Question

Find he value of k for which the quadratic equation is (k+1)^2x 2 -(k-1)x +1=0 has equal roots

Answer 1

(K+1)x2 - 2(K-1)x + 1 = 0
THIS EQUATION HAS EQUAL ROOTS (GIVEN)
THEREFORE, D = 0
A = K + 1
B = -2(K-1)
C = 1
D = 0
B2 - 4AC = 0
(2-2K)2 - 4(K+1)(1) = 0
4 + 4K2 - 8K - 4K - 4 = 0
4K​2 - 12K = 0
4K(K-3) = 0
4K = 0   OR   K - 3 = 0
K = 0   OR   K = 3
HOPE, THIS ANSWER HELPS!!!
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