What type of number is \(\sqrt{121}+\sqrt{2}\)?
Answer and explanation
Correct answer: Irrational number
\(\sqrt{121}=11\), which is rational, while \(\sqrt{2}\) is irrational. The sum of a rational number and an irrational number is always irrational. Hence, \(\sqrt{121}+\sqrt{2}=11+\sqrt{2}\) is irrational. Integers and natural numbers are rational, so options C and D cannot be correct. Exam tip: the square root of a perfect square is an integer, but \(2\) is not a perfect square.
Frequently asked questions
What is the correct answer to this question?
Irrational number
Why is this the correct answer?
\(\sqrt{121}=11\), which is rational, while \(\sqrt{2}\) is irrational. The sum of a rational number and an irrational number is always irrational. Hence, \(\sqrt{121}+\sqrt{2}=11+\sqrt{2}\) is irrational. Integers and natural numbers are rational, so options C and D cannot be correct. Exam tip: the square root of a perfect square is an integer, but \(2\) is not a perfect square.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Number Systems. Topic: Irrational numbers.
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