If a = (3+√7)/2 , then find the value of a² + 1/a²

1.	If a = (3+√7)/2 , then find the value of a² + 1/a²
Answer» $\blue{\bold{\underline{\underline{Answer:}}}}$ $\green{\tt{\therefore{a^{2}+\frac{1}{a^{2}}=\frac{40-9\sqrt{7}}{2}}}}\\$ $\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$ $\green{\underline \bold{<klux>GIVEN</klux> :}} \\ \tt: \implies a = \frac{3 + \sqrt{7} }{ 2} \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } =?$ • ACCORDING to given QUESTION : $\bold{As \: we \: know \: that} \\ \tt: \implies {a}^{2} = ( \frac{3 + \sqrt{7} }{2})^{2} \\ \\ \tt: \implies {a}^{2} = \frac{ {3}^{2} + {( \sqrt{7}) }^{2} + 2 \times 3 \times \sqrt{7} }{4} \\ \\ \tt: \implies {a}^{2} = \frac{9 + 7 + 6 \sqrt{7} }{4} \\ \\ \tt: \implies {a}^{2} = \frac{16 + 6 \sqrt{7} }{4} \\ \\ \tt: \implies {a}^{2} = \frac{8 + 3 \sqrt{7} }{2} - - - - - (1) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{1}{( { \frac{3 + \sqrt{7} }{2}) }^{2} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{4}{9 + 7 + 6 \sqrt{7} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{2}{8 + 3 \sqrt{7} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{2}{8 + 3 \sqrt{7} } \times \frac{8 - 3\sqrt{7} }{8 - 3 \sqrt{7} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{16 - 6 \sqrt{7} }{64 - 63} \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = 16 - 6 \sqrt{7} - - - - - (2) \\ \\ \bold{For \: finding \: value } \\ \tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } = \frac{8 + 3 \sqrt{7} }{2} + 16 - 6 \sqrt{7} \\ \\ \tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } = \frac{8 + 3 \sqrt{7} + 32 - 12 \sqrt{7} }{2} \\ \\ \green{\tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } = \frac{40 - 9 \sqrt{7} }{2} }$

Answer»

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{a^{2}+\frac{1}{a^{2}}=\frac{40-9\sqrt{7}}{2}}}}\\$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{<klux>GIVEN</klux> :}} \\ \tt: \implies a = \frac{3 + \sqrt{7} }{ 2} \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } =?$

• ACCORDING to given QUESTION :

$\bold{As \: we \: know \: that} \\ \tt: \implies {a}^{2} = ( \frac{3 + \sqrt{7} }{2})^{2} \\ \\ \tt: \implies {a}^{2} = \frac{ {3}^{2} + {( \sqrt{7}) }^{2} + 2 \times 3 \times \sqrt{7} }{4} \\ \\ \tt: \implies {a}^{2} = \frac{9 + 7 + 6 \sqrt{7} }{4} \\ \\ \tt: \implies {a}^{2} = \frac{16 + 6 \sqrt{7} }{4} \\ \\ \tt: \implies {a}^{2} = \frac{8 + 3 \sqrt{7} }{2} - - - - - (1) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{1}{( { \frac{3 + \sqrt{7} }{2}) }^{2} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{4}{9 + 7 + 6 \sqrt{7} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{2}{8 + 3 \sqrt{7} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{2}{8 + 3 \sqrt{7} } \times \frac{8 - 3\sqrt{7} }{8 - 3 \sqrt{7} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{16 - 6 \sqrt{7} }{64 - 63} \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = 16 - 6 \sqrt{7} - - - - - (2) \\ \\ \bold{For \: finding \: value } \\ \tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } = \frac{8 + 3 \sqrt{7} }{2} + 16 - 6 \sqrt{7} \\ \\ \tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } = \frac{8 + 3 \sqrt{7} + 32 - 12 \sqrt{7} }{2} \\ \\ \green{\tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } = \frac{40 - 9 \sqrt{7} }{2} }$

If a = (3+√7)/2 , then find the value of a² + 1/a²

• ACCORDING to given QUESTION :

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