                $endgroup$
                – Yves Daoust
                Sep 21 '16 at 13:07






              • 1




                $begingroup$
                I also like the brute force. Of course, not as elegant as other solutions, but very very straightforward. And you can only find elegant solutions with experience.
                $endgroup$
                – Adrian
                Oct 21 '16 at 10:10










              • $begingroup$
                @YvesDaoust, if $x$ is supposed to be real, then $42$ is another valid answer.
                $endgroup$
                – Carsten S
                Jan 10 '17 at 18:44










              • $begingroup$
                @CarstenS: of course. This exercise is about "completing the square" and can be solved without caring about $x$.
                $endgroup$
                – Yves Daoust
                Jan 10 '17 at 20:12










              1




              1




              $begingroup$
              I actually like the brute force approach for this one. Sometimes we need to get our hands "dirty"
              $endgroup$
              – Rolazaro Azeveires
              Aug 16 '15 at 22:43




              $begingroup$
              I actually like the brute force approach for this one. Sometimes we need to get our hands "dirty"
              $endgroup$
              – Rolazaro Azeveires
              Aug 16 '15 at 22:43




              3




              3




              $begingroup$
              A pretty inefficient approach, and what if the OP doesn't know about complex numbers ?
              $endgroup$
              – Yves Daoust
              Sep 21 '16 at 13:07




              $begingroup$
              A pretty inefficient approach, and what if the OP doesn't know about complex numbers ?
              $endgroup$
              – Yves Daoust
              Sep 21 '16 at 13:07




              1




              1




              $begingroup$
              I also like the brute force. Of course, not as elegant as other solutions, but very very straightforward. And you can only find elegant solutions with experience.
              $endgroup$
              – Adrian
              Oct 21 '16 at 10:10




              $begingroup$
              I also like the brute force. Of course, not as elegant as other solutions, but very very straightforward. And you can only find elegant solutions with experience.
              $endgroup$
              – Adrian
              Oct 21 '16 at 10:10












              $begingroup$
              @YvesDaoust, if $x$ is supposed to be real, then $42$ is another valid answer.
              $endgroup$
              – Carsten S
              Jan 10 '17 at 18:44




              $begingroup$
              @YvesDaoust, if $x$ is supposed to be real, then $42$ is another valid answer.
              $endgroup$
              – Carsten S
              Jan 10 '17 at 18:44












              $begingroup$
              @CarstenS: of course. This exercise is about "completing the square" and can be solved without caring about $x$.
              $endgroup$
              – Yves Daoust
              Jan 10 '17 at 20:12






              $begingroup$
              @CarstenS: of course. This exercise is about "completing the square" and can be solved without caring about $x$.
              $endgroup$
              – Yves Daoust
              Jan 10 '17 at 20:12













              4












              $begingroup$

              HINT:
              complete the square in $$3x^2-2x+7$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                barak manos has given good hint
                $endgroup$
                – Bhaskara-III
                Sep 11 '16 at 12:10










              • $begingroup$
                i have edited a little
                $endgroup$
                – Bhaskara-III
                Sep 21 '16 at 12:53










              • $begingroup$
                Obviously, the OP is precisely stuck because he is hasn't learnt how to complete the square.
                $endgroup$
                – Yves Daoust
                Sep 21 '16 at 12:55










              • $begingroup$
                @YvesDaoust why is that so obvious? If this is from a book, why couldn't the chapter this question is taken from be about completing the square? It might be that the OP just didn't know how to proceed while still being able to complete the square; he could have just not seen the advantage of this manipulation.
                $endgroup$
                – Hirshy
                Sep 21 '16 at 13:11
















              4












              $begingroup$

              HINT:
              complete the square in $$3x^2-2x+7$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                barak manos has given good hint
                $endgroup$
                – Bhaskara-III
                Sep 11 '16 at 12:10










              • $begingroup$
                i have edited a little
                $endgroup$
                – Bhaskara-III
                Sep 21 '16 at 12:53










              • $begingroup$
                Obviously, the OP is precisely stuck because he is hasn't learnt how to complete the square.
                $endgroup$
                – Yves Daoust
                Sep 21 '16 at 12:55










              • $begingroup$
                @YvesDaoust why is that so obvious? If this is from a book, why couldn't the chapter this question is taken from be about completing the square? It might be that the OP just didn't know how to proceed while still being able to complete the square; he could have just not seen the advantage of this manipulation.
                $endgroup$
                – Hirshy
                Sep 21 '16 at 13:11














              4












              4








              4





              $begingroup$

              HINT:
              complete the square in $$3x^2-2x+7$$






              share|cite|improve this answer











              $endgroup$



              HINT:
              complete the square in $$3x^2-2x+7$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Sep 21 '16 at 12:51









              Bhaskara-III

              1,1372827




              1,1372827










              answered Aug 16 '15 at 17:53









              HirshyHirshy

              4,26121339




              4,26121339












              • $begingroup$
                barak manos has given good hint
                $endgroup$
                – Bhaskara-III
                Sep 11 '16 at 12:10










              • $begingroup$
                i have edited a little
                $endgroup$
                – Bhaskara-III
                Sep 21 '16 at 12:53










              • $begingroup$
                Obviously, the OP is precisely stuck because he is hasn't learnt how to complete the square.
                $endgroup$
                – Yves Daoust
                Sep 21 '16 at 12:55










              • $begingroup$
                @YvesDaoust why is that so obvious? If this is from a book, why couldn't the chapter this question is taken from be about completing the square? It might be that the OP just didn't know how to proceed while still being able to complete the square; he could have just not seen the advantage of this manipulation.
                $endgroup$
                – Hirshy
                Sep 21 '16 at 13:11


















              • $begingroup$
                barak manos has given good hint
                $endgroup$
                – Bhaskara-III
                Sep 11 '16 at 12:10










              • $begingroup$
                i have edited a little
                $endgroup$
                – Bhaskara-III
                Sep 21 '16 at 12:53










              • $begingroup$
                Obviously, the OP is precisely stuck because he is hasn't learnt how to complete the square.
                $endgroup$
                – Yves Daoust
                Sep 21 '16 at 12:55










              • $begingroup$
                @YvesDaoust why is that so obvious? If this is from a book, why couldn't the chapter this question is taken from be about completing the square? It might be that the OP just didn't know how to proceed while still being able to complete the square; he could have just not seen the advantage of this manipulation.
                $endgroup$
                – Hirshy
                Sep 21 '16 at 13:11
















              $begingroup$
              barak manos has given good hint
              $endgroup$
              – Bhaskara-III
              Sep 11 '16 at 12:10




              $begingroup$
              barak manos has given good hint
              $endgroup$
              – Bhaskara-III
              Sep 11 '16 at 12:10












              $begingroup$
              i have edited a little
              $endgroup$
              – Bhaskara-III
              Sep 21 '16 at 12:53




              $begingroup$
              i have edited a little
              $endgroup$
              – Bhaskara-III
              Sep 21 '16 at 12:53












              $begingroup$
              Obviously, the OP is precisely stuck because he is hasn't learnt how to complete the square.
              $endgroup$
              – Yves Daoust
              Sep 21 '16 at 12:55




              $begingroup$
              Obviously, the OP is precisely stuck because he is hasn't learnt how to complete the square.
              $endgroup$
              – Yves Daoust
              Sep 21 '16 at 12:55












              $begingroup$
              @YvesDaoust why is that so obvious? If this is from a book, why couldn't the chapter this question is taken from be about completing the square? It might be that the OP just didn't know how to proceed while still being able to complete the square; he could have just not seen the advantage of this manipulation.
              $endgroup$
              – Hirshy
              Sep 21 '16 at 13:11




              $begingroup$
              @YvesDaoust why is that so obvious? If this is from a book, why couldn't the chapter this question is taken from be about completing the square? It might be that the OP just didn't know how to proceed while still being able to complete the square; he could have just not seen the advantage of this manipulation.
              $endgroup$
              – Hirshy
              Sep 21 '16 at 13:11











              2












              $begingroup$

              Starting from



              $$3x^2-2x+7=0\$$
              $$x^2-frac{2}{3}x+frac{7}{3}=0\$$
              $$x^2-2cdotfrac{1}{3}cdot x+left(frac{1}{3}right)^2-left(frac{1}{3}right)^2+frac{7}{3}=0\$$
              $$left(x-frac{1}{3}right)^2+frac{21-1}{9}=0\$$
              $$ left(x-frac{1}{3}right)^2=frac{-20}{9}$$






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Starting from



                $$3x^2-2x+7=0\$$
                $$x^2-frac{2}{3}x+frac{7}{3}=0\$$
                $$x^2-2cdotfrac{1}{3}cdot x+left(frac{1}{3}right)^2-left(frac{1}{3}right)^2+frac{7}{3}=0\$$
                $$left(x-frac{1}{3}right)^2+frac{21-1}{9}=0\$$
                $$ left(x-frac{1}{3}right)^2=frac{-20}{9}$$






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Starting from



                  $$3x^2-2x+7=0\$$
                  $$x^2-frac{2}{3}x+frac{7}{3}=0\$$
                  $$x^2-2cdotfrac{1}{3}cdot x+left(frac{1}{3}right)^2-left(frac{1}{3}right)^2+frac{7}{3}=0\$$
                  $$left(x-frac{1}{3}right)^2+frac{21-1}{9}=0\$$
                  $$ left(x-frac{1}{3}right)^2=frac{-20}{9}$$






                  share|cite|improve this answer











                  $endgroup$



                  Starting from



                  $$3x^2-2x+7=0\$$
                  $$x^2-frac{2}{3}x+frac{7}{3}=0\$$
                  $$x^2-2cdotfrac{1}{3}cdot x+left(frac{1}{3}right)^2-left(frac{1}{3}right)^2+frac{7}{3}=0\$$
                  $$left(x-frac{1}{3}right)^2+frac{21-1}{9}=0\$$
                  $$ left(x-frac{1}{3}right)^2=frac{-20}{9}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Sep 21 '16 at 15:51









                  Bhaskara-III

                  1,1372827




                  1,1372827










                  answered Aug 17 '15 at 9:23









                  MonKMonK

                  1,527515




                  1,527515























                      2












                      $begingroup$

                      HINT:



                      $$3x^2-2x+7=3left(x-frac13right)^2+6+frac23$$






                      share|cite|improve this answer











                      $endgroup$









                      • 3




                        $begingroup$
                        In general, I find the notation $6frac{2}{3}$ to be a source of confusion because it is unclear if one means $6+frac{2}{3}$ or $6cdotfrac{2}{3}$. I strongly suggest either improper fractions or explicitly writing out the operation instead of mixed fractions.
                        $endgroup$
                        – Michael Burr
                        Aug 16 '15 at 20:11










                      • $begingroup$
                        @MichaelBurr: I thought about it while writing the answer, but figured it could hardly be seen as $6cdotfrac23$ without explicitly using the dot. In any case, I followed your suggestion and added a plus in order to avoid any possible confusion. Thanks.
                        $endgroup$
                        – barak manos
                        Aug 16 '15 at 20:25
















                      2












                      $begingroup$

                      HINT:



                      $$3x^2-2x+7=3left(x-frac13right)^2+6+frac23$$






                      share|cite|improve this answer











                      $endgroup$









                      • 3




                        $begingroup$
                        In general, I find the notation $6frac{2}{3}$ to be a source of confusion because it is unclear if one means $6+frac{2}{3}$ or $6cdotfrac{2}{3}$. I strongly suggest either improper fractions or explicitly writing out the operation instead of mixed fractions.
                        $endgroup$
                        – Michael Burr
                        Aug 16 '15 at 20:11










                      • $begingroup$
                        @MichaelBurr: I thought about it while writing the answer, but figured it could hardly be seen as $6cdotfrac23$ without explicitly using the dot. In any case, I followed your suggestion and added a plus in order to avoid any possible confusion. Thanks.
                        $endgroup$
                        – barak manos
                        Aug 16 '15 at 20:25














                      2












                      2








                      2





                      $begingroup$

                      HINT:



                      $$3x^2-2x+7=3left(x-frac13right)^2+6+frac23$$






                      share|cite|improve this answer











                      $endgroup$



                      HINT:



                      $$3x^2-2x+7=3left(x-frac13right)^2+6+frac23$$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Sep 26 '16 at 13:16









                      Bhaskara-III

                      1,1372827




                      1,1372827










                      answered Aug 16 '15 at 17:53









                      barak manosbarak manos

                      37.8k74199




                      37.8k74199








                      • 3




                        $begingroup$
                        In general, I find the notation $6frac{2}{3}$ to be a source of confusion because it is unclear if one means $6+frac{2}{3}$ or $6cdotfrac{2}{3}$. I strongly suggest either improper fractions or explicitly writing out the operation instead of mixed fractions.
                        $endgroup$
                        – Michael Burr
                        Aug 16 '15 at 20:11










                      • $begingroup$
                        @MichaelBurr: I thought about it while writing the answer, but figured it could hardly be seen as $6cdotfrac23$ without explicitly using the dot. In any case, I followed your suggestion and added a plus in order to avoid any possible confusion. Thanks.
                        $endgroup$
                        – barak manos
                        Aug 16 '15 at 20:25














                      • 3




                        $begingroup$
                        In general, I find the notation $6frac{2}{3}$ to be a source of confusion because it is unclear if one means $6+frac{2}{3}$ or $6cdotfrac{2}{3}$. I strongly suggest either improper fractions or explicitly writing out the operation instead of mixed fractions.
                        $endgroup$
                        – Michael Burr
                        Aug 16 '15 at 20:11










                      • $begingroup$
                        @MichaelBurr: I thought about it while writing the answer, but figured it could hardly be seen as $6cdotfrac23$ without explicitly using the dot. In any case, I followed your suggestion and added a plus in order to avoid any possible confusion. Thanks.
                        $endgroup$
                        – barak manos
                        Aug 16 '15 at 20:25








                      3




                      3




                      $begingroup$
                      In general, I find the notation $6frac{2}{3}$ to be a source of confusion because it is unclear if one means $6+frac{2}{3}$ or $6cdotfrac{2}{3}$. I strongly suggest either improper fractions or explicitly writing out the operation instead of mixed fractions.
                      $endgroup$
                      – Michael Burr
                      Aug 16 '15 at 20:11




                      $begingroup$
                      In general, I find the notation $6frac{2}{3}$ to be a source of confusion because it is unclear if one means $6+frac{2}{3}$ or $6cdotfrac{2}{3}$. I strongly suggest either improper fractions or explicitly writing out the operation instead of mixed fractions.
                      $endgroup$
                      – Michael Burr
                      Aug 16 '15 at 20:11












                      $begingroup$
                      @MichaelBurr: I thought about it while writing the answer, but figured it could hardly be seen as $6cdotfrac23$ without explicitly using the dot. In any case, I followed your suggestion and added a plus in order to avoid any possible confusion. Thanks.
                      $endgroup$
                      – barak manos
                      Aug 16 '15 at 20:25




                      $begingroup$
                      @MichaelBurr: I thought about it while writing the answer, but figured it could hardly be seen as $6cdotfrac23$ without explicitly using the dot. In any case, I followed your suggestion and added a plus in order to avoid any possible confusion. Thanks.
                      $endgroup$
                      – barak manos
                      Aug 16 '15 at 20:25











                      2












                      $begingroup$

                      Starting from, $$3x^2-2x+7=0$$ Dividing by 3:$$x^2-frac{2x}{3}+frac{7}{3}=0$$ Completing the square: $$left(x-frac 13right)^2-frac{1}{9}+frac{7}{3}=0$$ $$left(x-frac 13right)^2=frac{1}{9}-frac{7}{3}$$ $$left(x-frac 13right)^2=frac{1-21}{9}$$
                      $$left(x-frac 13right)^2=-frac{20}{9}$$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        I don't think further editing here is needed.
                        $endgroup$
                        – Daniel Fischer♦
                        Jan 17 '17 at 10:40
















                      2












                      $begingroup$

                      Starting from, $$3x^2-2x+7=0$$ Dividing by 3:$$x^2-frac{2x}{3}+frac{7}{3}=0$$ Completing the square: $$left(x-frac 13right)^2-frac{1}{9}+frac{7}{3}=0$$ $$left(x-frac 13right)^2=frac{1}{9}-frac{7}{3}$$ $$left(x-frac 13right)^2=frac{1-21}{9}$$
                      $$left(x-frac 13right)^2=-frac{20}{9}$$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        I don't think further editing here is needed.
                        $endgroup$
                        – Daniel Fischer♦
                        Jan 17 '17 at 10:40














                      2












                      2








                      2





                      $begingroup$

                      Starting from, $$3x^2-2x+7=0$$ Dividing by 3:$$x^2-frac{2x}{3}+frac{7}{3}=0$$ Completing the square: $$left(x-frac 13right)^2-frac{1}{9}+frac{7}{3}=0$$ $$left(x-frac 13right)^2=frac{1}{9}-frac{7}{3}$$ $$left(x-frac 13right)^2=frac{1-21}{9}$$
                      $$left(x-frac 13right)^2=-frac{20}{9}$$






                      share|cite|improve this answer











                      $endgroup$



                      Starting from, $$3x^2-2x+7=0$$ Dividing by 3:$$x^2-frac{2x}{3}+frac{7}{3}=0$$ Completing the square: $$left(x-frac 13right)^2-frac{1}{9}+frac{7}{3}=0$$ $$left(x-frac 13right)^2=frac{1}{9}-frac{7}{3}$$ $$left(x-frac 13right)^2=frac{1-21}{9}$$
                      $$left(x-frac 13right)^2=-frac{20}{9}$$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 17 '17 at 5:33

























                      answered Oct 18 '16 at 8:56









                      Bhaskara-IIIBhaskara-III

                      1,1372827




                      1,1372827












                      • $begingroup$
                        I don't think further editing here is needed.
                        $endgroup$
                        – Daniel Fischer♦
                        Jan 17 '17 at 10:40


















                      • $begingroup$
                        I don't think further editing here is needed.
                        $endgroup$
                        – Daniel Fischer♦
                        Jan 17 '17 at 10:40
















                      $begingroup$
                      I don't think further editing here is needed.
                      $endgroup$
                      – Daniel Fischer♦
                      Jan 17 '17 at 10:40




                      $begingroup$
                      I don't think further editing here is needed.
                      $endgroup$
                      – Daniel Fischer♦
                      Jan 17 '17 at 10:40











                      1












                      $begingroup$

                      Expand the binomial and compare it to the trinomial.



                      $$3x^2-2x+7iffleft(x-frac13right)^2=x^2-frac23x+frac19.$$



                      If you divide the polynomial by $3$, you get closer, with two identical terms



                      $$frac{3x^2-2x+7}3=x^2-frac23x+frac73.$$



                      To get a perfect identity, it now suffices to add a well-chosen constant



                      $$frac{3x^2-2x+7}3-frac{20}9=x^2-frac23x+frac19.$$



                      Now as the polynomial is known to evaluate to $0$, you know the value of the RHS.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        As you removed my edit: $3x^2-2x+7$ can't be equivalent to anything, you are missing the $ldots =0$.
                        $endgroup$
                        – Hirshy
                        Sep 21 '16 at 13:28










                      • $begingroup$
                        @Hirshy: no, you missed the intent. The double arrow relates the two expressions that are to be compared. Appending $=0$ turns the left one in an equation, while the right one is an identity, and this would result in a false proposition.
                        $endgroup$
                        – Yves Daoust
                        Sep 21 '16 at 13:32












                      • $begingroup$
                        I did indeed miss that, sorry. You might want to use another arrow for your purpose?
                        $endgroup$
                        – Hirshy
                        Sep 21 '16 at 13:35










                      • $begingroup$
                        @Hirshy: I didn't find one that pleased me.
                        $endgroup$
                        – Yves Daoust
                        Sep 21 '16 at 13:54
















                      1












                      $begingroup$

                      Expand the binomial and compare it to the trinomial.



                      $$3x^2-2x+7iffleft(x-frac13right)^2=x^2-frac23x+frac19.$$



                      If you divide the polynomial by $3$, you get closer, with two identical terms



                      $$frac{3x^2-2x+7}3=x^2-frac23x+frac73.$$



                      To get a perfect identity, it now suffices to add a well-chosen constant



                      $$frac{3x^2-2x+7}3-frac{20}9=x^2-frac23x+frac19.$$



                      Now as the polynomial is known to evaluate to $0$, you know the value of the RHS.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        As you removed my edit: $3x^2-2x+7$ can't be equivalent to anything, you are missing the $ldots =0$.
                        $endgroup$
                        – Hirshy
                        Sep 21 '16 at 13:28










                      • $begingroup$
                        @Hirshy: no, you missed the intent. The double arrow relates the two expressions that are to be compared. Appending $=0$ turns the left one in an equation, while the right one is an identity, and this would result in a false proposition.
                        $endgroup$
                        – Yves Daoust
                        Sep 21 '16 at 13:32












                      • $begingroup$
                        I did indeed miss that, sorry. You might want to use another arrow for your purpose?
                        $endgroup$
                        – Hirshy
                        Sep 21 '16 at 13:35










                      • $begingroup$
                        @Hirshy: I didn't find one that pleased me.
                        $endgroup$
                        – Yves Daoust
                        Sep 21 '16 at 13:54














                      1












                      1








                      1





                      $begingroup$

                      Expand the binomial and compare it to the trinomial.



                      $$3x^2-2x+7iffleft(x-frac13right)^2=x^2-frac23x+frac19.$$



                      If you divide the polynomial by $3$, you get closer, with two identical terms



                      $$frac{3x^2-2x+7}3=x^2-frac23x+frac73.$$



                      To get a perfect identity, it now suffices to add a well-chosen constant



                      $$frac{3x^2-2x+7}3-frac{20}9=x^2-frac23x+frac19.$$



                      Now as the polynomial is known to evaluate to $0$, you know the value of the RHS.






                      share|cite|improve this answer











                      $endgroup$



                      Expand the binomial and compare it to the trinomial.



                      $$3x^2-2x+7iffleft(x-frac13right)^2=x^2-frac23x+frac19.$$



                      If you divide the polynomial by $3$, you get closer, with two identical terms



                      $$frac{3x^2-2x+7}3=x^2-frac23x+frac73.$$



                      To get a perfect identity, it now suffices to add a well-chosen constant



                      $$frac{3x^2-2x+7}3-frac{20}9=x^2-frac23x+frac19.$$



                      Now as the polynomial is known to evaluate to $0$, you know the value of the RHS.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Sep 21 '16 at 13:15

























                      answered Sep 21 '16 at 13:03









                      Yves DaoustYves Daoust

                      125k671223




                      125k671223












                      • $begingroup$
                        As you removed my edit: $3x^2-2x+7$ can't be equivalent to anything, you are missing the $ldots =0$.
                        $endgroup$
                        – Hirshy
                        Sep 21 '16 at 13:28










                      • $begingroup$
                        @Hirshy: no, you missed the intent. The double arrow relates the two expressions that are to be compared. Appending $=0$ turns the left one in an equation, while the right one is an identity, and this would result in a false proposition.
                        $endgroup$
                        – Yves Daoust
                        Sep 21 '16 at 13:32












                      • $begingroup$
                        I did indeed miss that, sorry. You might want to use another arrow for your purpose?
                        $endgroup$
                        – Hirshy
                        Sep 21 '16 at 13:35










                      • $begingroup$
                        @Hirshy: I didn't find one that pleased me.
                        $endgroup$
                        – Yves Daoust
                        Sep 21 '16 at 13:54


















                      • $begingroup$
                        As you removed my edit: $3x^2-2x+7$ can't be equivalent to anything, you are missing the $ldots =0$.
                        $endgroup$
                        – Hirshy
                        Sep 21 '16 at 13:28










                      • $begingroup$
                        @Hirshy: no, you missed the intent. The double arrow relates the two expressions that are to be compared. Appending $=0$ turns the left one in an equation, while the right one is an identity, and this would result in a false proposition.
                        $endgroup$
                        – Yves Daoust
                        Sep 21 '16 at 13:32












                      • $begingroup$
                        I did indeed miss that, sorry. You might want to use another arrow for your purpose?
                        $endgroup$
                        – Hirshy
                        Sep 21 '16 at 13:35










                      • $begingroup$
                        @Hirshy: I didn't find one that pleased me.
                        $endgroup$
                        – Yves Daoust
                        Sep 21 '16 at 13:54
















                      $begingroup$
                      As you removed my edit: $3x^2-2x+7$ can't be equivalent to anything, you are missing the $ldots =0$.
                      $endgroup$
                      – Hirshy
                      Sep 21 '16 at 13:28




                      $begingroup$
                      As you removed my edit: $3x^2-2x+7$ can't be equivalent to anything, you are missing the $ldots =0$.
                      $endgroup$
                      – Hirshy
                      Sep 21 '16 at 13:28












                      $begingroup$
                      @Hirshy: no, you missed the intent. The double arrow relates the two expressions that are to be compared. Appending $=0$ turns the left one in an equation, while the right one is an identity, and this would result in a false proposition.
                      $endgroup$
                      – Yves Daoust
                      Sep 21 '16 at 13:32






                      $begingroup$
                      @Hirshy: no, you missed the intent. The double arrow relates the two expressions that are to be compared. Appending $=0$ turns the left one in an equation, while the right one is an identity, and this would result in a false proposition.
                      $endgroup$
                      – Yves Daoust
                      Sep 21 '16 at 13:32














                      $begingroup$
                      I did indeed miss that, sorry. You might want to use another arrow for your purpose?
                      $endgroup$
                      – Hirshy
                      Sep 21 '16 at 13:35




                      $begingroup$
                      I did indeed miss that, sorry. You might want to use another arrow for your purpose?
                      $endgroup$
                      – Hirshy
                      Sep 21 '16 at 13:35












                      $begingroup$
                      @Hirshy: I didn't find one that pleased me.
                      $endgroup$
                      – Yves Daoust
                      Sep 21 '16 at 13:54




                      $begingroup$
                      @Hirshy: I didn't find one that pleased me.
                      $endgroup$
                      – Yves Daoust
                      Sep 21 '16 at 13:54





                      protected by Alex M. Nov 19 '16 at 19:43



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