Question

The value of $a$ for which one root of the quadratic equation $(a^2-5a+3)x^2+(3a-1)x+2=0$ is twice as large as the other, is

• A. $-\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $-\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{2}{3}$

$(a^2-5a+3)x^2+(3a-1)x+2=0$
Let the roots be $\alpha$ and $2\alpha .$
Then, $\alpha +2\alpha =\frac{1-3a}{a^2-5a+3}$ and $\alpha \cdot 2\alpha =\frac{2}{a^2-5a+3}$
$\Rightarrow \alpha =\frac{1-3a}{3(a^2-5a+3)}$ and ${\alpha }^2=\frac{1}{a^2-5a+3}$

$\therefore {\left(\frac{1-3a}{3(a^2-5a+3)}\right)}^2=\frac{1}{a^2-5a+3}$
$\Rightarrow 1+9a^2-6a=9a^2-45a+27$
$\Rightarrow 39a=26$
$\Rightarrow a=\frac{2}{3}$