Question

Find the value of k for which (k-4) x2 + 2(k-4) x + 20 = 0 has equal roots​

Answer 1

Your Answer is -1.

Step-by-step explanation:

(k-4)*2+2*(k-4)+20=0

k-4)*2+2*(k-4)+20=02k-8+2k-8+20=0

k-4)*2+2*(k-4)+20=02k-8+2k-8+20=04k-8+12=0

k-4)*2+2*(k-4)+20=02k-8+2k-8+20=04k-8+12=04k+4=0

k-4)*2+2*(k-4)+20=02k-8+2k-8+20=04k-8+12=04k+4=04k= -4

k-4)*2+2*(k-4)+20=02k-8+2k-8+20=04k-8+12=04k+4=04k= -4k= -4/4

k-4)*2+2*(k-4)+20=02k-8+2k-8+20=04k-8+12=04k+4=04k= -4k= -4/4k =-1

Hope it helped.