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 C

$\frac{2}{3}$

Let the roots are $\alpha$ and $2\alpha$

$\Rightarrow \alpha +2\alpha =\frac{1-3a}{a^2-5a+3}$ and $\alpha .2\alpha =\frac{2}{a^2-5a+3}\Rightarrow 2\left[\frac{1}{9}\frac{(1-3a{)}^2}{(a^2-5a+3{)}^2}\right]=\frac{2}{a^2-5a+3}$

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