If p(x) = x² + 6x + 10, which statement about its real zeroes is correct?
Answer and explanation
Correct answer: It has no real zero
The governing concept is that the real zeroes of p(x) are the x-values where the graph meets the x-axis. For the quadratic p(x) = x² + 6x + 10, complete the square: p(x) = (x + 3)² + 1. Since (x + 3)² is always at least 0, p(x) is always at least 1, so it can never equal 0 for any real x. Equivalently, its discriminant is b² − 4ac = 36 − 40 = −4, which is negative, confirming that there are no real roots. Therefore option C is correct. Option A would require a positive discriminant, option B would require a zero discriminant, and option D is impossible for a nonzero quadratic.
Frequently asked questions
What is the correct answer to this question?
It has no real zero
Why is this the correct answer?
The governing concept is that the real zeroes of p(x) are the x-values where the graph meets the x-axis. For the quadratic p(x) = x² + 6x + 10, complete the square: p(x) = (x + 3)² + 1. Since (x + 3)² is always at least 0, p(x) is always at least 1, so it can never equal 0 for any real x. Equivalently, its discriminant is b² − 4ac = 36 − 40 = −4, which is negative, confirming that there are no real roots. Therefore option C is correct. Option A would require a positive discriminant, option B would require a zero discriminant, and option D is impossible for a nonzero quadratic.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Geometrical meaning of the zeroes of a polynomial..
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