Question
The roots of x² − (k+3)x + (3k − 1) = 0 are real
and distinct, and the larger root exceeds the smaller by 5. Find k.Solution
ATQ,
Let roots a, b with a − b = 5 and a + b = k+3, ab = 3k − 1. Then (a − b)² = (a + b)² − 4ab ⇒ 25 = (k+3)² − 4(3k − 1). ⇒ 25 = k² + 6k + 9 − 12k + 4 ⇒ 0 = k² − 6k − 12. k = [6 ± √(36 + 48)]/2 = [6 ± √84]/2 = [6 ± 2√21]/2 = 3 ± √21. Both make distinct real roots. Answer: k = 3 ± √21.
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