Question

kindly solve the attached question ​

Answer 1

Hope it helps you plz mark as brilliant answer

Answer 2

GIVEN :

• $\sf{x=3+2\sqrt{2}}$

TO FIND :

• Value of :

     $\;\;\;\;\;\sf{\bigg(\sqrt{x}+\dfrac{1}{\sqrt{x}}\bigg)\bigg(\sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg)=?= m }$

SOLUTION :

We need to find,

$\implies\sf{\bigg(\sqrt{x}+\dfrac{1}{\sqrt{x}}\bigg)\bigg(\sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg)=?}$

Using algebraic identity

( a + b ) ( a - b ) = a² - b²

$\implies\sf{\bigg(\sqrt{x}+\dfrac{1}{\sqrt{x}}\bigg)\bigg(\sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg)=(\sqrt{x})^2 - \left(\dfrac{1}{\sqrt{x}}\right)^2}$

$\implies\sf{m=x - \dfrac{1}{x}}$

$\implies\sf{m=\dfrac{x^2-1}{x}}$

putting value of x

$\implies\sf{m=\dfrac{(3+2\sqrt2)^2-1}{3+2\sqrt2}}$

using trigonometric identity

( a + b )² = a² + b² + 2 ab

in numerator

$\implies\sf{m=\dfrac{(3)^2+(2\sqrt2)^2+2(3)(2\sqrt2)-1}{3+2\sqrt2}}$

$\implies\sf{m=\dfrac{9+8+12\sqrt2-1}{3+2\sqrt2}}$

$\implies\sf{m=\dfrac{16+12\sqrt2}{3+2\sqrt2}}$

Rationalising the denominator

$\implies\sf{m=\dfrac{16+12\sqrt2}{3+2\sqrt2}\times \dfrac{3-2\sqrt2}{3-2\sqrt2}}$

$\implies\sf{m=\dfrac{16(3)+12\sqrt2(3)+16(-2\sqrt{2})+12\sqrt{2}(-2\sqrt{2})}{(3+2\sqrt2)\;\;(3-2\sqrt2)}}$

Using trigonometric identity

( a + b ) ( a - b ) = a² - b²

In denominator

$\implies\sf{m=\dfrac{48+36\sqrt2-32\sqrt{2}-48}{9-8}}$

$\implies\sf{m=\dfrac{0+4\sqrt2}{1}}$

$\implies\sf{m=4\sqrt2}$

therefore,

$\implies\underbrace{\boxed{\sf{\bigg(\sqrt{x}+\dfrac{1}{\sqrt{x}}\bigg)\bigg(\sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg)=4\sqrt2}}}\;\;\;\purple{\bigstar}$