If p ≠ 1 in (p − 1)x² − 2(p + 1)x + (p + 3) = 0, what is the condition on p for real roots?
Answer and explanation
Correct answer: Every p ≠ 1
Because p ≠ 1, the coefficient of x² is nonzero, so the equation is genuinely quadratic. Its discriminant is D = [−2(p + 1)]² − 4(p − 1)(p + 3). Now (p + 1)² = p² + 2p + 1 and (p − 1)(p + 3) = p² + 2p − 3. Therefore D = 4[(p² + 2p + 1) − (p² + 2p − 3)] = 16, which is positive for every real p. Thus two distinct real roots exist for every p except p = 1, where the equation ceases to be quadratic. Hence option D is correct; option A incorrectly introduces an unnecessary upper bound.
Frequently asked questions
What is the correct answer to this question?
Every p ≠ 1
Why is this the correct answer?
Because p ≠ 1, the coefficient of x² is nonzero, so the equation is genuinely quadratic. Its discriminant is D = [−2(p + 1)]² − 4(p − 1)(p + 3). Now (p + 1)² = p² + 2p + 1 and (p − 1)(p + 3) = p² + 2p − 3. Therefore D = 4[(p² + 2p + 1) − (p² + 2p − 3)] = 16, which is positive for every real p. Thus two distinct real roots exist for every p except p = 1, where the equation ceases to be quadratic. Hence option D is correct; option A incorrectly introduces an unnecessary upper bound.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Nature of Roots.
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