How to solve quadratic equation problem having a prepositional logic?

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When $alpha,beta$ are roots of $x^2+bx+c=0$, Find the equation whose roots are $p$ and $q$ where, $p=alpha +beta^2,,q=beta+alpha^2$. Also when $alpha,beta$ are imaginary show that, $b=-1,$if and only if $p,q$ are real.




So far I have found the required equation as (say g(x)); $$g(x)=x^2-(b^2-b-2c)x+(b^2+c^2+c+3bc)=0$$



For the second part, it is given that $alpha,beta$ are imaginary. i.e $b^2-4c<0$



First I assumed $b=-1$ then $$Delta g(x)=(b^2-b-2c)^2-4(b^2+c^2+c+3bc) =0$$
which gives $p,q$ are real (also coincidental)



Then I assumed $p,q$ are real which gives,
$$Delta g(x)geq0$$
$$(b^2-b-2c)^2geq4(b^2+c^2+c+3bc)$$



after few simplifications I ended up with,
$$(b^2-4c)(b+1)^2-4b^2(b+1)geq0$$



How can I say that $b=-1$ at this stage, Do I need to simplify this further? Please Help. Thanks.




quadratics




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    I think there's an error in $g(x)$: the constant term should be $;colorred-b^colorred3+c+c^2+3bc$.
    – Bernard
    Jul 18 at 19:15















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When $alpha,beta$ are roots of $x^2+bx+c=0$, Find the equation whose roots are $p$ and $q$ where, $p=alpha +beta^2,,q=beta+alpha^2$. Also when $alpha,beta$ are imaginary show that, $b=-1,$if and only if $p,q$ are real.




So far I have found the required equation as (say g(x)); $$g(x)=x^2-(b^2-b-2c)x+(b^2+c^2+c+3bc)=0$$



For the second part, it is given that $alpha,beta$ are imaginary. i.e $b^2-4c<0$



First I assumed $b=-1$ then $$Delta g(x)=(b^2-b-2c)^2-4(b^2+c^2+c+3bc) =0$$
which gives $p,q$ are real (also coincidental)



Then I assumed $p,q$ are real which gives,
$$Delta g(x)geq0$$
$$(b^2-b-2c)^2geq4(b^2+c^2+c+3bc)$$



after few simplifications I ended up with,
$$(b^2-4c)(b+1)^2-4b^2(b+1)geq0$$



How can I say that $b=-1$ at this stage, Do I need to simplify this further? Please Help. Thanks.




quadratics




share|cite|improve this question















  • 1




    I think there's an error in $g(x)$: the constant term should be $;colorred-b^colorred3+c+c^2+3bc$.
    – Bernard
    Jul 18 at 19:15













up vote
0
down vote

favorite









up vote
0
down vote

favorite












When $alpha,beta$ are roots of $x^2+bx+c=0$, Find the equation whose roots are $p$ and $q$ where, $p=alpha +beta^2,,q=beta+alpha^2$. Also when $alpha,beta$ are imaginary show that, $b=-1,$if and only if $p,q$ are real.




So far I have found the required equation as (say g(x)); $$g(x)=x^2-(b^2-b-2c)x+(b^2+c^2+c+3bc)=0$$



For the second part, it is given that $alpha,beta$ are imaginary. i.e $b^2-4c<0$



First I assumed $b=-1$ then $$Delta g(x)=(b^2-b-2c)^2-4(b^2+c^2+c+3bc) =0$$
which gives $p,q$ are real (also coincidental)



Then I assumed $p,q$ are real which gives,
$$Delta g(x)geq0$$
$$(b^2-b-2c)^2geq4(b^2+c^2+c+3bc)$$



after few simplifications I ended up with,
$$(b^2-4c)(b+1)^2-4b^2(b+1)geq0$$



How can I say that $b=-1$ at this stage, Do I need to simplify this further? Please Help. Thanks.




quadratics




share|cite|improve this question












When $alpha,beta$ are roots of $x^2+bx+c=0$, Find the equation whose roots are $p$ and $q$ where, $p=alpha +beta^2,,q=beta+alpha^2$. Also when $alpha,beta$ are imaginary show that, $b=-1,$if and only if $p,q$ are real.




So far I have found the required equation as (say g(x)); $$g(x)=x^2-(b^2-b-2c)x+(b^2+c^2+c+3bc)=0$$



For the second part, it is given that $alpha,beta$ are imaginary. i.e $b^2-4c<0$



First I assumed $b=-1$ then $$Delta g(x)=(b^2-b-2c)^2-4(b^2+c^2+c+3bc) =0$$
which gives $p,q$ are real (also coincidental)



Then I assumed $p,q$ are real which gives,
$$Delta g(x)geq0$$
$$(b^2-b-2c)^2geq4(b^2+c^2+c+3bc)$$



after few simplifications I ended up with,
$$(b^2-4c)(b+1)^2-4b^2(b+1)geq0$$



How can I say that $b=-1$ at this stage, Do I need to simplify this further? Please Help. Thanks.





quadratics





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asked Jul 18 at 18:55









emil

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  • 1




    I think there's an error in $g(x)$: the constant term should be $;colorred-b^colorred3+c+c^2+3bc$.
    – Bernard
    Jul 18 at 19:15













  • 1




    I think there's an error in $g(x)$: the constant term should be $;colorred-b^colorred3+c+c^2+3bc$.
    – Bernard
    Jul 18 at 19:15








1




1




I think there's an error in $g(x)$: the constant term should be $;colorred-b^colorred3+c+c^2+3bc$.
– Bernard
Jul 18 at 19:15





I think there's an error in $g(x)$: the constant term should be $;colorred-b^colorred3+c+c^2+3bc$.
– Bernard
Jul 18 at 19:15











1 Answer
1






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up vote
2
down vote



accepted










There's an error in your polynomial $g(x)$: indeed



  • $ p+q=alpha+beta+alpha^2+beta^2=-b+(-b)^2-2c=b^2-b-2c$,


  • $pq=alpha^3+beta^3+alphabeta+alpha^2beta^2=(alpha+beta)^3-3alphabeta(alpha+beta)+alphabeta+alpha^2beta^2=-b^3+3bc+c+c^2$,


so $qquadqquad g(x)=x^2-(b^2-b-2c)x-b^3+3bc+c+c^2$.



Now $p$ and $q$ are real if and only if $;(b^2-b-2c)^2+4(b^3-3bc-c-c^2)ge 0$.
Let's factorise the l.h.s.:
beginalign
(b^2-b&-2c)^2+4(b^3-3bc-c-c^2)\
&=(b^2-b)^2-4(b^2-b)c+4c^2+4(b^3-3bc-c-c^2)\
&=(b^2-b)^2+4b^3+4c(b-b^2+c-3b-1-c)=b^2(b+1)^2-4c(b+1)^2\&=(b+1)^2(b^2-4c)ge 0.
endalign
If $alpha$ and $beta$ are imaginary, $b^2-4c<0$ so the only possibility for $(b+1)^2(b^2-4c)$ to be $ge 0$ is that
$$(b+1)^2=0.$$






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    There's an error in your polynomial $g(x)$: indeed



    • $ p+q=alpha+beta+alpha^2+beta^2=-b+(-b)^2-2c=b^2-b-2c$,


    • $pq=alpha^3+beta^3+alphabeta+alpha^2beta^2=(alpha+beta)^3-3alphabeta(alpha+beta)+alphabeta+alpha^2beta^2=-b^3+3bc+c+c^2$,


    so $qquadqquad g(x)=x^2-(b^2-b-2c)x-b^3+3bc+c+c^2$.



    Now $p$ and $q$ are real if and only if $;(b^2-b-2c)^2+4(b^3-3bc-c-c^2)ge 0$.
    Let's factorise the l.h.s.:
    beginalign
    (b^2-b&-2c)^2+4(b^3-3bc-c-c^2)\
    &=(b^2-b)^2-4(b^2-b)c+4c^2+4(b^3-3bc-c-c^2)\
    &=(b^2-b)^2+4b^3+4c(b-b^2+c-3b-1-c)=b^2(b+1)^2-4c(b+1)^2\&=(b+1)^2(b^2-4c)ge 0.
    endalign
    If $alpha$ and $beta$ are imaginary, $b^2-4c<0$ so the only possibility for $(b+1)^2(b^2-4c)$ to be $ge 0$ is that
    $$(b+1)^2=0.$$






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      There's an error in your polynomial $g(x)$: indeed



      • $ p+q=alpha+beta+alpha^2+beta^2=-b+(-b)^2-2c=b^2-b-2c$,


      • $pq=alpha^3+beta^3+alphabeta+alpha^2beta^2=(alpha+beta)^3-3alphabeta(alpha+beta)+alphabeta+alpha^2beta^2=-b^3+3bc+c+c^2$,


      so $qquadqquad g(x)=x^2-(b^2-b-2c)x-b^3+3bc+c+c^2$.



      Now $p$ and $q$ are real if and only if $;(b^2-b-2c)^2+4(b^3-3bc-c-c^2)ge 0$.
      Let's factorise the l.h.s.:
      beginalign
      (b^2-b&-2c)^2+4(b^3-3bc-c-c^2)\
      &=(b^2-b)^2-4(b^2-b)c+4c^2+4(b^3-3bc-c-c^2)\
      &=(b^2-b)^2+4b^3+4c(b-b^2+c-3b-1-c)=b^2(b+1)^2-4c(b+1)^2\&=(b+1)^2(b^2-4c)ge 0.
      endalign
      If $alpha$ and $beta$ are imaginary, $b^2-4c<0$ so the only possibility for $(b+1)^2(b^2-4c)$ to be $ge 0$ is that
      $$(b+1)^2=0.$$






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        There's an error in your polynomial $g(x)$: indeed



        • $ p+q=alpha+beta+alpha^2+beta^2=-b+(-b)^2-2c=b^2-b-2c$,


        • $pq=alpha^3+beta^3+alphabeta+alpha^2beta^2=(alpha+beta)^3-3alphabeta(alpha+beta)+alphabeta+alpha^2beta^2=-b^3+3bc+c+c^2$,


        so $qquadqquad g(x)=x^2-(b^2-b-2c)x-b^3+3bc+c+c^2$.



        Now $p$ and $q$ are real if and only if $;(b^2-b-2c)^2+4(b^3-3bc-c-c^2)ge 0$.
        Let's factorise the l.h.s.:
        beginalign
        (b^2-b&-2c)^2+4(b^3-3bc-c-c^2)\
        &=(b^2-b)^2-4(b^2-b)c+4c^2+4(b^3-3bc-c-c^2)\
        &=(b^2-b)^2+4b^3+4c(b-b^2+c-3b-1-c)=b^2(b+1)^2-4c(b+1)^2\&=(b+1)^2(b^2-4c)ge 0.
        endalign
        If $alpha$ and $beta$ are imaginary, $b^2-4c<0$ so the only possibility for $(b+1)^2(b^2-4c)$ to be $ge 0$ is that
        $$(b+1)^2=0.$$






        share|cite|improve this answer













        There's an error in your polynomial $g(x)$: indeed



        • $ p+q=alpha+beta+alpha^2+beta^2=-b+(-b)^2-2c=b^2-b-2c$,


        • $pq=alpha^3+beta^3+alphabeta+alpha^2beta^2=(alpha+beta)^3-3alphabeta(alpha+beta)+alphabeta+alpha^2beta^2=-b^3+3bc+c+c^2$,


        so $qquadqquad g(x)=x^2-(b^2-b-2c)x-b^3+3bc+c+c^2$.



        Now $p$ and $q$ are real if and only if $;(b^2-b-2c)^2+4(b^3-3bc-c-c^2)ge 0$.
        Let's factorise the l.h.s.:
        beginalign
        (b^2-b&-2c)^2+4(b^3-3bc-c-c^2)\
        &=(b^2-b)^2-4(b^2-b)c+4c^2+4(b^3-3bc-c-c^2)\
        &=(b^2-b)^2+4b^3+4c(b-b^2+c-3b-1-c)=b^2(b+1)^2-4c(b+1)^2\&=(b+1)^2(b^2-4c)ge 0.
        endalign
        If $alpha$ and $beta$ are imaginary, $b^2-4c<0$ so the only possibility for $(b+1)^2(b^2-4c)$ to be $ge 0$ is that
        $$(b+1)^2=0.$$







        share|cite|improve this answer













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        answered Jul 18 at 19:59









        Bernard

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