Question

If a=3+√5÷2,then find the value of a^2+1÷a^2

Answer 1

Question:

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

To find:

$\bigstar To \: find \: {a}^{2} + \dfrac{1}{ {a}^{2} }$

Answer:

$The \: value \: is \: \underline{ \: \underline{ \: \bold{ \sf \pink{ {a}^{2} - \dfrac{1}{ {a}^{2} } \: is \: \dfrac{3 \sqrt{5} }{2} }} \: } \: }$

Given:

$a = \dfrac{3 + \sqrt{5} }{2}$

Step-by-step explanation:

$a = \dfrac{3 + \sqrt{5} }{2}$

$\implies \dfrac{1}{a} = \dfrac{2}{3 + \sqrt{5} }$

Now rationalising the denominator,

$\implies \dfrac{1}{a} = \dfrac{2(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5}) }$

It's of the form, ( a² - b² ) = (a+b) (a-b)

$\implies \dfrac{2(3 - \sqrt{5})}{ {(3)}^{2} - {( \sqrt{5}) }^{2} }$

$\implies \dfrac{2(3 - \sqrt{5})}{ 9 - 5}$

$\implies \dfrac{ \not{2}(3 - \sqrt{5})}{ \not{ 4} _{2} }$

$\implies \boxed{a = \dfrac{3 + \sqrt{5} }{2}}$

$\implies \boxed{ \dfrac{1}{a} = \frac{3 - \sqrt{5}}{2} }$

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Now finding the value,

$\bigstar {a}^{2} - \dfrac{1}{ {a}^{2} }$

$\hookrightarrow (a + \dfrac{1}{a} )(a - \dfrac{1}{a} )$

$\hookrightarrow \big( \dfrac{3 + \sqrt{5} }{2} + \dfrac{3 - \sqrt{5} }{2} \big) \big( \dfrac{3 + \sqrt{5} }{2} - \dfrac{3 - \sqrt{5} }{2} \big)$

$\hookrightarrow \big( \dfrac{3 \bold{ + \sqrt{5}} + 3\bold{ - \sqrt{5}} }{2} \big)\big( \dfrac{ \bold{3} + \sqrt{5} \bold{ - 3} + \sqrt{5} }{2}\big)$

$\hookrightarrow \big( \dfrac{6}{2} \big) \big( \dfrac{ \sqrt{5} }{2} \big)$

$\hookrightarrow \boxed{ {a}^{2} - \dfrac{1}{ {a}^{2} } = \dfrac{3 \sqrt{5} }{2} }$

$\therefore The \: value \: is \: \underline{ \: \bold{ {a}^{2} - \dfrac{1}{ {a}^{2} } \: is \: \dfrac{3 \sqrt{5} }{2} } \: }$

Formula used:

$\bigstar {a}^{2} - {b}^{2} = (a + b)(a - b)$