Question : If $\sqrt{11-3 \sqrt{8}}=a+b \sqrt{2}$, then what is the value of (2a + 3b) ?
Option 1: 7
Option 2: 9
Option 3: 3
Option 4: 5
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Correct Answer: 3
Solution : Given: $\sqrt{11-3 \sqrt{8}}=a+b \sqrt{2}$ ⇒ $\sqrt{11-3 \sqrt{2\times2\times2}}=a+b \sqrt{2}$ ⇒ $\sqrt{9+2-2×3\sqrt{2}}=a+b \sqrt{2}$ ⇒ $\sqrt{(3)^2+(\sqrt2)^2-2\times3\sqrt{2}}=a+b \sqrt{2}$ ⇒ $\sqrt{(3-\sqrt{2})^2}=a+b \sqrt{2}$ ⇒ $3-\sqrt{2}=a+b \sqrt{2}$ Comparing both sides, we get, $a=3,b=-1$ Now, $(2a+3b)$ = $(2\times3+3\times(-1))$ = $3$ Hence, the correct answer is 3.
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