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. $-\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

Answer: (D)

$\frac{2}{3}$

let us rewrite the equation as $a{x}^2+bx+c=0$ having roots as $\alpha$ and $2\alpha$
$2{\alpha }^2=\frac{c}{a}$ ...(i)
And $3\alpha =-\frac{b}{a}$ ...(ii)
From (i) and (ii) by elimination of $\alpha$, we get
$2b^2=9ac$
Substituting the actual values of $a,b,c$, we obtain
$2(3a-1{)}^2=9(2\left)\right(a^2-5a+3)$
$9a^2-6a+1=9a^2-45a+27$ ...(iii)
Therefore from (iii), we have
$a=\frac{2}{3}$