Question

For what value of
k does
(k-12) x2 + 2(k-12)x+2=0
have equal roots?​

Answer 1

Step-by-step explanation:

k-12x²+(2k-24)x+2

b=2k-24

a=k-12

c=2

d=0

b²-4ac=0

(2k-24) ²-4(k-12) (2)=0

4k²+576(-96-4k+48)(2)=0

4k²+576-192-8k+96=0

4k²-8k+480=0

divide by 4

k²-2k+120=0

k²+10k-12k+120=0

k(k+10) -12(k-10)=0

(k+10)(k-10)=0,k-12=0

k²-10²=0,k=12

k²=100,k=12

k=10,k=12