Question

# Find the values of k for which the following equation has equal roots : (k-12) x2 +2 (k-12)x+2 = 0

Solution

(K-12) x² + 2(K-12) nt Q=0
equation has real roots it
6-49c=0
[2CK-12)]- 4 (k-12]2=0
qCk_12) {(k-12) - 2}=0
K-1470
K=14
for k=12,14 equation has
K=12
or
real root