(k + 4)x2 + (k + 1)x + 1 = 0 has equal roots if
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k = 5
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k = 3
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k = -5
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k > 0
Reveal answer
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A
Correct answer
Explanation
Since the equation has equal roots, D = 0. b2 - 4ac = 0 (k + 1)2 - 4 (k + 4) (1) = 0 k2 + 2k + 1 - 4k - 16 = 0 $\Rightarrow$ k2 - 2k - 15 = 0 k2 - 5k + 3k - 15 = 0 k(k - 5) +3(k - 5) = 0 $\Rightarrow$ (k - 5) (k + 3) = 0 k = 5, -3, so option (1) is the answer.
AI explanation
A quadratic has equal (repeated) roots when its discriminant b²–4ac equals zero. Here a=(k+4), b=(k+1), c=1, so (k+1)²–4(k+4)=0 simplifies to k²–2k–15=0, which factors as (k–5)(k+3)=0. This gives k=5 (or k=–3), so k=5 is a valid value that produces equal roots.