If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root,then $a+b+c$ can be...












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If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root, then $a+b+c$ can be equal to(where $a,b,cin R,ane0$)

$(1);5a$
$(2);3b$
$(3);2c$
$(4);0$




As $x=1$ is the root of the equation $2x^3+x^2+2x-5=0$ and its other two roots are complex in nature. So the common root is $x=1$


Put $x=1$ in $ax^3+(a+b)x^2+(b+c)x+c=0$ we get $2a+2b+2c=0$ so $(4)$ option is correct. But $(1)$ and $(3)$ options are also correct. I don't know how they are correct.






quadratics







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  • $begingroup$
    Can't they have complex common roots?
    $endgroup$
    – Bernard
    Dec 26 '18 at 10:46










  • $begingroup$
    Have you checked with the roots of 2x²+3x+5=0 too? Your second equation gives 1 real root (=1) and 2 complex roots coming from the equation I wrote.
    $endgroup$
    – Epsilon zero
    Dec 26 '18 at 10:46


















2












$begingroup$


If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root, then $a+b+c$ can be equal to(where $a,b,cin R,ane0$)

$(1);5a$
$(2);3b$
$(3);2c$
$(4);0$




As $x=1$ is the root of the equation $2x^3+x^2+2x-5=0$ and its other two roots are complex in nature. So the common root is $x=1$


Put $x=1$ in $ax^3+(a+b)x^2+(b+c)x+c=0$ we get $2a+2b+2c=0$ so $(4)$ option is correct. But $(1)$ and $(3)$ options are also correct. I don't know how they are correct.






quadratics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can't they have complex common roots?
    $endgroup$
    – Bernard
    Dec 26 '18 at 10:46










  • $begingroup$
    Have you checked with the roots of 2x²+3x+5=0 too? Your second equation gives 1 real root (=1) and 2 complex roots coming from the equation I wrote.
    $endgroup$
    – Epsilon zero
    Dec 26 '18 at 10:46
















2












2








2


1



$begingroup$


If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root, then $a+b+c$ can be equal to(where $a,b,cin R,ane0$)

$(1);5a$
$(2);3b$
$(3);2c$
$(4);0$




As $x=1$ is the root of the equation $2x^3+x^2+2x-5=0$ and its other two roots are complex in nature. So the common root is $x=1$


Put $x=1$ in $ax^3+(a+b)x^2+(b+c)x+c=0$ we get $2a+2b+2c=0$ so $(4)$ option is correct. But $(1)$ and $(3)$ options are also correct. I don't know how they are correct.






quadratics







share|cite|improve this question











$endgroup$




If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root, then $a+b+c$ can be equal to(where $a,b,cin R,ane0$)

$(1);5a$
$(2);3b$
$(3);2c$
$(4);0$




As $x=1$ is the root of the equation $2x^3+x^2+2x-5=0$ and its other two roots are complex in nature. So the common root is $x=1$


Put $x=1$ in $ax^3+(a+b)x^2+(b+c)x+c=0$ we get $2a+2b+2c=0$ so $(4)$ option is correct. But $(1)$ and $(3)$ options are also correct. I don't know how they are correct.






quadratics




quadratics






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edited Dec 26 '18 at 10:48









Bernard

123k741116




123k741116










asked Dec 26 '18 at 10:25









user984325user984325

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  • $begingroup$
    Can't they have complex common roots?
    $endgroup$
    – Bernard
    Dec 26 '18 at 10:46










  • $begingroup$
    Have you checked with the roots of 2x²+3x+5=0 too? Your second equation gives 1 real root (=1) and 2 complex roots coming from the equation I wrote.
    $endgroup$
    – Epsilon zero
    Dec 26 '18 at 10:46




















  • $begingroup$
    Can't they have complex common roots?
    $endgroup$
    – Bernard
    Dec 26 '18 at 10:46










  • $begingroup$
    Have you checked with the roots of 2x²+3x+5=0 too? Your second equation gives 1 real root (=1) and 2 complex roots coming from the equation I wrote.
    $endgroup$
    – Epsilon zero
    Dec 26 '18 at 10:46


















$begingroup$
Can't they have complex common roots?
$endgroup$
– Bernard
Dec 26 '18 at 10:46




$begingroup$
Can't they have complex common roots?
$endgroup$
– Bernard
Dec 26 '18 at 10:46












$begingroup$
Have you checked with the roots of 2x²+3x+5=0 too? Your second equation gives 1 real root (=1) and 2 complex roots coming from the equation I wrote.
$endgroup$
– Epsilon zero
Dec 26 '18 at 10:46






$begingroup$
Have you checked with the roots of 2x²+3x+5=0 too? Your second equation gives 1 real root (=1) and 2 complex roots coming from the equation I wrote.
$endgroup$
– Epsilon zero
Dec 26 '18 at 10:46












2 Answers
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Note that
$$ax^3+(a+b)x^2+(b+c)x+c=(x+1)(ax^2+bx+c)$$
and
$$2x^3+x^2+2x-5=(x-1)(2x^2+3x+5)$$
So we can also have the case when the roots in common are the two complex roots: $a=2k$, $b=3k$ and $c=5k$, that is $a+b+c=5a=2c$.






share|cite|improve this answer









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    $begingroup$

    The equation $2x^3+x^2+2x−5=0$ has 3 solutions, $$1, quad frac14 left(-3-isqrt{31} right), quad mbox{and} quad frac14 left(-3+isqrt{31} right).$$ If you put $x=1$ (as you did) you obtained $a+b+c=0$. In the same way if you put $x=frac14 left(-3-isqrt{31} right)$ (to check your computation maybe use wolframaplha) you get $2b=3a$ and $2c=5a$, which is equivalent to $$a+b+c=5a.$$ I let you do on your own the last case.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      $begingroup$

      Note that
      $$ax^3+(a+b)x^2+(b+c)x+c=(x+1)(ax^2+bx+c)$$
      and
      $$2x^3+x^2+2x-5=(x-1)(2x^2+3x+5)$$
      So we can also have the case when the roots in common are the two complex roots: $a=2k$, $b=3k$ and $c=5k$, that is $a+b+c=5a=2c$.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Note that
        $$ax^3+(a+b)x^2+(b+c)x+c=(x+1)(ax^2+bx+c)$$
        and
        $$2x^3+x^2+2x-5=(x-1)(2x^2+3x+5)$$
        So we can also have the case when the roots in common are the two complex roots: $a=2k$, $b=3k$ and $c=5k$, that is $a+b+c=5a=2c$.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Note that
          $$ax^3+(a+b)x^2+(b+c)x+c=(x+1)(ax^2+bx+c)$$
          and
          $$2x^3+x^2+2x-5=(x-1)(2x^2+3x+5)$$
          So we can also have the case when the roots in common are the two complex roots: $a=2k$, $b=3k$ and $c=5k$, that is $a+b+c=5a=2c$.






          share|cite|improve this answer









          $endgroup$



          Note that
          $$ax^3+(a+b)x^2+(b+c)x+c=(x+1)(ax^2+bx+c)$$
          and
          $$2x^3+x^2+2x-5=(x-1)(2x^2+3x+5)$$
          So we can also have the case when the roots in common are the two complex roots: $a=2k$, $b=3k$ and $c=5k$, that is $a+b+c=5a=2c$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 26 '18 at 10:50









          Robert ZRobert Z

          101k1069142




          101k1069142























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              $begingroup$

              The equation $2x^3+x^2+2x−5=0$ has 3 solutions, $$1, quad frac14 left(-3-isqrt{31} right), quad mbox{and} quad frac14 left(-3+isqrt{31} right).$$ If you put $x=1$ (as you did) you obtained $a+b+c=0$. In the same way if you put $x=frac14 left(-3-isqrt{31} right)$ (to check your computation maybe use wolframaplha) you get $2b=3a$ and $2c=5a$, which is equivalent to $$a+b+c=5a.$$ I let you do on your own the last case.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                The equation $2x^3+x^2+2x−5=0$ has 3 solutions, $$1, quad frac14 left(-3-isqrt{31} right), quad mbox{and} quad frac14 left(-3+isqrt{31} right).$$ If you put $x=1$ (as you did) you obtained $a+b+c=0$. In the same way if you put $x=frac14 left(-3-isqrt{31} right)$ (to check your computation maybe use wolframaplha) you get $2b=3a$ and $2c=5a$, which is equivalent to $$a+b+c=5a.$$ I let you do on your own the last case.






                share|cite|improve this answer









                $endgroup$
















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                  2








                  2





                  $begingroup$

                  The equation $2x^3+x^2+2x−5=0$ has 3 solutions, $$1, quad frac14 left(-3-isqrt{31} right), quad mbox{and} quad frac14 left(-3+isqrt{31} right).$$ If you put $x=1$ (as you did) you obtained $a+b+c=0$. In the same way if you put $x=frac14 left(-3-isqrt{31} right)$ (to check your computation maybe use wolframaplha) you get $2b=3a$ and $2c=5a$, which is equivalent to $$a+b+c=5a.$$ I let you do on your own the last case.






                  share|cite|improve this answer









                  $endgroup$



                  The equation $2x^3+x^2+2x−5=0$ has 3 solutions, $$1, quad frac14 left(-3-isqrt{31} right), quad mbox{and} quad frac14 left(-3+isqrt{31} right).$$ If you put $x=1$ (as you did) you obtained $a+b+c=0$. In the same way if you put $x=frac14 left(-3-isqrt{31} right)$ (to check your computation maybe use wolframaplha) you get $2b=3a$ and $2c=5a$, which is equivalent to $$a+b+c=5a.$$ I let you do on your own the last case.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 26 '18 at 11:03









                  antoineantoine

                  211




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