Question

If roots of the quadratic equation
3y²+k y +12=0
are real and equal
then find k​

Answer 1

Let $a=3$, $b=k$, $c=12$ in $3y^2+ky+12=0$ .

we are given that  $b^2-4ac=0$.

so,  $k^2-2(3)(12)=0$

now, $k^2=6*6*2=6^2*2$

       $k=6\sqrt{2}$

∴ equation is as follows:-

$3y^2+6\sqrt{2} y+12=0$