Question

find the roots of the quadratic equation X^2– x + k =0 and the value of the constant

k given that one root of the equation is twice the other.

Answer 1

$$x^2-x+k=0$$

By Viet Theorem

$$x_1+x_2=-\dfrac{-1}{1}=1$$

$$x_1x_2=\dfrac{k}{1}=k$$

Given $x_2=2x_1.$ Then

$$x_1+2x_1=1$$

$$3x_1=1$$

$$x_1=\dfrac{1}{3}, x_2=\dfrac{2}{3}$$

$$k=\dfrac{1}{3}(\dfrac{2}{3})=\dfrac{2}{9}$$

$\{\dfrac{1}{3}, \dfrac{2}{3}\}$

$k=\dfrac{2}{9}$