How does this equation always work? $sqrt{x times (x+2)+1} = x + 1$ for non-negative $x$, and $=|x|-1$ for...












3












$begingroup$


When I was playing with new calculator's functions, somehow I managed to get a formula, which works with all real numbers (negative number have a slight change). I had asked my teacher about it, but never figured out why it works:



Non-Negative:
$$forall x in Bbb R_0^+: sqrt{x times (x+2)+1} = x + 1$$



Negative:
$$forall x in Bbb R^-: sqrt{x times (x+2)+1} = |x| - 1$$






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $x(x+2)+1=x^2+2x+1=(x+1)^2$.
    $endgroup$
    – A.Γ.
    Jan 2 at 22:27






  • 1




    $begingroup$
    Your second formula is not true for $x=-1/2$.
    $endgroup$
    – farruhota
    Jan 2 at 22:34










  • $begingroup$
    Check your edited second equation. Are you sure it works for $-1 < x < 0$?
    $endgroup$
    – Deepak
    Jan 2 at 22:37
















3












$begingroup$


When I was playing with new calculator's functions, somehow I managed to get a formula, which works with all real numbers (negative number have a slight change). I had asked my teacher about it, but never figured out why it works:



Non-Negative:
$$forall x in Bbb R_0^+: sqrt{x times (x+2)+1} = x + 1$$



Negative:
$$forall x in Bbb R^-: sqrt{x times (x+2)+1} = |x| - 1$$






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $x(x+2)+1=x^2+2x+1=(x+1)^2$.
    $endgroup$
    – A.Γ.
    Jan 2 at 22:27






  • 1




    $begingroup$
    Your second formula is not true for $x=-1/2$.
    $endgroup$
    – farruhota
    Jan 2 at 22:34










  • $begingroup$
    Check your edited second equation. Are you sure it works for $-1 < x < 0$?
    $endgroup$
    – Deepak
    Jan 2 at 22:37














3












3








3





$begingroup$


When I was playing with new calculator's functions, somehow I managed to get a formula, which works with all real numbers (negative number have a slight change). I had asked my teacher about it, but never figured out why it works:



Non-Negative:
$$forall x in Bbb R_0^+: sqrt{x times (x+2)+1} = x + 1$$



Negative:
$$forall x in Bbb R^-: sqrt{x times (x+2)+1} = |x| - 1$$






algebra-precalculus







share|cite|improve this question











$endgroup$




When I was playing with new calculator's functions, somehow I managed to get a formula, which works with all real numbers (negative number have a slight change). I had asked my teacher about it, but never figured out why it works:



Non-Negative:
$$forall x in Bbb R_0^+: sqrt{x times (x+2)+1} = x + 1$$



Negative:
$$forall x in Bbb R^-: sqrt{x times (x+2)+1} = |x| - 1$$






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 22:48









Blue

48.7k870156




48.7k870156










asked Jan 2 at 22:25









Dia. FrostDia. Frost

213




213








  • 2




    $begingroup$
    $x(x+2)+1=x^2+2x+1=(x+1)^2$.
    $endgroup$
    – A.Γ.
    Jan 2 at 22:27






  • 1




    $begingroup$
    Your second formula is not true for $x=-1/2$.
    $endgroup$
    – farruhota
    Jan 2 at 22:34










  • $begingroup$
    Check your edited second equation. Are you sure it works for $-1 < x < 0$?
    $endgroup$
    – Deepak
    Jan 2 at 22:37














  • 2




    $begingroup$
    $x(x+2)+1=x^2+2x+1=(x+1)^2$.
    $endgroup$
    – A.Γ.
    Jan 2 at 22:27






  • 1




    $begingroup$
    Your second formula is not true for $x=-1/2$.
    $endgroup$
    – farruhota
    Jan 2 at 22:34










  • $begingroup$
    Check your edited second equation. Are you sure it works for $-1 < x < 0$?
    $endgroup$
    – Deepak
    Jan 2 at 22:37








2




2




$begingroup$
$x(x+2)+1=x^2+2x+1=(x+1)^2$.
$endgroup$
– A.Γ.
Jan 2 at 22:27




$begingroup$
$x(x+2)+1=x^2+2x+1=(x+1)^2$.
$endgroup$
– A.Γ.
Jan 2 at 22:27




1




1




$begingroup$
Your second formula is not true for $x=-1/2$.
$endgroup$
– farruhota
Jan 2 at 22:34




$begingroup$
Your second formula is not true for $x=-1/2$.
$endgroup$
– farruhota
Jan 2 at 22:34












$begingroup$
Check your edited second equation. Are you sure it works for $-1 < x < 0$?
$endgroup$
– Deepak
Jan 2 at 22:37




$begingroup$
Check your edited second equation. Are you sure it works for $-1 < x < 0$?
$endgroup$
– Deepak
Jan 2 at 22:37










3 Answers
3






active

oldest

votes


















2












$begingroup$

Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
$$
sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
$$
You may also note that:




  • For $x geq -1$, we have $|x+1| = x+1$

  • For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)


This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
    $endgroup$
    – Dia. Frost
    Jan 2 at 22:42










  • $begingroup$
    Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
    $endgroup$
    – Sambo
    Jan 2 at 23:51






  • 1




    $begingroup$
    Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
    $endgroup$
    – Sambo
    Jan 2 at 23:52



















3












$begingroup$

The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Note that you have a perfect square trinomial.



    $$x(x+2)+1 = x^2+2x+1$$



    You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.



    $$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$



    Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get



    $$x^2+2(x)(1)+1^2 = (x+1)^2$$



    So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so



    $$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
      $$
      sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
      $$
      You may also note that:




      • For $x geq -1$, we have $|x+1| = x+1$

      • For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)


      This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
        $endgroup$
        – Dia. Frost
        Jan 2 at 22:42










      • $begingroup$
        Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
        $endgroup$
        – Sambo
        Jan 2 at 23:51






      • 1




        $begingroup$
        Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
        $endgroup$
        – Sambo
        Jan 2 at 23:52
















      2












      $begingroup$

      Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
      $$
      sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
      $$
      You may also note that:




      • For $x geq -1$, we have $|x+1| = x+1$

      • For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)


      This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
        $endgroup$
        – Dia. Frost
        Jan 2 at 22:42










      • $begingroup$
        Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
        $endgroup$
        – Sambo
        Jan 2 at 23:51






      • 1




        $begingroup$
        Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
        $endgroup$
        – Sambo
        Jan 2 at 23:52














      2












      2








      2





      $begingroup$

      Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
      $$
      sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
      $$
      You may also note that:




      • For $x geq -1$, we have $|x+1| = x+1$

      • For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)


      This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).






      share|cite|improve this answer









      $endgroup$



      Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
      $$
      sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
      $$
      You may also note that:




      • For $x geq -1$, we have $|x+1| = x+1$

      • For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)


      This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 2 at 22:35









      SamboSambo

      2,2562532




      2,2562532












      • $begingroup$
        Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
        $endgroup$
        – Dia. Frost
        Jan 2 at 22:42










      • $begingroup$
        Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
        $endgroup$
        – Sambo
        Jan 2 at 23:51






      • 1




        $begingroup$
        Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
        $endgroup$
        – Sambo
        Jan 2 at 23:52


















      • $begingroup$
        Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
        $endgroup$
        – Dia. Frost
        Jan 2 at 22:42










      • $begingroup$
        Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
        $endgroup$
        – Sambo
        Jan 2 at 23:51






      • 1




        $begingroup$
        Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
        $endgroup$
        – Sambo
        Jan 2 at 23:52
















      $begingroup$
      Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
      $endgroup$
      – Dia. Frost
      Jan 2 at 22:42




      $begingroup$
      Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
      $endgroup$
      – Dia. Frost
      Jan 2 at 22:42












      $begingroup$
      Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
      $endgroup$
      – Sambo
      Jan 2 at 23:51




      $begingroup$
      Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
      $endgroup$
      – Sambo
      Jan 2 at 23:51




      1




      1




      $begingroup$
      Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
      $endgroup$
      – Sambo
      Jan 2 at 23:52




      $begingroup$
      Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
      $endgroup$
      – Sambo
      Jan 2 at 23:52











      3












      $begingroup$

      The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$






          share|cite|improve this answer









          $endgroup$



          The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 22:27









          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

          76.7k42866




          76.7k42866























              1












              $begingroup$

              Note that you have a perfect square trinomial.



              $$x(x+2)+1 = x^2+2x+1$$



              You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.



              $$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$



              Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get



              $$x^2+2(x)(1)+1^2 = (x+1)^2$$



              So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so



              $$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Note that you have a perfect square trinomial.



                $$x(x+2)+1 = x^2+2x+1$$



                You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.



                $$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$



                Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get



                $$x^2+2(x)(1)+1^2 = (x+1)^2$$



                So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so



                $$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Note that you have a perfect square trinomial.



                  $$x(x+2)+1 = x^2+2x+1$$



                  You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.



                  $$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$



                  Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get



                  $$x^2+2(x)(1)+1^2 = (x+1)^2$$



                  So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so



                  $$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$






                  share|cite|improve this answer









                  $endgroup$



                  Note that you have a perfect square trinomial.



                  $$x(x+2)+1 = x^2+2x+1$$



                  You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.



                  $$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$



                  Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get



                  $$x^2+2(x)(1)+1^2 = (x+1)^2$$



                  So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so



                  $$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 2 at 22:43









                  KM101KM101

                  6,0601525




                  6,0601525






























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                      Quarter-circle Tiles

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                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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                      Mont Emei

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                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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