Question

What is 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. $-2/3$
• B. $1/3$
• C. $-1/3$
• D. $2/3$

Answer 1

The correct option is D $2/3$

Let $\alpha ,2\alpha$ be the roots of the given equation

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

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

By $\left(1\right)$ and $\left(2\right)$ we have

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

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

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

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

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

$\Rightarrow -39a=-26$

$\Rightarrow a=\frac{-26}{-39}=\frac{2}{3}$

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