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

Answer 1

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

Given equation is $(a^2-5a+3)x^2+(3a-1)x+2=0$
Let the roots be $p$ and $2p$
$\Rightarrow p+2p=\frac{1-3a}{a^2-5a+3},$

$\&p\times 2p=\frac{2}{a^2-5a+3}$

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

$\Rightarrow \frac{(1-3a{)}^2}{a^2-5a+3}=9\Rightarrow 9a^2-6a+1-9a^2+45a-27=0\Rightarrow 39a=26\Rightarrow a=\frac{2}{3}$