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If $\alpha$ and $\beta$ are the roots of the equation $x^2 + 8x - 5 = 0$, find the quadratic equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.

My working out so far: I know that $\alpha + \beta = -8$ and $\alpha \beta = -5$ (from the roots) and thenIi go on to work out that $\alpha= -8-\beta$ and $\beta= -8-\alpha$, then I substitute into what the question asks me.

$\frac{-8-\beta}{-8-\alpha}$ and $\frac{-8-\alpha}{-8-\beta}$ however I do not know how to proceed further. I might be doing this completely wrong and my apologies for that.

Another solution came to me that if $\alpha$ and $\beta$ are roots of the other unknown equation. I can somehow manipulate that to find the answer. But I don't think that will work. All help is appreciated thank you.

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    $\begingroup$ I know this has already been answered below, but I think it's even simpler to note that the equation you are looking for is just $(x-\alpha/\beta)(x-\beta/\alpha)=0$. $\endgroup$ Aug 19, 2013 at 10:40
  • $\begingroup$ Much simpler indeed. But I think the answers below provide a much more reasonable and understandable answer (what you do and why). However this is much simpler. $\endgroup$
    – MATHSUSER
    Aug 19, 2013 at 10:43
  • $\begingroup$ You know $\alpha$ and $\beta$ from the problem statement, so their product and sum is irrelevant. Just substitute them into the above and you have your answer. The steps described below are not germane to the problem and only complicate things. $\endgroup$ Aug 19, 2013 at 10:46
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    $\begingroup$ Yes. All you have to do is write $\alpha = -4 - \sqrt{21}$, $\beta = -4 + \sqrt{21}$, and the desired quadratic equation is $(x-\alpha/\beta)(x-\beta/\alpha)$. That's it. The most important thing is to note that if you are asked to find the quadratic equation with roots $a$ and $b$, the answer is just $(x-a)(x-b)$. You don't have to look for relationships between $a$ and $b$ to find it. $\endgroup$ Aug 19, 2013 at 11:02
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    $\begingroup$ @Greebo Point taken. I don't why I couldn't see this before; I had the blinders on. $\endgroup$ Aug 20, 2013 at 0:33

5 Answers 5

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The product of the roots $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$ is $1$. That part was easy! The sum will be more work.

The sum $\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}$ of the roots simplifies to $\dfrac{\alpha^2+\beta^2}{\alpha\beta}$.

But $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$. Thus the sum of the roots is $\dfrac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$.

Substituting the known values of $\alpha+\beta$ and $\alpha\beta$ we find that the sum of the roots is $\dfrac{(-8)^2-2(-5)}{-5}$.

This simplifies to $-\dfrac{74}{5}$.

Thus the equation is $$x^2+\frac{74}{5}x+1=0.$$ We can multiply through by $5$ if we wish.

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    $\begingroup$ Why go through all of this? The OP is given $\alpha$ and $\beta$ in the problem statement by virtue of $x^2 +8x - 5 =0$. Thus, the quadratic equation with the two desired roots is just $(x-\alpha/\beta)(x-\beta/\alpha)$. What does it matter what their sum and product are? $\endgroup$ Aug 19, 2013 at 10:48
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    $\begingroup$ I agree that one could solve explicitly for $\alpha$ and $\beta$. But the relationship between the elementary symmetric functions and the roots is very important. Expressing symmetric functions like $s/t+t/s$ in terms of elementary symmetric functions has theoretical importance. So the method we used gives some practice at that. Also, in a similar problem with say cubics, getting the roots explicitly would be painful, but the above approach still works smoothly. $\endgroup$ Aug 19, 2013 at 10:58
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    $\begingroup$ Ok. But if one were asked to find the cubic equation with roots $a$, $b$, and $c$, wouldn't you want them to recognize that the answer is just $(x-a)(x-b)(x-c)$? $\endgroup$ Aug 19, 2013 at 11:05
  • $\begingroup$ @AnonSubmitter85 That is exactly what is done. The relationships between the roots and the coefficients is exactly what you get when you compute this formula. $\endgroup$
    – Greebo
    Aug 19, 2013 at 11:10
  • $\begingroup$ That too. But suppose we are given a cubic equation, and the roots are being called $\alpha,\beta,\gamma$ but they are not findable in any nice way. Suppose we want a cubic whose roots are $\alpha^2$, $\beta^2$, and $\gamma^2$. Using symmetric functions, we can find the equation. But again, mainly the exercise was about practice with symmetric functions. $\endgroup$ Aug 19, 2013 at 11:16
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Let the new roots be $\gamma = \frac {\alpha}{\beta}$ and $\delta = \frac {\beta}{\alpha}$.

Compute $\gamma\delta =1$ and $\gamma+\delta = \frac {\alpha^2+\beta^2}{\alpha\beta}$

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

Can you put the pieces together now?

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

As $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$, using the Vieta formulas, the equation is

$$-5x^2-(8^2+2\cdot5)x-4=0,$$

$$5x^2+74x+5=0.$$

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If $α$ and $β$ are the roots of the quadratic equation $ax^2+bx+c=0$ ,that is, $x^2+\frac{b}{a}x+\frac{c}{a}=0$ ($a$ is not equal to zero), then the equation can be written as $x^2-(α+β)x+αβ=0$.

The equation whose roots are $\frac{α}{β}$ and $\frac{β}{α}$ is: $x^2-(\frac{α}{β}+\frac{β}{α})x+(\frac{α}{β})(\frac{β}{α})=0$.

$\implies x^2-(\frac{α^2+b^2}{αβ})x+1=0$

$\implies x^2-(\frac{(α+b)^2-2αb}{αβ})x+1=0$

$α$ and $β$ are the roots of the equation $x² + 8x - 5 = 0$

$α+β=-8$

$αβ=-5$

Therefore the required equation is $x^2-(\frac{(-8)^2-2(-5)}{-5})x+1=0$

$\implies x^2-(\frac{74}{-5})x+1=0$

$\implies x^2+\frac{74}{5}x+1=0$

$\implies 5x^2+74x+5=0$

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Since we want the zeroes of the new quadratic polynomial to be the mutual reciprocals $ \ \frac{\alpha}{\beta} \ $ and $ \ \frac{\beta}{\alpha} \ \ , $ it must be palindromic, that is, of the form $ \ Ax^2 + Bx + A \ \ $ (the sort of symmetry André Nicolas was referring to in his comments). Viete tells us that $ \ \alpha + \beta \ = \ -8 \ $ and that $ \ \alpha·\beta \ = \ -5 \ \ , $ as other posters have found. So we may conclude that the sum of the reciprocals of these roots is $$ \frac{\alpha \ + \ \beta}{\alpha · \beta} \ \ = \ \ \frac{1}{\alpha} \ + \ \frac{1}{\beta} \ \ = \ \ \frac{-8}{-5} \ \ = \ \ \frac85 \ \ . $$ We may also divide the sum-of-roots equation through to find that $$ \frac{\alpha \ + \ \beta}{\alpha} \ \ = \ \ 1 \ + \ \frac{ \beta}{\alpha} \ \ = \ \ \frac{-8}{\alpha} \ \ \ \text{and} \ \ \ \frac{\alpha \ + \ \beta}{\beta} \ \ = \ \ 1 \ + \ \frac{\alpha}{ \beta} \ \ = \ \ \frac{-8}{\beta} \ \ . $$ We thus obtain $$ \left(1 \ + \ \frac{ \beta}{\alpha} \right) \ + \ \left(1 \ + \ \frac{\alpha}{ \beta} \right) \ = \ -8·\left(\frac{1}{\alpha} + \frac{1}{\beta} \right) \ \ = \ \ -\frac{64}{5} $$ $$ \Rightarrow \ \ \frac{\alpha}{ \beta} \ + \ \frac{ \beta}{\alpha} \ \ = \ \ -\frac{64}{5} \ - \ 2 \ \ = \ \ -\frac{74}{5} \ \ . $$ Our monic palindromic polynomial is therefore $ \ x^2 \ + \ \frac{74}{5} · x \ + \ 1 \ \ . $ [In answer to the question, a corresponding quadratic equation is $ \ A·\left(x^2 + \frac{74}{5} · x + 1 \right) \ = \ 0 \ \ , $ where any non-zero value for $ \ A \ $ will do.]

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It isn't overly difficult to work with the roots themselves: the discriminant of the original quadratic polynomial is $ \ 8^2 + 4·5 \ = \ 84 \ \ , $ so the zeroes are $ \ \alpha \ , \ \beta \ = \ -4 \ \pm \ \sqrt{21} \ \ . $ From this, we have reasonably convenient calculation with "conjugate" numbers: $$ \frac{\alpha}{\beta} \ + \ \frac{\beta}{\alpha} \ \ = \ \ \frac{\alpha^2 \ + \ \beta^2}{\alpha·\beta} \ \ = \ \ \frac{(-4 \ + \ \sqrt{21})^2 \ + \ (-4 \ - \ \sqrt{21})^2}{(-4 \ + \ \sqrt{21})·(-4 \ - \ \sqrt{21})} \ \ = \ \ \frac{2·[ \ (-4)^2 \ + \ (\sqrt{21})^2 \ ]}{(-4)^2 \ - \ (\sqrt{21})^2} $$ $$ = \ \ \frac{2·( 16 \ + \ 21 )}{16 \ - \ 21} \ = \ -\frac{74}{5} \ \ . $$

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