1. ## roots of quadratic equation

Question: a)Prove that, if the sum of the reciprocals of the roots of the equation
$\displaystyle ax^2+bx+c=0$
is 1 then $\displaystyle b+ c = 0.$
If, in addition, one root of the equation is twice the other use the result of $\displaystyle 2b^2=9ac$ to find one set values of a, b, c. Solve the equation.

My attempt

a) $\displaystyle ax^2+bx+c=0$
$\displaystyle \alpha + \beta = -\frac{b}{a}$
$\displaystyle \alpha \beta = \frac{c}{a}$
$\displaystyle \frac{1}{\alpha} +\frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = -\frac{b}{c} = 1$
$\displaystyle b+c=0$
b)
$\displaystyle \alpha_{1} = \frac{1}{\alpha}$
$\displaystyle \beta_{1}=\frac{1}{\beta} =\frac{2}{\alpha}$
sum of roots
$\displaystyle sum =\frac{1}{\alpha} +\frac{2}{\alpha} = \frac{3}{\alpha}$
product of roots
$\displaystyle product =\frac{1}{\alpha} \times \frac{2}{\alpha} = \frac{2}{\alpha^2}$

$\displaystyle \frac{3}{\alpha} =1$

$\displaystyle \alpha =3$
therefore

$\displaystyle \beta_{1} = \frac{2}{3}$
since

sum of roots = $\displaystyle 1$

products of roots = $\displaystyle \frac{2}{\alpha^2} = \frac{2}{3^2} = \frac{2}{9}$

$\displaystyle x^2 - (1)x + \frac{2}{9} = 0$
$\displaystyle 9x^2 - 9x + 2 =0$
from this; a=9, b = -9, c = 2 and x = 2/3 or 1/3

the answer given by the book is a =2, b= -9 c= 9 x = 3, 3/2
I know how to find $\displaystyle ax^2+bx+c=0$ but i dont know how to use it to find the answer

I think your error is at the start of part b) It says one root is twice the other, so $\alpha = 2\beta$, $\frac1\alpha = \frac1{2\beta}$, $\alpha + \beta = 3\beta = 1$, and $\alpha\beta = 2\beta^2$