Question

For what value of k does (k-12)xsq+(k-12)x+2=0 have equal root?

Answer 1

(k-12)x^2+(k-12)x+2=0  

above equation has equal roots

b^2-4ac=0

(k-12)x^2+(k-12)x+2=0  

b^2-4ac=0

(k-2)^2-4(k-12)*2=0

k^2-4k+4-8k+96=0

k^2-12k+100=0

k^2-10k-2k+100=0

k(k-10)-2(k-10)=0

(k-10)(k-2)=0

k-10=0

k-2=0

k=2

k=10

SO that k is 10 and k is 2.

These both are the real roots of above equations!!  

It can also be solved by using factorization method and quadratic method too