Question

Find the value of $k$ for the quadratic equation $3y^2+ky+12=0$ has two equal roots.

Answer 1

$3y^2+ky+12=0$

Comparing equation with $a{x}^2+bx+c=0$

$a=3$,$x=y$,$b=k$ and $c=12$

Since the equation has $2$ equal roots,$D=0$

$\Rightarrow b^2-4ac=0$

$\Rightarrow k^2-4\times 3\times 12=0$

$\Rightarrow k^2=144$

$\therefore k=\pm 12$