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 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

$\frac{2}{3}$

$(a^2-5a+3a)x^2+(3a-1)x+2=\mathrm{0......}\left(1\right)$

Let $\alpha$ and $2\alpha$ be the roots of $\left(1\right)$, then

$(a^2-5a+3){\alpha }^2+(3a-1)\alpha +2=0........\left(2\right)and(a^2-5a+3)\left(4{\alpha }^2\right)+(3a-1)\left(2\alpha \right)+2=0.......\left(3\right)$

Multiplying equation $\left(2\right)$ by $4$ and subtracting it form $\left(3\right)$, we get $(3a-1)\left(2\alpha \right)+6=0$

Clearly $a\ne \frac{1}{3}$

$\therefore \alpha =\frac{-3}{(3a-1)}$

Putting this value in $\left(2\right)$, we get

$(a^2-5a+3)\left(9\right)-(3a-1{)}^2(3)+2(3a-1{)}^2=0\Rightarrow 9a^2-45a+27-(9a^2-6a+1)=0\Rightarrow -39a+26=0\Rightarrow a=\frac{2}{3}Fora=\frac{2}{3},$ the equation becomes $x^2+9x+18=0,$ whose roots are $-3,-6.$