If x² + 2(t + 1)x + (3t + 7) = 0 has no real roots, in which interval will t lie?
Answer and explanation
Correct answer: −2 < t < 3
For no real roots, the discriminant must be negative. In this equation a = 1, b = 2(t + 1), and c = 3t + 7. Therefore D = [2(t + 1)]² − 4(3t + 7) = 4(t + 1)² − 4(3t + 7) = 4(t² + 2t + 1 − 3t − 7) = 4(t² − t − 6). Factoring gives D = 4(t − 3)(t + 2). The product is negative between the two roots −2 and 3. Hence −2 < t < 3. At the endpoints D = 0, giving equal real roots, while outside the interval D > 0.
Frequently asked questions
What is the correct answer to this question?
−2 < t < 3
Why is this the correct answer?
For no real roots, the discriminant must be negative. In this equation a = 1, b = 2(t + 1), and c = 3t + 7. Therefore D = [2(t + 1)]² − 4(3t + 7) = 4(t + 1)² − 4(3t + 7) = 4(t² + 2t + 1 − 3t − 7) = 4(t² − t − 6). Factoring gives D = 4(t − 3)(t + 2). The product is negative between the two roots −2 and 3. Hence −2 < t < 3. At the endpoints D = 0, giving equal real roots, while outside the interval D > 0.
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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