Question

$if \: \frac{1}{a} = 2 \\ \\ sho w \: that$
i)
${a}^{2} + \frac{1}{ {a}^{2} } = 6$
ii)
${a}^{4} + \frac{1}{ {a}^{4} } = 34$
iii)
$(a + \frac{1}{a} ) ^{2} = 8$
please show step by step instructions
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Answer 1

Question:-

$\sf{If \: a - \dfrac{1}{a} = 2 \: show \: that}$

$\sf{ i) \: {a}^{2} + \dfrac{1}{ {a}^{2} } = 2}$

$\sf{ ii) \: {a}^{4} + \dfrac{1}{ {a}^{4} } = 34} \\ \sf{iii) \: {(a + \frac{1}{a}) }^{2} } = 8$

Required Answer:-

Given:-

$\\ \sf{ \rightarrow{a - \dfrac{1}{a} = 2}}$

Formula Applied:-

$\\ \sf{ \rightarrow{ {a}^{2} + {b}^{2} = (a - b) {}^{2} + 2ab}} \\ \\ \sf{ \rightarrow{ {a}^{2} + {b}^{2} = (a + b) {}^{2} - 2ab}} \\ \\ \sf{ \rightarrow{(a + b) {}^{2} = (a - b) {}^{2} + 4ab}}$

Solution:-

$\sf{ i) \: {a}^{2} + \dfrac{1}{ {a}^{2} } = 2} \\ \\ \sf{ \bold{{a}^{2} + \dfrac{1}{ {a}^{2} }}} \\ \\ \sf{ \implies{ {(a)}^{2} + {( \dfrac{1}{a}} )}^{2} }$

Applying formula:-

$\\ \sf{ \implies{ {(a - \dfrac{1}{a} ) {}^{2} + 2 \times \cancel{a } \times \frac{1}{ \cancel{a}} }}}$

Substituting the value:-

$\\ \sf{ \implies{(2) {}^{2} +2}}$

$\\ \sf{ \implies{ \red{6}}}$

Hence, L. S. = R. S. (Showed!)

$\sf{ ii) \: {a}^{4} + \dfrac{1}{ {a}^{4} } = 34} \\ \\ \sf{ \bold{ {a}^{4} + \dfrac{1}{ {a}^{4} }}}\\ \\ \sf{ \implies{ ({a}^{2} ) {}^{2} + ( \frac{1}{ {a}^{2} } ) {}^{2} }}$

Applying formula:-

$\\ \sf{ \implies{( {a}^{2} + \dfrac{1}{ {a}^{2} } ) {}^{2} - 2 \times \cancel{ {a}^{2} } \times \dfrac{1}{ \cancel{ {a}^{2} }}}}$

$\\ \sf{ \implies{ \{( {a} - \dfrac{1}{ {a} } ) {}^{2} + 2\times \cancel{ {a} } \times \dfrac{1}{ \cancel{ a}}}} \} {}^{2} - 2$

Substituting the value:-

$\\ \sf{ \implies{( {2}^{2} + 2) {}^{2} - 2}}$

$\\ \sf{ \implies{ {6}^{2} - 2}}$

$\\ \sf{ \implies{ \red{34}}}$

Hence, L. S. = R. S. (Showed!)

$\sf{iii) \: {(a + \frac{1}{a}) }^{2} } = 8$

$\\ \sf{ \bold{ {(a + \frac{1}{a}) }^{2} }}$

Applying formula:-

$\\ \sf{ \implies{(a - \frac{1}{a} ) {}^{2} + 4 \times \cancel{a} \times \dfrac{1}{ \cancel{a}}}}$

Substituting the value:-

$\\ \sf{ \implies{ {2}^{2} + 4}}$

$\\ \sf{ \implies{4 + 4}}$

$\\ \sf{ \implies{ \red{8}}}$

Hence, L. S. = R. S. (Showed!)