Question

$a + \frac{1}{a} = 4 \\ find \: ( {a + \frac{1}{a}) }^{2}$
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Answer 1

Answer:

$Using \: \: the \: \: identity : \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

Step-by-step explanation:

$a + \frac{1}{a} = 4 \\ \\ {(a + \frac{1}{a}) }^{2} = {4}^{2} \\ {a}^{2} + \frac{1}{ {a}^{2} } +( 2 \times a \times \frac{1}{a} ) = 16 \\ {a}^{2} + \frac{1}{ {a}^{2} } +( 2 \times \cancel a \times \frac{1}{ \cancel a} ) = 16 \\ {a}^{2} + \frac{1}{ {a}^{2} } + 2 = 16 \\ {a}^{2} + \frac{1}{ {a}^{2} } = 16 - 2 \\ \large \boxed{ {a}^{2} + \frac{1}{ {a}^{2} } = \underline{ \underline{ 14}} \: \: or \: \: {(a + \frac{1}{a}) }^{2} = \underline{ \underline{14}}}$