Question

Find k if one root of the equation $x^2-6kx+8=0$ is twice the other

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

Answer 1

The correct option is B $\pm 1$

since the one root is twice the others,

Let the one root be a and other root will be 2a

Now sum of roots $a+2a=6k$

and product of roots $2a^2=8$

$\therefore a=2,-2$

Now,

$3a=6k$

$k=\frac{a}{2}$

Now putting the obtained value of a, we get

$k=1,-1$