Question

Which of the following is rational number ?

• A. \(\sqrt{8}\)
• B. \(\sqrt{8} - \sqrt{4}\)
• C. \(\sqrt{2} + \sqrt{2}\)
• D. \(\sqrt{63} \times \sqrt{7}\)

Hint:

When multiplying square roots, combine them: \(\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}\).
Check if \(63 \times 7\) is a perfect square.
\(63 = 9 \times 7 \implies 63 \times 7 = 9 \times 7^2\).
Since both \(9\) and \(7^2\) are perfect squares, their product must also be a perfect square, making the result a rational number instantly.

Answer 1

The Correct Option is D

Step 1: Simplify option D by pulling out a factor first.
Write $\sqrt{63}$ as $\sqrt{9 \times 7} = 3\sqrt7$. Then \[ \sqrt{63} \times \sqrt7 = 3\sqrt7 \times \sqrt7 = 3 \times (\sqrt7)^2 = 3 \times 7 = 21 \] Since 21 is a whole number, it is rational.
Step 2: Quickly rule out the others.
$\sqrt8 = 2\sqrt2$ stays irrational since $\sqrt2$ never cancels. $\sqrt8 - \sqrt4 = 2\sqrt2 - 2$ is still irrational because subtracting a whole number from an irrational number keeps it irrational. $\sqrt2 + \sqrt2 = 2\sqrt2$ is irrational for the same reason.
Step 3: Confirm option D is the only rational result.
Only in option D does the surd part cancel out completely, using $\sqrt7 \times \sqrt7 = 7$, leaving a pure integer.
Step 4: State the answer.
So $\sqrt{63} \times \sqrt7$ is the rational number among the choices.
\[ \boxed{21} \]