If a = frac{3+sqrt{5}}{2},then find the value | vedclass

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(B) Given,$a = \frac{3+\sqrt{5}}{2}$.First,calculate $a^{2}$:$a^{2} = \left(\frac{3+\sqrt{5}}{2}\right)^{2} = \frac{9 + 5 + 6\sqrt{5}}{4} = \frac{14 +

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$7$

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(B) Given,$a = \frac{3+\sqrt{5}}{2}$.First,calculate $a^{2}$:$a^{2} = \left(\frac{3+\sqrt{5}}{2}\right)^{2} = \frac{9 + 5 + 6\sqrt{5}}{4} = \frac{14 + 6\sqrt{5}}{4} = \frac{7 + 3\sqrt{5}}{2}$.Next,calculate $\frac{1}{a^{2}}$ by rationalizing the denominator:$\frac{1}{a^{2}} = \frac{2}{7 + 3\sqrt{5}} = \frac{2(7 - 3\sqrt{5})}{(7 + 3\sqrt{5})(7 - 3\sqrt{5})} = \frac{2(7 - 3\sqrt{5})}{49 - 45} = \frac{2(7 - 3\sqrt{5})}{4} = \frac{7 - 3\sqrt{5}}{2}$.Finally,add $a^{2}$ and $\frac{1}{a^{2}}$:$a^{2} + \frac{1}{a^{2}} = \frac{7 + 3\sqrt{5}}{2} + \frac{7 - 3\sqrt{5}}{2} = \frac{7 + 3\sqrt{5} + 7 - 3\sqrt{5}}{2} = \frac{14}{2} = 7$.

If $a = \frac{3+\sqrt{5}}{2}$,then find the value of $a^{2} + \frac{1}{a^{2}}$.

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