Question

find the value of k for which the roots of the quadratic equation $[k-4]x^{2} + 2[k-4]x + 2=0$ are equal

Answer 1

(k-4)x²+2(k-4)x+2=0

⇒b²=4ac

then a=k-4; b=2(k-4) ;c=2; substituting the values

(2k-8)²=4×(k-4)×2

4k²+64-32k=8k-32

4k²-40k+96=0

k²-10k+24=0

(k-4)(k-6)=0

k=4,6

thank you