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The roots of $2x^2 − 8x − 1$ are $\alpha$ and $\beta$. Suppose another quadratic, $x^2 + qx + r$, has roots $1/(\alpha^3\beta)$ and $1/(\beta^3\alpha)$. What are $q$ and $r$?

What I did to solve this question was $-q = (\beta^2 + \alpha^2)/(\alpha^3\beta^3)$

I sub in for the denominator $-1/2$, but I'm not sure what to do with the numerator?

Thank you.

egreg
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user639649
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2 Answers2

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Guide:

Notice that

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2-2\alpha\beta$$

and you already know the sum and product of $\alpha$ and $\beta$.

Siong Thye Goh
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  • After i did everything I get r=16, and q= 136, but I dont get the same roots as the original equation, I was wondering if you knew where I went wrong and what the roots should be – user639649 Apr 24 '19 at 11:03
  • hmmm... i don't quite get the meaning of you don't get the same roots as the original equation. – Siong Thye Goh Apr 24 '19 at 11:11
  • I dont get the same roots as the original equation, do you know what r and q would be? Because i keep going around in circles and keep getting it wrong. – user639649 Apr 24 '19 at 11:12
  • you mean you are given the solution and your answer does not match it? – Siong Thye Goh Apr 24 '19 at 11:12
  • Yes ! I was wondering why and what you would do – user639649 Apr 24 '19 at 11:29
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    the first move that i would take in such circumstances is to investigate if there is a possibility that the answer is wrong. Upon verifying that it is correct, I would then inspect my steps to look out for careless mistakes. You might want to share the expected answer and your working. – Siong Thye Goh Apr 24 '19 at 12:12
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The issue may be one of checking. I do get $q=136, r=16$ also and it seems to check out. Checking requires rationalizing the denominator, and a hidden trick lies there.

From the quadratic formula for the original quadratic equation we get

$\alpha=(4+3\sqrt{2})/2$

$\beta=(4-3\sqrt{2})/2$

For the quadratic equation derived in the problem, $x^2+136x+16=0$ the roots, purposely left unlabeled, are

$-68+48\sqrt{2}$

$-68-48\sqrt{2}$

We are to prove that $1/(\alpha^3\beta)$ and $1/(\alpha\beta^3)$ match the latter roots, allowing either order.

Look at $1/(\alpha^3\beta)$. A convenient way to proceed is to first render

$1/(\alpha^3\beta)=\beta^2/(\alpha^3\beta^3)=-8\beta^2$

where we have rendered $\alpha\beta=-1/2$ from the coefficients of the given equation, thereby we have rationalized the denominator. If we put in the value if $\beta$ given above we obtain

$\color{blue}{1/(\alpha^3\beta)=-68+48\sqrt{2}}$

That checks out as one of the target roots. The reader should now verify that

$\color{blue}{1/(\alpha\beta^3)=-68-48\sqrt{2}}$

by the same method, thus both roots are matched and the claimed values of $q$ and $r$ are correct.

Oscar Lanzi
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