If a = 5 + 2sqrt{6} and b = frac{1}{a},then w | vedclass

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(B) Given $a = 5 + 2\sqrt{6}$.Since $b = \frac{1}{a}$,we rationalize the denominator:$b = \frac{1}{5 + 2\sqrt{6}} 	imes \frac{5 - 2\sqrt{6}}{5 - 2\sq

Vedclass · Accepted Answer

$98$

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(B) Given $a = 5 + 2\sqrt{6}$.Since $b = \frac{1}{a}$,we rationalize the denominator:$b = \frac{1}{5 + 2\sqrt{6}} 	imes \frac{5 - 2\sqrt{6}}{5 - 2\sqrt{6}} = \frac{5 - 2\sqrt{6}}{5^2 - (2\sqrt{6})^2} = \frac{5 - 2\sqrt{6}}{25 - 24} = 5 - 2\sqrt{6}$.We need to find $a^2 + b^2$. Using the identity $a^2 + b^2 = (a + b)^2 - 2ab$:First,calculate $a + b = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 10$.Next,calculate $ab = (5 + 2\sqrt{6})(5 - 2\sqrt{6}) = 5^2 - (2\sqrt{6})^2 = 25 - 24 = 1$.Substituting these values into the identity:$a^2 + b^2 = (10)^2 - 2(1) = 100 - 2 = 98$.

If $a = 5 + 2\sqrt{6}$ and $b = \frac{1}{a}$,then what will be the value of $a^2 + b^2$?

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