Q. Given that $\alpha$ and $\beta$ are roots of the equation $x^2+3x-6=0$, find a quadratic equation with integer coefficients whose roots are $\frac{2}{\alpha}$ and $\frac{2}{\beta}$
My working for this question basically gave me a wrong answer although i see it as perfectly ok. I basically used the quadratic formula and got $\frac{-3\pm\sqrt{33}}{2}$ and then i divided each one by 2 again giving the same answer but over 4. Then i just made brackets using the answer and got: $(x+\frac{3+\sqrt{33}}{4})(x+\frac{3-\sqrt{33}}{4})$ which expands to give $x^2+\frac{3x}{2}-\frac{3}{2}$ But the answer im supposed to get is $3x^2-3x-2=0$ which i can't get. I f anyone could see whats wron with my method and show me how to get the right answer that'd be great unfortunatley my notes arent eith me atm thnx

2. Originally Posted by oxrigby
Q. Given that $\alpha$ and $\beta$ are roots of the equation $x^2+3x-6=0$, find a quadratic equation with integer coefficients whose roots are $\frac{2}{\alpha}$ and $\frac{2}{\beta}$
My working for this question basically gave me a wrong answer although i see it as perfectly ok. I basically used the quadratic formula and got $\frac{-3\pm\sqrt{33}}{2}$ and then i divided each one by 2 again giving the same answer but over 4. Then i just made brackets using the answer and got: $(x+\frac{3+\sqrt{33}}{4})(x+\frac{3-\sqrt{33}}{4})$ which expands to give $x^2+\frac{3x}{2}-\frac{3}{2}$ But the answer im supposed to get is $3x^2-3x-2=0$ which i can't get. I f anyone could see whats wron with my method and show me how to get the right answer that'd be great unfortunatley my notes arent eith me atm thnx
You do not have to solve the given equation.
You have to apply the given condition which is in question. See here,

Since $\alpha$ and $\beta$ are the roots of the equation $x^2+3x-6=0$

from the equation, a = 1, b = 3 and c = -6

so, sum of roots $= \alpha + \beta = \frac{-b}{a}= \frac{-3}{2}$

Product of roots $= \alpha.\beta=\frac{c}{a}=\frac{-6}{1}=-6$

Now, we have to make an equation whose roots are $\frac{2}{\alpha}$ and $\frac{2}{\beta}$

now, SUM of roots = $\frac{2}{\alpha}+\frac{2}{\beta}$

$= \frac{2\alpha +2\beta}{\alpha \beta}= \frac{2(\alpha +\beta)}{\alpha \beta}$

$=\frac{2\left(\frac{-3}{2}\right)}{-6}=\frac{-3}{-6}=\frac{1}{2}$

PRODUCT of roots = $\frac{2}{\alpha}\times \frac{2}{\beta}$

$= \frac{4}{\alpha \beta}= \frac{4}{-6}=\frac{-2}{3}$

So, the required equation is

$x^2 - (SUM)x + (PRODUCT) = 0$

$x^2-\left(\frac{1}{2}\right)x + \left(\frac{-2}{3}\right)=0$

$6x^2-3x-4=0$