Question : If $a=\sqrt{6}+\sqrt{5},b=\sqrt{6}-\sqrt{5}$, then $2a^{2}-5\; ab+2b^{2} =?$
Option 1: 38
Option 2: 39
Option 3: 40
Option 4: 41
Correct Answer: 39
Solution :
Given: $a=\sqrt{6}+\sqrt{5}$ and $b=\sqrt{6}-\sqrt{5}$
Then,
$a^{2} = (\sqrt{6}+\sqrt{5})^{2} = 6 + 2\sqrt{30} + 5 = 11 + 2\sqrt{30}$
$b^{2} = (\sqrt{6}-\sqrt{5})^{2} = 6 - 2\sqrt{30} + 5 = 11 - 2\sqrt{30}$
Now, the product of $a$ and $b$:
$ab = (\sqrt{6}+\sqrt{5})(\sqrt{6}-\sqrt{5}) = 6 - 5 = 1$
Putting the values into the given expression:
$2a^{2}-5ab+2b^{2} = 2(11 + 2\sqrt{30}) - 5(1) + 2(11 - 2\sqrt{30})$
Simplifying this, we get:
$= 22 + 4\sqrt{30}- 5 + 22 - 4\sqrt{30}$
$=44-5 = 39$
Hence, the correct answer is 39.
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