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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, $$\Delta g(x)\geq0$$ $$(b^2-b-2c)^2\geq4(b^2+c^2+c+3bc)$$

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.

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

1 Answers1

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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$,

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

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

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.: \begin{align} (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. \end{align} 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 $$(b+1)^2=0.$$

Bernard
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