Question
There is a number \( x \) such that \( x^2 \) is irrational but \( x^4 \) is rational. Then \( x \) can be:
(a) \( \sqrt{5} \)
(b) \( \sqrt{2} \)
(c) \( 3\sqrt{2} \)
(d) \( 4\sqrt{5} \)
Solution
We need a number \( x \) such that:
- \( x^2 \) is irrational
- \( x^4 \) is rational
Option (a): \( \sqrt{5} \)
\[ x^2 = 5 \quad (\text{rational}) \]
✘ Not suitable
Option (b): \( \sqrt{2} \)
\[ x^2 = 2 \quad (\text{rational}) \]
✘ Not suitable
Option (c): \( 3\sqrt{2} \)
\[ x^2 = 9 \times 2 = 18 \quad (\text{rational}) \]
✘ Not suitable
Option (d): \( 4\sqrt{5} \)
\[ x^2 = 16 \times 5 = 80 \quad (\text{rational}) \]
✘ Not suitable
None of the given options satisfy the condition.
Correct Idea:
Let \( x = \sqrt[4]{2} \)
\[ x^2 = \sqrt{2} \quad (\text{irrational}) \]
\[ x^4 = 2 \quad (\text{rational}) \]
Final Answer
✔ None of the options are correct