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Find the values of k, for which the given equation has real roots:x2 + k(4x + k - 1) + 2 = 0 |
Answer» Roots are equal ∴ d = 0 d = b2 – 4ac d = (4k)2 – 4 (k2 – k + 2) (1) d = 16k2 – 4k2 + 4k – 8 Put d = 0 0 = 16k2 – 4k2 + 4k – 8 12k2 + 4k2 + 4k – 8 = 0 (divide by 4) 3k2 + k – 2 = 0 3k2 + 3k – 2k – 2 = 0 3k (k + 1) – 2k (k + 1) = 0 (3k–2k)(k+1) = 0 (3k–2k)=0 or (k+1) = 0 k = 0 k = –1 |
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