If x² - 2rx + (r² - 16) = 0, what is the nature of the roots for any real value of r?
Answer and explanation
Correct answer: Two real rational and distinct
For a quadratic ax² + bx + c = 0, the discriminant is D = b² - 4ac. In this equation, a = 1, b = -2r, and c = r² - 16. Therefore D = (-2r)² - 4(1)(r² - 16) = 4r² - 4r² + 64 = 64. This value is positive for every real r, so the roots are always real and distinct. Moreover, 64 is a perfect square, so the quadratic formula produces rational roots: x = (2r ± 8)/2 = r ± 4. Hence option A is correct. Equal roots would require D = 0, no real roots require D < 0, and irrational roots would require a positive but non-square discriminant.
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What is the correct answer to this question?
Two real rational and distinct
Why is this the correct answer?
For a quadratic ax² + bx + c = 0, the discriminant is D = b² - 4ac. In this equation, a = 1, b = -2r, and c = r² - 16. Therefore D = (-2r)² - 4(1)(r² - 16) = 4r² - 4r² + 64 = 64. This value is positive for every real r, so the roots are always real and distinct. Moreover, 64 is a perfect square, so the quadratic formula produces rational roots: x = (2r ± 8)/2 = r ± 4. Hence option A is correct. Equal roots would require D = 0, no real roots require D < 0, and irrational roots would require a positive but non-square discriminant.
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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