If p(x)=x^2−16, which statement about the type of zeroes is correct?
Answer and explanation
Correct answer: Both are rational real
To find the zeroes, set p(x)=0: x²−16=0. This is a difference of two squares, so (x−4)(x+4)=0. Therefore x=4 or x=−4. Both roots are integers, and integers are rational numbers; they are also real because they lie on the real number line. Hence both zeroes are rational real numbers, as stated in option A. The presence of a square does not automatically make a root irrational: √16 simplifies to 4. Option B incorrectly leaves √16 unsimplified, option C would require a negative discriminant, and option D wrongly assigns different types to the two roots.
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What is the correct answer to this question?
Both are rational real
Why is this the correct answer?
To find the zeroes, set p(x)=0: x²−16=0. This is a difference of two squares, so (x−4)(x+4)=0. Therefore x=4 or x=−4. Both roots are integers, and integers are rational numbers; they are also real because they lie on the real number line. Hence both zeroes are rational real numbers, as stated in option A. The presence of a square does not automatically make a root irrational: √16 simplifies to 4. Option B incorrectly leaves √16 unsimplified, option C would require a negative discriminant, and option D wrongly assigns different types to the two roots.
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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