Condition in terms of b and a if $ax^2+bx+c=0$ has two consecutive odd positive integers as roots











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The roots of the equation$$ax^2+bx+c=0$$, where $a geq 0$, are two consecutive odd positive integers, then




(A) $|b|leq 4a$



(B) $|b|geq 4a$



(C) $|b|=2a$



(D) None of these



My attempt



Let p and q be the roots then if they are consecutive positive integers (q>p) then $$ pq=frac{c}{a} geq 0$$
So, $$c geq 0$$ and $$q-p=2$$
So, $$frac{sqrt{b^2-4ac}}{a}=2$$
So,$$|b|>2a$$
$(Since, a>0,c>0)$



But I know that 4ac should be taken in consideration since its not equal to zero. But I don't know how to use it.



Any hints and suggestions are welcome!










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    The roots of the equation$$ax^2+bx+c=0$$, where $a geq 0$, are two consecutive odd positive integers, then




    (A) $|b|leq 4a$



    (B) $|b|geq 4a$



    (C) $|b|=2a$



    (D) None of these



    My attempt



    Let p and q be the roots then if they are consecutive positive integers (q>p) then $$ pq=frac{c}{a} geq 0$$
    So, $$c geq 0$$ and $$q-p=2$$
    So, $$frac{sqrt{b^2-4ac}}{a}=2$$
    So,$$|b|>2a$$
    $(Since, a>0,c>0)$



    But I know that 4ac should be taken in consideration since its not equal to zero. But I don't know how to use it.



    Any hints and suggestions are welcome!










    share|cite|improve this question


























      up vote
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      down vote

      favorite
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      up vote
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      The roots of the equation$$ax^2+bx+c=0$$, where $a geq 0$, are two consecutive odd positive integers, then




      (A) $|b|leq 4a$



      (B) $|b|geq 4a$



      (C) $|b|=2a$



      (D) None of these



      My attempt



      Let p and q be the roots then if they are consecutive positive integers (q>p) then $$ pq=frac{c}{a} geq 0$$
      So, $$c geq 0$$ and $$q-p=2$$
      So, $$frac{sqrt{b^2-4ac}}{a}=2$$
      So,$$|b|>2a$$
      $(Since, a>0,c>0)$



      But I know that 4ac should be taken in consideration since its not equal to zero. But I don't know how to use it.



      Any hints and suggestions are welcome!










      share|cite|improve this question
















      The roots of the equation$$ax^2+bx+c=0$$, where $a geq 0$, are two consecutive odd positive integers, then




      (A) $|b|leq 4a$



      (B) $|b|geq 4a$



      (C) $|b|=2a$



      (D) None of these



      My attempt



      Let p and q be the roots then if they are consecutive positive integers (q>p) then $$ pq=frac{c}{a} geq 0$$
      So, $$c geq 0$$ and $$q-p=2$$
      So, $$frac{sqrt{b^2-4ac}}{a}=2$$
      So,$$|b|>2a$$
      $(Since, a>0,c>0)$



      But I know that 4ac should be taken in consideration since its not equal to zero. But I don't know how to use it.



      Any hints and suggestions are welcome!







      algebra-precalculus quadratics self-learning






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      edited Nov 21 at 21:34









      Micah

      29.5k1363104




      29.5k1363104










      asked Nov 21 at 21:20









      jayant98

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          Simply, If $2n-1$ and $2n+1$ are the roots (with $nge 1$) then
          $$ax^2+bx+c=a(x-2n+1)(x-2n-1)=a((x-2n)^2-1)=a(x^2-4nx+4n^2-1) $$
          so $b=-4na$ and hence $|b|=4nage 4a$.






          share|cite|improve this answer





















          • Oh, you have the easier and more simple way! Thanks.
            – jayant98
            Nov 21 at 21:34


















          up vote
          1
          down vote













          Using the quadratic formula, we have that (using $p<q$ as the roots):
          $$p+2=q$$
          $$frac{-b-sqrt{b^2-4ac}}{2a}+2=frac{-b+sqrt{b^2-4ac}}{2a}$$
          $$frac{-b}{2a}-frac{sqrt{b^2-4ac}}{2a}+2=frac{-b}{2a}+frac{sqrt{b^2-4ac}}{2a}$$
          $$2=2frac{sqrt{b^2-4ac}}{2a}$$
          $$2a=sqrt{b^2-4ac}$$
          $$4a^2=b^2-4ac$$
          $$b^2=4a^2+4ac$$
          $$frac{b^2}{a^2}=4(1+frac{c}{a})=4(1+pq)$$
          $$frac{|b|}{a}=2sqrt{1+pq}$$
          Then, given that $p,q$ are different odd integers, you can show that $pqgeq3$, so that $2sqrt{1+pq}geq2sqrt{1+3}=4$






          share|cite|improve this answer




























            up vote
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            Since we are given that there are two distinct roots and that $a ge 0$, we must have



            $a > 0, tag 0$



            since otherwise (i.e., with $a = 0$), the "quadratic" $bx + c$ has at most one root.



            Let



            $n ge 0, tag 1$



            $r = 2n + 1, tag 2$



            $s = 2n + 3; tag 3$



            suppose for the moment



            $a = 1; tag 4$



            then



            $(x - r)(x - s) = (x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (2n + 1)(2n + 3) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = 0; tag 5$



            here we have



            $b = -(4n + 4), tag 6$



            whence



            $vert b vert = 4n + 4 ge 4 = 4a; tag 7$



            thus (B) binds when $a = 1$; now if



            $a ne 1, tag 8$



            the quadratic of the form



            $ax^2 + bx + c = a(x^2 + dfrac{b}{a} x + dfrac{c}{a}) tag 9$



            has zeroes $2n + 1$, $2n + 3$ provided



            $(x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = x^2 + dfrac{b}{a} x + dfrac{c}{a}; tag{10}$



            thus,



            $dfrac{b}{a} = -(4n + 4), tag{11}$



            or



            $b = -(4n + 4)a, tag{12}$



            whence, with $a > 0$,



            $vert b vert = (4n + 4)a ge 4a, tag{13}$



            and we see that the correct choice is (B) here as well.






            share|cite|improve this answer




























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              Easy way to check:



              Take a quadratic equation which has 1 and 3 as roots. $(x-1)(x-3)=x^2-4x+4x$, so you see that it is possible for $|b|=4|a|$, ruling out choice (C).



              Do the problem again choosing the roots 3 and 5. $(x-3)(x-5)=x^2-8x+15$, so you can conclude that $|b|>=4a$, and (B) is correct. (If you continue choosing larger pairs of numbers as roots, $frac{|b|}{a}$ will only increase.)






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              • Sorry @MoKo19 but we have to give answer in subjective way...So we can't just do that...This question given just has the options...but we have to solve it 'properly'.
                – jayant98
                Nov 21 at 21:32










              • @jayant98: Not sure what you mean by 'properly'. If you mean analytically, then generate the quadratic equation where $a=1$ with roots $p,p+2$. Because This will give you $y=(x-p)(x-p-2)=x^2-2px-2x+p^2+2p$, and $frac{|b|}{a}=|b|=|2p+2|=2p+2$. For odd integer roots, $p=1,3,5,...$, so the ratio will be $4,8,12,16,...$. I will write a different answer showing how to solve via the quadratic formula.
                – MoKo19
                Nov 21 at 21:42











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              4 Answers
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              active

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              4 Answers
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              active

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              active

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              active

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              up vote
              5
              down vote



              accepted










              Simply, If $2n-1$ and $2n+1$ are the roots (with $nge 1$) then
              $$ax^2+bx+c=a(x-2n+1)(x-2n-1)=a((x-2n)^2-1)=a(x^2-4nx+4n^2-1) $$
              so $b=-4na$ and hence $|b|=4nage 4a$.






              share|cite|improve this answer





















              • Oh, you have the easier and more simple way! Thanks.
                – jayant98
                Nov 21 at 21:34















              up vote
              5
              down vote



              accepted










              Simply, If $2n-1$ and $2n+1$ are the roots (with $nge 1$) then
              $$ax^2+bx+c=a(x-2n+1)(x-2n-1)=a((x-2n)^2-1)=a(x^2-4nx+4n^2-1) $$
              so $b=-4na$ and hence $|b|=4nage 4a$.






              share|cite|improve this answer





















              • Oh, you have the easier and more simple way! Thanks.
                – jayant98
                Nov 21 at 21:34













              up vote
              5
              down vote



              accepted







              up vote
              5
              down vote



              accepted






              Simply, If $2n-1$ and $2n+1$ are the roots (with $nge 1$) then
              $$ax^2+bx+c=a(x-2n+1)(x-2n-1)=a((x-2n)^2-1)=a(x^2-4nx+4n^2-1) $$
              so $b=-4na$ and hence $|b|=4nage 4a$.






              share|cite|improve this answer












              Simply, If $2n-1$ and $2n+1$ are the roots (with $nge 1$) then
              $$ax^2+bx+c=a(x-2n+1)(x-2n-1)=a((x-2n)^2-1)=a(x^2-4nx+4n^2-1) $$
              so $b=-4na$ and hence $|b|=4nage 4a$.







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              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 21 at 21:33









              Hagen von Eitzen

              275k21268495




              275k21268495












              • Oh, you have the easier and more simple way! Thanks.
                – jayant98
                Nov 21 at 21:34


















              • Oh, you have the easier and more simple way! Thanks.
                – jayant98
                Nov 21 at 21:34
















              Oh, you have the easier and more simple way! Thanks.
              – jayant98
              Nov 21 at 21:34




              Oh, you have the easier and more simple way! Thanks.
              – jayant98
              Nov 21 at 21:34










              up vote
              1
              down vote













              Using the quadratic formula, we have that (using $p<q$ as the roots):
              $$p+2=q$$
              $$frac{-b-sqrt{b^2-4ac}}{2a}+2=frac{-b+sqrt{b^2-4ac}}{2a}$$
              $$frac{-b}{2a}-frac{sqrt{b^2-4ac}}{2a}+2=frac{-b}{2a}+frac{sqrt{b^2-4ac}}{2a}$$
              $$2=2frac{sqrt{b^2-4ac}}{2a}$$
              $$2a=sqrt{b^2-4ac}$$
              $$4a^2=b^2-4ac$$
              $$b^2=4a^2+4ac$$
              $$frac{b^2}{a^2}=4(1+frac{c}{a})=4(1+pq)$$
              $$frac{|b|}{a}=2sqrt{1+pq}$$
              Then, given that $p,q$ are different odd integers, you can show that $pqgeq3$, so that $2sqrt{1+pq}geq2sqrt{1+3}=4$






              share|cite|improve this answer

























                up vote
                1
                down vote













                Using the quadratic formula, we have that (using $p<q$ as the roots):
                $$p+2=q$$
                $$frac{-b-sqrt{b^2-4ac}}{2a}+2=frac{-b+sqrt{b^2-4ac}}{2a}$$
                $$frac{-b}{2a}-frac{sqrt{b^2-4ac}}{2a}+2=frac{-b}{2a}+frac{sqrt{b^2-4ac}}{2a}$$
                $$2=2frac{sqrt{b^2-4ac}}{2a}$$
                $$2a=sqrt{b^2-4ac}$$
                $$4a^2=b^2-4ac$$
                $$b^2=4a^2+4ac$$
                $$frac{b^2}{a^2}=4(1+frac{c}{a})=4(1+pq)$$
                $$frac{|b|}{a}=2sqrt{1+pq}$$
                Then, given that $p,q$ are different odd integers, you can show that $pqgeq3$, so that $2sqrt{1+pq}geq2sqrt{1+3}=4$






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Using the quadratic formula, we have that (using $p<q$ as the roots):
                  $$p+2=q$$
                  $$frac{-b-sqrt{b^2-4ac}}{2a}+2=frac{-b+sqrt{b^2-4ac}}{2a}$$
                  $$frac{-b}{2a}-frac{sqrt{b^2-4ac}}{2a}+2=frac{-b}{2a}+frac{sqrt{b^2-4ac}}{2a}$$
                  $$2=2frac{sqrt{b^2-4ac}}{2a}$$
                  $$2a=sqrt{b^2-4ac}$$
                  $$4a^2=b^2-4ac$$
                  $$b^2=4a^2+4ac$$
                  $$frac{b^2}{a^2}=4(1+frac{c}{a})=4(1+pq)$$
                  $$frac{|b|}{a}=2sqrt{1+pq}$$
                  Then, given that $p,q$ are different odd integers, you can show that $pqgeq3$, so that $2sqrt{1+pq}geq2sqrt{1+3}=4$






                  share|cite|improve this answer












                  Using the quadratic formula, we have that (using $p<q$ as the roots):
                  $$p+2=q$$
                  $$frac{-b-sqrt{b^2-4ac}}{2a}+2=frac{-b+sqrt{b^2-4ac}}{2a}$$
                  $$frac{-b}{2a}-frac{sqrt{b^2-4ac}}{2a}+2=frac{-b}{2a}+frac{sqrt{b^2-4ac}}{2a}$$
                  $$2=2frac{sqrt{b^2-4ac}}{2a}$$
                  $$2a=sqrt{b^2-4ac}$$
                  $$4a^2=b^2-4ac$$
                  $$b^2=4a^2+4ac$$
                  $$frac{b^2}{a^2}=4(1+frac{c}{a})=4(1+pq)$$
                  $$frac{|b|}{a}=2sqrt{1+pq}$$
                  Then, given that $p,q$ are different odd integers, you can show that $pqgeq3$, so that $2sqrt{1+pq}geq2sqrt{1+3}=4$







                  share|cite|improve this answer












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                  share|cite|improve this answer










                  answered Nov 21 at 21:57









                  MoKo19

                  1914




                  1914






















                      up vote
                      1
                      down vote













                      Since we are given that there are two distinct roots and that $a ge 0$, we must have



                      $a > 0, tag 0$



                      since otherwise (i.e., with $a = 0$), the "quadratic" $bx + c$ has at most one root.



                      Let



                      $n ge 0, tag 1$



                      $r = 2n + 1, tag 2$



                      $s = 2n + 3; tag 3$



                      suppose for the moment



                      $a = 1; tag 4$



                      then



                      $(x - r)(x - s) = (x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (2n + 1)(2n + 3) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = 0; tag 5$



                      here we have



                      $b = -(4n + 4), tag 6$



                      whence



                      $vert b vert = 4n + 4 ge 4 = 4a; tag 7$



                      thus (B) binds when $a = 1$; now if



                      $a ne 1, tag 8$



                      the quadratic of the form



                      $ax^2 + bx + c = a(x^2 + dfrac{b}{a} x + dfrac{c}{a}) tag 9$



                      has zeroes $2n + 1$, $2n + 3$ provided



                      $(x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = x^2 + dfrac{b}{a} x + dfrac{c}{a}; tag{10}$



                      thus,



                      $dfrac{b}{a} = -(4n + 4), tag{11}$



                      or



                      $b = -(4n + 4)a, tag{12}$



                      whence, with $a > 0$,



                      $vert b vert = (4n + 4)a ge 4a, tag{13}$



                      and we see that the correct choice is (B) here as well.






                      share|cite|improve this answer

























                        up vote
                        1
                        down vote













                        Since we are given that there are two distinct roots and that $a ge 0$, we must have



                        $a > 0, tag 0$



                        since otherwise (i.e., with $a = 0$), the "quadratic" $bx + c$ has at most one root.



                        Let



                        $n ge 0, tag 1$



                        $r = 2n + 1, tag 2$



                        $s = 2n + 3; tag 3$



                        suppose for the moment



                        $a = 1; tag 4$



                        then



                        $(x - r)(x - s) = (x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (2n + 1)(2n + 3) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = 0; tag 5$



                        here we have



                        $b = -(4n + 4), tag 6$



                        whence



                        $vert b vert = 4n + 4 ge 4 = 4a; tag 7$



                        thus (B) binds when $a = 1$; now if



                        $a ne 1, tag 8$



                        the quadratic of the form



                        $ax^2 + bx + c = a(x^2 + dfrac{b}{a} x + dfrac{c}{a}) tag 9$



                        has zeroes $2n + 1$, $2n + 3$ provided



                        $(x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = x^2 + dfrac{b}{a} x + dfrac{c}{a}; tag{10}$



                        thus,



                        $dfrac{b}{a} = -(4n + 4), tag{11}$



                        or



                        $b = -(4n + 4)a, tag{12}$



                        whence, with $a > 0$,



                        $vert b vert = (4n + 4)a ge 4a, tag{13}$



                        and we see that the correct choice is (B) here as well.






                        share|cite|improve this answer























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          Since we are given that there are two distinct roots and that $a ge 0$, we must have



                          $a > 0, tag 0$



                          since otherwise (i.e., with $a = 0$), the "quadratic" $bx + c$ has at most one root.



                          Let



                          $n ge 0, tag 1$



                          $r = 2n + 1, tag 2$



                          $s = 2n + 3; tag 3$



                          suppose for the moment



                          $a = 1; tag 4$



                          then



                          $(x - r)(x - s) = (x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (2n + 1)(2n + 3) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = 0; tag 5$



                          here we have



                          $b = -(4n + 4), tag 6$



                          whence



                          $vert b vert = 4n + 4 ge 4 = 4a; tag 7$



                          thus (B) binds when $a = 1$; now if



                          $a ne 1, tag 8$



                          the quadratic of the form



                          $ax^2 + bx + c = a(x^2 + dfrac{b}{a} x + dfrac{c}{a}) tag 9$



                          has zeroes $2n + 1$, $2n + 3$ provided



                          $(x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = x^2 + dfrac{b}{a} x + dfrac{c}{a}; tag{10}$



                          thus,



                          $dfrac{b}{a} = -(4n + 4), tag{11}$



                          or



                          $b = -(4n + 4)a, tag{12}$



                          whence, with $a > 0$,



                          $vert b vert = (4n + 4)a ge 4a, tag{13}$



                          and we see that the correct choice is (B) here as well.






                          share|cite|improve this answer












                          Since we are given that there are two distinct roots and that $a ge 0$, we must have



                          $a > 0, tag 0$



                          since otherwise (i.e., with $a = 0$), the "quadratic" $bx + c$ has at most one root.



                          Let



                          $n ge 0, tag 1$



                          $r = 2n + 1, tag 2$



                          $s = 2n + 3; tag 3$



                          suppose for the moment



                          $a = 1; tag 4$



                          then



                          $(x - r)(x - s) = (x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (2n + 1)(2n + 3) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = 0; tag 5$



                          here we have



                          $b = -(4n + 4), tag 6$



                          whence



                          $vert b vert = 4n + 4 ge 4 = 4a; tag 7$



                          thus (B) binds when $a = 1$; now if



                          $a ne 1, tag 8$



                          the quadratic of the form



                          $ax^2 + bx + c = a(x^2 + dfrac{b}{a} x + dfrac{c}{a}) tag 9$



                          has zeroes $2n + 1$, $2n + 3$ provided



                          $(x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = x^2 + dfrac{b}{a} x + dfrac{c}{a}; tag{10}$



                          thus,



                          $dfrac{b}{a} = -(4n + 4), tag{11}$



                          or



                          $b = -(4n + 4)a, tag{12}$



                          whence, with $a > 0$,



                          $vert b vert = (4n + 4)a ge 4a, tag{13}$



                          and we see that the correct choice is (B) here as well.







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                          answered Nov 22 at 8:41









                          Robert Lewis

                          42.7k22862




                          42.7k22862






















                              up vote
                              0
                              down vote













                              Easy way to check:



                              Take a quadratic equation which has 1 and 3 as roots. $(x-1)(x-3)=x^2-4x+4x$, so you see that it is possible for $|b|=4|a|$, ruling out choice (C).



                              Do the problem again choosing the roots 3 and 5. $(x-3)(x-5)=x^2-8x+15$, so you can conclude that $|b|>=4a$, and (B) is correct. (If you continue choosing larger pairs of numbers as roots, $frac{|b|}{a}$ will only increase.)






                              share|cite|improve this answer





















                              • Sorry @MoKo19 but we have to give answer in subjective way...So we can't just do that...This question given just has the options...but we have to solve it 'properly'.
                                – jayant98
                                Nov 21 at 21:32










                              • @jayant98: Not sure what you mean by 'properly'. If you mean analytically, then generate the quadratic equation where $a=1$ with roots $p,p+2$. Because This will give you $y=(x-p)(x-p-2)=x^2-2px-2x+p^2+2p$, and $frac{|b|}{a}=|b|=|2p+2|=2p+2$. For odd integer roots, $p=1,3,5,...$, so the ratio will be $4,8,12,16,...$. I will write a different answer showing how to solve via the quadratic formula.
                                – MoKo19
                                Nov 21 at 21:42















                              up vote
                              0
                              down vote













                              Easy way to check:



                              Take a quadratic equation which has 1 and 3 as roots. $(x-1)(x-3)=x^2-4x+4x$, so you see that it is possible for $|b|=4|a|$, ruling out choice (C).



                              Do the problem again choosing the roots 3 and 5. $(x-3)(x-5)=x^2-8x+15$, so you can conclude that $|b|>=4a$, and (B) is correct. (If you continue choosing larger pairs of numbers as roots, $frac{|b|}{a}$ will only increase.)






                              share|cite|improve this answer





















                              • Sorry @MoKo19 but we have to give answer in subjective way...So we can't just do that...This question given just has the options...but we have to solve it 'properly'.
                                – jayant98
                                Nov 21 at 21:32










                              • @jayant98: Not sure what you mean by 'properly'. If you mean analytically, then generate the quadratic equation where $a=1$ with roots $p,p+2$. Because This will give you $y=(x-p)(x-p-2)=x^2-2px-2x+p^2+2p$, and $frac{|b|}{a}=|b|=|2p+2|=2p+2$. For odd integer roots, $p=1,3,5,...$, so the ratio will be $4,8,12,16,...$. I will write a different answer showing how to solve via the quadratic formula.
                                – MoKo19
                                Nov 21 at 21:42













                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote









                              Easy way to check:



                              Take a quadratic equation which has 1 and 3 as roots. $(x-1)(x-3)=x^2-4x+4x$, so you see that it is possible for $|b|=4|a|$, ruling out choice (C).



                              Do the problem again choosing the roots 3 and 5. $(x-3)(x-5)=x^2-8x+15$, so you can conclude that $|b|>=4a$, and (B) is correct. (If you continue choosing larger pairs of numbers as roots, $frac{|b|}{a}$ will only increase.)






                              share|cite|improve this answer












                              Easy way to check:



                              Take a quadratic equation which has 1 and 3 as roots. $(x-1)(x-3)=x^2-4x+4x$, so you see that it is possible for $|b|=4|a|$, ruling out choice (C).



                              Do the problem again choosing the roots 3 and 5. $(x-3)(x-5)=x^2-8x+15$, so you can conclude that $|b|>=4a$, and (B) is correct. (If you continue choosing larger pairs of numbers as roots, $frac{|b|}{a}$ will only increase.)







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 21 at 21:29









                              MoKo19

                              1914




                              1914












                              • Sorry @MoKo19 but we have to give answer in subjective way...So we can't just do that...This question given just has the options...but we have to solve it 'properly'.
                                – jayant98
                                Nov 21 at 21:32










                              • @jayant98: Not sure what you mean by 'properly'. If you mean analytically, then generate the quadratic equation where $a=1$ with roots $p,p+2$. Because This will give you $y=(x-p)(x-p-2)=x^2-2px-2x+p^2+2p$, and $frac{|b|}{a}=|b|=|2p+2|=2p+2$. For odd integer roots, $p=1,3,5,...$, so the ratio will be $4,8,12,16,...$. I will write a different answer showing how to solve via the quadratic formula.
                                – MoKo19
                                Nov 21 at 21:42


















                              • Sorry @MoKo19 but we have to give answer in subjective way...So we can't just do that...This question given just has the options...but we have to solve it 'properly'.
                                – jayant98
                                Nov 21 at 21:32










                              • @jayant98: Not sure what you mean by 'properly'. If you mean analytically, then generate the quadratic equation where $a=1$ with roots $p,p+2$. Because This will give you $y=(x-p)(x-p-2)=x^2-2px-2x+p^2+2p$, and $frac{|b|}{a}=|b|=|2p+2|=2p+2$. For odd integer roots, $p=1,3,5,...$, so the ratio will be $4,8,12,16,...$. I will write a different answer showing how to solve via the quadratic formula.
                                – MoKo19
                                Nov 21 at 21:42
















                              Sorry @MoKo19 but we have to give answer in subjective way...So we can't just do that...This question given just has the options...but we have to solve it 'properly'.
                              – jayant98
                              Nov 21 at 21:32




                              Sorry @MoKo19 but we have to give answer in subjective way...So we can't just do that...This question given just has the options...but we have to solve it 'properly'.
                              – jayant98
                              Nov 21 at 21:32












                              @jayant98: Not sure what you mean by 'properly'. If you mean analytically, then generate the quadratic equation where $a=1$ with roots $p,p+2$. Because This will give you $y=(x-p)(x-p-2)=x^2-2px-2x+p^2+2p$, and $frac{|b|}{a}=|b|=|2p+2|=2p+2$. For odd integer roots, $p=1,3,5,...$, so the ratio will be $4,8,12,16,...$. I will write a different answer showing how to solve via the quadratic formula.
                              – MoKo19
                              Nov 21 at 21:42




                              @jayant98: Not sure what you mean by 'properly'. If you mean analytically, then generate the quadratic equation where $a=1$ with roots $p,p+2$. Because This will give you $y=(x-p)(x-p-2)=x^2-2px-2x+p^2+2p$, and $frac{|b|}{a}=|b|=|2p+2|=2p+2$. For odd integer roots, $p=1,3,5,...$, so the ratio will be $4,8,12,16,...$. I will write a different answer showing how to solve via the quadratic formula.
                              – MoKo19
                              Nov 21 at 21:42


















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