Question

if a = 3-2√2 then find a²+1/a²​

Answer 1

Given

$\rm \: : \implies \: a = 3 - 2 \sqrt{2}$

To Find

$\rm \: : \implies \: \: {a}^{2} + \dfrac{1}{ {a}^{2} }$

If

$\rm \: : \implies \: a = 3 - 2 \sqrt{2}$

Then

$\rm \: : \implies \dfrac{1}{a} = \dfrac{1}{3 - 2 \sqrt{2} }$

Rationalize The Denominator

$\rm \: : \implies \: \: \dfrac{1}{a} = \dfrac{1}{3 - 2\sqrt{2} } \times \dfrac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} }$

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

$\rm \: : \implies \: \: \dfrac{1}{a} = \dfrac{3 + 2 \sqrt{2} }{9 - 8} = 3 + 2 \sqrt{2}$

We know that

$\rm \: : \implies \: \: (a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab$

we get

$\rm \: : \implies \: \: \bigg(a + \dfrac{1}{a} \bigg)^{2} = {a}^{2} + \dfrac{1}{ {a}^{2} } + 2 \times a \times \dfrac{1}{a}$

$\rm \: : \implies \: \: \bigg(a + \dfrac{1}{a} \bigg)^{2} = {a}^{2} + \dfrac{1}{ {a}^{2} } + 2$

$\rm \: : \implies \: \: \bigg(a + \dfrac{1}{a} \bigg)^{2} - 2 = {a}^{2} + \dfrac{1}{ {a}^{2} }$

Put the value

$\rm \: : \implies \: \: \dfrac{1}{a} = 3 + 2 \sqrt{2} \: \: and \: a = 3 - 2 \sqrt{2}$

we get

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

$\rm \: : \implies \: {a}^{2} + \dfrac{1}{ {a}^{2} } = (3 + 3 ) {}^{2} - 2$

$\rm \: : \implies \: {a}^{2} + \dfrac{1}{ {a}^{2} } = (6) {}^{2} - 2$

$\rm \: : \implies \: {a}^{2} + \dfrac{1}{ {a}^{2} } = 36 - 2$

$\rm \: : \implies \: {a}^{2} + \dfrac{1}{ {a}^{2} } = 34$

Answer

$\rm \: : \implies \: {a}^{2} + \dfrac{1}{ {a}^{2} } = 34$

Answer 2

★GIVEN:-

• a = 3 - 2✓2

★TO FIND:-

• a²+1/a²

★FORMULA USED:-

• $\sf{ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab}$

★SOLUTION:-

First, we will find the value of 1/a

$\sf : \longrightarrow{a = 3 - 2 \sqrt{2} } \\ \\ \sf : \longrightarrow{ \frac{1}{a} = \frac{1}{3 - 2 \sqrt{2} } } \\ \\ \rm{rationalising \: the \: denominator} \\ \\ \sf : \longrightarrow{ \frac{1}{a} = \frac{1}{3 - 2 \sqrt{2} } \times \frac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} } } \\ \\ \sf : \longrightarrow{ \frac{1}{a} = \frac{3 + 2 \sqrt{2} }{(3 - 2 \sqrt{2)}{(3 + 2 \sqrt{2} }) } } \\ \\ \sf : \longrightarrow{ \frac{1}{a} = \frac{3 + 2 \sqrt{2} }{9 + \cancel{6 \sqrt{2}} - \cancel{ 6 \sqrt{2}} - 8} } \\ \\ \sf : \longrightarrow \pink{ \frac{1}{a} = 3 + 2 \sqrt{2} }$

Now, We know that:-

$\displaystyle \sf{ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab } \\ \\ : \longrightarrow\sf{ {(a + \frac{1}{a}) }^{2} = {a}^{2} + \frac{1}{ {a}^{2} } + 2 \times \cancel{a} \times \frac{1}{ \cancel{a}} } \\ \\ : \longrightarrow\sf{{(a + \frac{1}{a}) }^{2} = {a}^{2} + \frac{1}{ {a}^{2} } + 2} \\ \\ : \longrightarrow\sf{{(3 - 2 \sqrt{2} + 3 + 2 \sqrt{2} ) }^{2} = {a}^{2} + \frac{1}{ {a}^{2} } + 2} \\ \\ : \longrightarrow\sf{ {(6)}^{2} = {a}^{2} + \frac{1}{ {a}^{2} } + 2 } \\ \\ : \longrightarrow\sf{ {a}^{2} + \frac{1}{ {a}^{2} } = 36 - 2 } \\ \\ : \implies{ \boxed{ \underline{ \huge{ \blue{ \sf{ \rm{ {a}^{2} + \frac{1}{ {a}^{2} } = 34}}}}}}}$

So, the value of a²+1/a² is 34.