If x² − 2rx + (r² − 49) = 0, what is the nature of the roots for any real value of r?
Answer and explanation
Correct answer: Two real, rational and distinct
The governing test is the discriminant D = b² − 4ac for ax² + bx + c = 0. In this equation, a = 1, b = −2r and c = r² − 49. Thus D = (−2r)² − 4(1)(r² − 49) = 4r² − 4r² + 196 = 196. Since D is positive, the equation has two real and distinct roots. Since 196 = 14² is a perfect square, the roots have the rational form x = [2r ± 14]/2 = r ± 7. Consequently, the correct classification is two real, rational and distinct roots. Equal roots would require D = 0, and no real roots would require D < 0. Option D is unsuitable because the discriminant is a perfect square, not a non-square. Therefore option A is the only correct answer.
Frequently asked questions
What is the correct answer to this question?
Two real, rational and distinct
Why is this the correct answer?
The governing test is the discriminant D = b² − 4ac for ax² + bx + c = 0. In this equation, a = 1, b = −2r and c = r² − 49. Thus D = (−2r)² − 4(1)(r² − 49) = 4r² − 4r² + 196 = 196. Since D is positive, the equation has two real and distinct roots. Since 196 = 14² is a perfect square, the roots have the rational form x = [2r ± 14]/2 = r ± 7. Consequently, the correct classification is two real, rational and distinct roots. Equal roots would require D = 0, and no real roots would require D < 0. Option D is unsuitable because the discriminant is a perfect square, not a non-square. Therefore option A is the only correct answer.
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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