Question

please help ,answer it ​

Answer 1

Given :

• a = 2 + √3

To find :

• a² + 1/a²

According to the question,

Step 1 :

At first we will rationalise 1/a

$: \implies \bf{ \dfrac{1}{a} = \dfrac{1 }{2 + \sqrt{3}} = \dfrac{2 - \sqrt{3} }{2 + \sqrt{3} } } \\ \\ : \implies \bf{ \dfrac{2 - \sqrt{3} }{4 - 3} = 2 - \sqrt{3} \: \pink \bigstar }$

Step 2 :

We have to find the value of a + 1/a

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

$:\implies \bf{4} \: \:\orange \bigstar$

Step 3 :

Squaring on both the sides

$: \implies \bf{ {a}^{2} + \dfrac{1}{ {a}^{2} } = {4}^{2} } \\ \\ : \implies \bf{ {a}^{2} + \dfrac{1}{ {a}^{2} } + 2 \times a \times \frac{1}{a} = 16} \\ \\ : \implies \bf{ {a}^{2} + \dfrac{1}{ {a}^{2} } = 16 - 2 } \\ \\ { \boxed{\underline{\implies \bf{ {a}^{2} + \dfrac{1}{ {a}^{2} } = 14 } \: \blue \bigstar}}}$

So,the answer is 14.

____________________

$\: \: \: \: \: \: \:$ Formula used : $\: \: \: \: \: \: \:$

• (a + b)² = a² + 2ab + b²

Answer 2

Given -

• a = 2 + $\sf{\sqrt{3}}$

To find -

• $\sf{a^2 + \dfrac{1}{a^2}}$

Solution -

• Firstly, find $\bold{a^2}$

• Identity to be used -

$\large\boxed{\sf{\green{(a + b)^2 = a^2 + b^2 + 2ab}}}$

↬ $\bold{(2 + \sqrt{3})^2}$

↬ $\bold{(2)^2 + (\sqrt{3})^2 + 2 \times 2 \times \sqrt{3}}$

↬ $\bold{4 + 3 + 4\sqrt{3}}$

$\large\overbrace{\underbrace{\bold{a^2 = 11 + 4 \sqrt{3}}}}$

• Now, Find 1/a.

↬ $\bold{\dfrac{1}{2 + \sqrt{3}}}$

• Rationalize the denominator.

↬ $\bold{\dfrac{2 - \sqrt{3}}{2 + \sqrt{3} }\times 2 - \sqrt{3}}$

↬$\bold{\dfrac{2 - \sqrt{3}}{(2)^2 - (\sqrt{3})^2}}$

↬ $\bold{\dfrac{2 - \sqrt{3}}{4 - 3}}$

↬ $\bold{2 - \sqrt{3}}$

$\large\overbrace{\underbrace{\bold{ \dfrac{1}{a} = 2 - \sqrt{3}}}}$

• Now, find the square of 1/a

Identity to be used -

$\large{\boxed{\sf{\green{(a - b)^2 = a^2 + b^2 - 2ab}}}}$

↬ $\bold{2 - \sqrt{3}}$

↬ $\bold{4 + 3 - 2 \times 2 \times \sqrt{3}}$

$\large\overbrace{\underbrace{\bold{ \dfrac{1}{a^{2} } = 7 - 4\sqrt{3}}}}$

• Now, add $\bold{a^2}$ and $\sf{\dfrac{1}{a^2}}$

↬ $\bold{7 + 4 \sqrt{3} + 7 - 4 \sqrt{3}}$

$\bold{7 + 4 \sqrt{3}~and~ - 4 \sqrt{3}}$ will get cancelled because they are having different signs.

↬ $\bold{7 + \cancel{4 \sqrt{3}} + 7 - \cancel{4 \sqrt{3}}}$

$\overbrace{\underbrace{\sf{\orange{a^2 + \dfrac{1}{a^2} = 14}}}}$