Question

[SIMPLE]Please solve:
(for values of k for which the given equation has real and equal roots):
$(k-12) x^{2} +2(k-12)x+2=0$

Answer 1

D= b² - 4ac =0
  
[2(k-12)]²  - 4×( k-12)×(2) =0 

4 ( k² - 24k + 144 )  - 8k + 96 =0 

4k²  - 96k + 576 - 8k + 96 = 0 

4k²  - 104 k +672 = 0

k²  -2 6k +168 =0 

By factorising method
 
k² - 14k - 12k +168 = 0 
 k×( k-14) -12× (k-14) =0 
(k-12)(k-14)=0 
k=12  or  k=14 

But k = 12 cannot satisfy the equation . So the value for k is 14.

K = 14.