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

Answer 1

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

Given equation $(a^2-5a+3)x^2+(3a-1)x+2=0$

Let $\alpha$ be the root of the equation.

Hence, other root will be $2\alpha$.

Now,

$3\alpha =\frac{1-3a}{a^2-5a+3}$

and

$2{\alpha }^2=\frac{2}{a^2-5a+3}$

${\alpha }^2=\frac{1}{a^2-5a+3}$

Hence,

$\frac{(1-3a{)}^2}{9(a^2-5a+3{)}^2}=\frac{1}{a^2-5a+3}$

$9(a^2-5a+3)=9a^2-6a+1$

$-45a+27=-6a+1$

$26=39a$

$a=\frac{2}{3}$.