If p(x) = x² + 6x + 7, what type of zeroes does it have?
Answer and explanation
Correct answer: Two distinct irrational real zeroes
For x² + 6x + 7, the coefficients are a = 1, b = 6 and c = 7. Its discriminant is D = b² − 4ac = 6² − 4(1)(7) = 36 − 28 = 8. Since D is positive, the polynomial has two distinct real zeroes. However, 8 is not a perfect square, so the square root appearing in the quadratic formula is irrational. Consequently both zeroes are irrational and distinct. Option A is correct. Option B would require a positive perfect-square discriminant, option C would require D = 0, and option D would require a negative discriminant.
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What is the correct answer to this question?
Two distinct irrational real zeroes
Why is this the correct answer?
For x² + 6x + 7, the coefficients are a = 1, b = 6 and c = 7. Its discriminant is D = b² − 4ac = 6² − 4(1)(7) = 36 − 28 = 8. Since D is positive, the polynomial has two distinct real zeroes. However, 8 is not a perfect square, so the square root appearing in the quadratic formula is irrational. Consequently both zeroes are irrational and distinct. Option A is correct. Option B would require a positive perfect-square discriminant, option C would require D = 0, and option D would require a negative discriminant.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Irrational numbers and real numbers.
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