Question

If one root of quadratic equation $x^2-x+3m=0$ is 4 times the root of the equation $x^2-x+m=,0$ where m$\ne 0,$ then m equals

• A. $\frac{12}{196}$
• B. $\frac{12}{169}$
• C. $\frac{12}{256}$
• D. $\frac{12}{189}$

Answer 1

The correct option is B $\frac{12}{169}$

Let $a,4a$ be the of the equations,

Since,

$x^2-x+3m=0$ and $x^2-x+m=0$

Therefore,

$16a^2-4a+3m=0$.......(1)............using $\left(4a\right)$

$a^2-a+m=0$............(2)...............using $\left(a\right)$

$\left(1\right)-3\times \left(2\right)$ gives,

$13a^2-a=0$

$\Rightarrow a=\frac{1}{13}$

From (2)

${\left(\frac{1}{13}\right)}^2-\frac{1}{13}+m=0$

$m=\frac{13-1}{{13}^2}$

$\therefore m=\frac{12}{169}$

Hence, this is the answer.