Question : If $a$ and $b$ are rationals and $a\sqrt{2}+b\sqrt{3}=\sqrt{98}+\sqrt{108}-\sqrt{48}-\sqrt{72}$ then, the values of $a$ and $b$ respectively, are:
Option 1: 1, 2
Option 2: 1, 3
Option 3: 2, 1
Option 4: 2, 3
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Correct Answer: 1, 2
Solution :
$a\sqrt{2}+b\sqrt{3} = \sqrt{98}+\sqrt{108}-\sqrt{48}-\sqrt{72}$
$=\sqrt{2\times49}+\sqrt{36\times3}-\sqrt{16\times3}-\sqrt{36\times2}$
$= 7\sqrt{2}+6\sqrt{3}-4\sqrt{3}-6\sqrt{2}$
⇒ $a\sqrt{2}+b\sqrt{3} = \sqrt{2}+2\sqrt{3}$
⇒ $a=1$ and $b=2$
Hence, the correct answer is 1, 2.
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