Question

Find the value of k if 2x^2+(k-6)x+8=0 has equal roots

Answer 1

IF IT HAS EQUAL ROOTS THEN DISCRIMINANT MUST BE O

=> D= O

HERE,,

$(k - 6) {}^{2} - 4(2)(8) = 0 \\ = > k {}^{2} + 36 - 12k \: - 64 = 0 \\ = > k {}^{2} - 12k - 28 = 0 \\ = > k {}^{2} - 14k + 2k - 28 = 0 \\ = > k(k - 14) + 2(k - 14) = 0 \\ = > (k + 2)(k - 14) = 0 \\ = > k = 14 \: \: \: \: or \: \: \: k \: = - 2$

Answer 2

if it has equal root then b^2 = 4ac

(k-6)^2= 4(2)(8)

k^2+36-12k=64

k^2-12k-28=0

k= (12+ 16)/2 and (12-16)/2

k=14, -2