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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 $(x-\alpha/\beta)(x-\beta/\alpha)$. $\endgroup$ – AnonSubmitter85 Aug 19 '13 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 '13 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. $\endgroup$ – AnonSubmitter85 Aug 19 '13 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. $\endgroup$ – AnonSubmitter85 Aug 19 '13 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. $\endgroup$ – AnonSubmitter85 Aug 20 '13 at 0:33
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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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  • $\begingroup$ Oh thank you and also Mark. Well I'm stupid because you guys make it look simple. $\endgroup$ – MATHSUSER Aug 19 '13 at 10:33
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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? $\endgroup$ – AnonSubmitter85 Aug 19 '13 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. $\endgroup$ – André Nicolas Aug 19 '13 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 $(x-a)(x-b)(x-c)$? $\endgroup$ – AnonSubmitter85 Aug 19 '13 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 '13 at 11:10
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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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If $α$ and $β$ are the roots of the quadratic equation $x^2+bx+c=0$ ($a$ is not equal to zero), then equation whose roots are $α$ and $β$ is $x^2-(α+β)x+αβ=0$.

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

=>$x^2-(\frac{α^2+b^2}{αβ})x+1=0$

=>$x^2-(\frac{(α+b)^2-2αb}{αβ})x+1=0$

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

$α+β=-8$

$αβ=-5$

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

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

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

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

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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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Relation between roots and coefficient we have read that $\alpha$ plus $\beta$ ( sum of the two roots equals to $-\frac{b}{a}$) that of products of roots equals to $\frac{b}{a}$)

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    $\begingroup$ This does not answer the question $\endgroup$ – Shailesh Jul 4 '17 at 1:06

protected by Community♦ Jul 4 '17 at 9:47

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