Question

The value of 'a' for which one root of the quadratic equation $\left(a^2-5a+3\right)x^2+\left(3a-1\right)x+2=0$ 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

Answer: (B)

$\frac{2}{3}$

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

Let the roots be ${\alpha }_1\beta$, given $\beta =2\alpha$

$\Rightarrow 3\alpha =\frac{-(3a-1)}{(a^2-5a+3)}\dots \left(1\right)$ (Sum of roots)

$\Rightarrow 2{\alpha }^2=\frac{2}{a^2-5a+3}\dots \left(2\right)$ (Product of roots)

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

$\Rightarrow \frac{(3a-1{)}^2-9(a^2-5a+3)}{(a^2-5a+3{)}^2}=0$

$\Rightarrow 9a^2-6a+1-9a^2+45a-27=0$

$\Rightarrow 39a-26=0$

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