If \(x=\sqrt{17}\), what type of number is \(x^2-5\)?
Answer and explanation
Correct answer: Rational number
With \(x=\sqrt{17}\), we have \(x^2=(\sqrt{17})^2=17\). Therefore \(x^2-5=17-5=12\). The number 12 is an integer and can be written as \(12/1\), so it is rational. Option B (irrational) is incorrect because irrational numbers cannot be expressed as a ratio of integers, whereas 12 can. Option C (non-real) is wrong because 12 is a real number. Option D (non-repeating decimal) is wrong because 12 is a terminating decimal (12.0), not a non-repeating non-terminating decimal. Exam tip: simplify powers of radicals first — squaring a square root removes the root, then evaluate the remaining arithmetic to classify the result.
Frequently asked questions
What is the correct answer to this question?
Rational number
Why is this the correct answer?
With \(x=\sqrt{17}\), we have \(x^2=(\sqrt{17})^2=17\). Therefore \(x^2-5=17-5=12\). The number 12 is an integer and can be written as \(12/1\), so it is rational. Option B (irrational) is incorrect because irrational numbers cannot be expressed as a ratio of integers, whereas 12 can. Option C (non-real) is wrong because 12 is a real number. Option D (non-repeating decimal) is wrong because 12 is a terminating decimal (12.0), not a non-repeating non-terminating decimal. Exam tip: simplify powers of radicals first — squaring a square root removes the root, then evaluate the remaining arithmetic to classify the result.
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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