Question

The roots of the quadratic equation are real and equal,find k. 3y^2+ky+12=0

Answer 1

Since the roots are real and equal then b
$b {}^{2} - 4ac = 0$
Therefore in the above equation
a=3
b=k
c=12
Substituting the values we get,
$k {2} - 4(3)(12) = 0$
$k ^{2} - 144 = 0$
$k {}^{2} = 144$
k = +12 or - 12