Question

The value of \(\frac{\sqrt{32} + \sqrt{48}}{\sqrt{8} + \sqrt{12}}\) is equal to :

• A. \(\sqrt{2}\)
• B. 2
• C. \(4\)
• D. \(8\)

Hint:

1. Simplify each surd first: \(\sqrt{32} = 4\sqrt{2}\), \(\sqrt{48} = 4\sqrt{3}\) \(\sqrt{8} = 2\sqrt{2}\), \(\sqrt{12} = 2\sqrt{3}\) 2. Substitute into the fraction: \(\frac{4\sqrt{2} + 4\sqrt{3}}{2\sqrt{2} + 2\sqrt{3}}\). 3. Factor out common numbers: \(\frac{4(\sqrt{2} + \sqrt{3})}{2(\sqrt{2} + \sqrt{3})}\). 4. Cancel the common term \((\sqrt{2} + \sqrt{3})\). 5. Result: \(\frac{4}{2} = 2\).

Answer 1

The Correct Option is B

Concept: To simplify expressions involving square roots (surds), we look for perfect square factors within the numbers under the square root. The property \(\sqrt{ab} = \sqrt{a}\sqrt{b}\) is used. Step 1: Simplify each square root term
\(\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2}\)
\(\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3}\)
\(\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}\)
\(\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}\) Step 2: Substitute the simplified terms back into the expression The expression is \(\frac{\sqrt{32} + \sqrt{48}}{\sqrt{8} + \sqrt{12}}\). Substituting the simplified terms: \[ \frac{4\sqrt{2} + 4\sqrt{3}}{2\sqrt{2} + 2\sqrt{3}} \] Step 3: Factor out common terms from the numerator and denominator
Numerator: \(4\sqrt{2} + 4\sqrt{3} = 4(\sqrt{2} + \sqrt{3})\)
Denominator: \(2\sqrt{2} + 2\sqrt{3} = 2(\sqrt{2} + \sqrt{3})\) So the expression becomes: \[ \frac{4(\sqrt{2} + \sqrt{3})}{2(\sqrt{2} + \sqrt{3})} \] Step 4: Cancel out the common factor \((\sqrt{2} + \sqrt{3})\) Since \((\sqrt{2} + \sqrt{3})\) is a common factor in both the numerator and the denominator, and it is not zero, we can cancel it out: \[ \frac{4}{2} \] Step 5: Calculate the final value \[ \frac{4}{2} = 2 \] The value of the expression is 2. This matches option (2).