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Find the values of k for which the following equation has equal roots : (k-12) x2 +2 (k-12)x+2 = 0

JEE/Engineering Exams
Maths
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(K-12) x² + 2(K-12) nt Q=0 equation has real roots it 6-49c=0 [2CK-12)]- 4 (k-12]2=0 qCk_12) {(k-12) - 2}=0 K-1470 K=14 for k=12,14 equation has K=12 or real root
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