Question

If one of the roots of the equation $16x^2-20kx+6=0$ is $\frac{3}{4}$, then find the value of $k$.

• A. $2$
• B. $-2$
• C. $1$
• D. $-1$

Answer 1

The correct option is C $1$

We know that if $\alpha$ is the root of an equation $a{x}^2+bx+c=0$, then $a{\alpha }^2+b\alpha +c=0$.
Substituting $x=\frac{3}{4}$ in the equation $16x^2-20kx+6=0$, we get,
$⟹16(\frac{3}{4}{)}^2-20k(\frac{3}{4})+6=0$
$⟹9-15k+6=0$
$⟹15-15k=0$
$⟹k=1$

$\therefore$ The value of $k$ is $1$.