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The set of all vales of m for which both the roots of the equation x2 - (m+1)x + m + 4 = 0 are real and negative, is A. (−,−3] [5, ∞)B. [−3, 5] C. (−4, −3] D. (−3, −1] |
Answer» For roots to be real its D ≥ 0 \(\sqrt{(m+1)^2 - 4(1)(m+4)}≥ 0\) (m + 1)2 – 4(m + 4) ≥ 0 m2 – 2m – 15 ≥ 0 (m – 1)2 – 16 ≥ 0 (m – 1)2 ≥ 16 m – 1 ≤ -4 or m – 1 ≥ 4 m ≤ -3 or m ≥ 5 For both roots to be negative product of roots should be positive and sum of roots should be negative. Product of roots = m + 4 > 0 ⇒ m > -4 Sum of roots = m + 1 < 0 ⇒ m < -1 After taking intersection of D ≥ 0, Product of roots > 0 and sum of roots < 0. We can say that the final answer is m ∈ (-4, -3] |
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