Which of the following statements about \(\sqrt[3]{12}\) is correct?
Answer and explanation
Correct answer: It is an irrational number
If \(\sqrt[3]{12}\) were rational it could be written as a reduced fraction \(\dfrac{p}{q}\) with integers p,q; cubing gives \(p^3=12q^3\). That forces the prime exponents in \(p^3\) to match those of 12q^3, so the exponents of primes coming from 12 must be multiples of 3. But 12=2^2·3^1 has exponents 2 and 1, which are not multiples of 3, a contradiction. Hence \(\sqrt[3]{12}\) is irrational. Distractors fail: 3 and 4 are wrong because 3^3=27 and 4^3=64 (not 12); a terminating decimal would be rational. Exam tip: first check whether the radicand is a perfect cube; use prime-power exponents to test rationality of integer roots.
Frequently asked questions
What is the correct answer to this question?
It is an irrational number
Why is this the correct answer?
If \(\sqrt[3]{12}\) were rational it could be written as a reduced fraction \(\dfrac{p}{q}\) with integers p,q; cubing gives \(p^3=12q^3\). That forces the prime exponents in \(p^3\) to match those of 12q^3, so the exponents of primes coming from 12 must be multiples of 3. But 12=2^2·3^1 has exponents 2 and 1, which are not multiples of 3, a contradiction. Hence \(\sqrt[3]{12}\) is irrational. Distractors fail: 3 and 4 are wrong because 3^3=27 and 4^3=64 (not 12); a terminating decimal would be rational. Exam tip: first check whether the radicand is a perfect cube; use prime-power exponents to test rationality of integer 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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