★ Find the value of k. So that, the Quadratic equation have real roots
=> y² + k² = 2(k+1)y ㅤ
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Answered by
2
Answer:
Answer to your question,
Y2 + K2-2 (K-1) Y=0
Y2 - 2(K-1) Y + K2 = 0
A = 1, B = -2(K-1), C = K2
GIVEN ROOTS ARE EQUAL, THEN B2 - 4AC = 0
{-2 (K-1)} SQUARE -4*1* K2 = 0
4 (K2 - 2K+1) - 4K2 = 0
4 [ K2 - 2K +1 - K2} = 0
4 {1-2K} = 0
1-2K = 0/4 = 0
1 = 2K
1/2=K
K=1/2
Answered by
1
Answer:
Y2 + K2-2 (K-1) Y=0
Y22(K-1) Y + K2 = 0
A = 1, B = -2(K-1), C = K2
GIVEN ROOTS ARE EQUAL, THEN B2 -
4AC = 0
(-2 (K-1)) SQUARE -4*1* K2 = 0
4 (K22K+1) - 4K2 = 0
4 [ K2 - 2K+1-K2} = 0
4 [1-2K) = 0
1-2K = 0/4 0
1 = 2K
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