Question

Given $x_1>0$ and $x_2=4x_1$ are solutions of the quadratic equation $a{x}^2+bx+c=0$. If $3a=2(c-b)$, then the value of $x_1$ equals to:

• A. $\frac{1}{4}$
• B. $\frac{3}{2}$
• C. $\frac{2}{3}$
• D. $\frac{1}{9}$

Answer 1

The correct option is A $\frac{1}{4}$

Sum of roots $=x_1+x_2=-\frac{b}{a}$

$\therefore x_1+4x_1=-\frac{b}{a}$

$\therefore 5x_1=-\frac{b}{a}$ ...(1)

Product of roots $=x_1{x}_2=\frac{c}{a}$

$\therefore x_{1}4x_1=\frac{c}{a}$

$\therefore 4x_1^2=\frac{c}{a}$ ...(2)

It is given that $3a=2(c-b)$

$\therefore 3=2(\frac{c}{a}-\frac{b}{a})$

Substituting (1) and (2) in above equation, we get

$\therefore 3=2(4x_1^2+5x_1)$

$\therefore 8x_1^2+10x_1-3=0$

$\therefore (2x_1+3)(4x_1-1)=0$

$\therefore x_1=-\frac{3}{2}$ or $x=\frac{1}{4}$

Since $x_1>0$

$\therefore x_1=\frac{1}{4}$