Question

if x = 3 + 2 root 2 find X square + 1 / X square​

Answer 1

Answer

• x² + 1/x² = 34

Given

$\bullet\ \; \rm x=3+2\sqrt{2}$

To Find

$\bullet\ \; \rm x^2+\dfrac{1}{x^2}$

Key Point

• While solving these type of questions , we need to find the value of reciprocal of x . For that , we need to rationalize the denominator .

Solution

$\bf x=3+2\sqrt{2}\\\\\to \rm \dfrac{1}{x}=\dfrac{1}{3+2\sqrt{2}}$

Rationalize the denominator ,

$\to \rm \dfrac{1}{x}=\dfrac{1}{3+2\sqrt{2}}\times \dfrac{3-2\sqrt{2}}{3-2\sqrt{2}}\\\\\to \rm \dfrac{1}{x}=\dfrac{3-2\sqrt{3}}{(3+2\sqrt{2})(3-2\sqrt{2})}\\\\\to \rm \dfrac{1}{x}=\dfrac{3-2\sqrt{2}}{(3)^2-(2\sqrt{2})^2}$

∵ ( a + b ) ( a - b ) = a² - b²

$\to \rm \dfrac{1}{x}=\dfrac{3-2\sqrt{2}}{9-8}\\\\\to \rm \dfrac{1}{x}=\dfrac{3-2\sqrt{2}}{1}\\\\\to \bf \dfrac{1}{x}=3-2\sqrt{2}$

Let's solve $\rm x+\dfrac{1}{x}$ ,

$\to \rm x+\dfrac{1}{x}=(3+2\sqrt{2})+(3-2\sqrt{2})\\\\\to \rm x+\dfrac{1}{x}=6$

Squaring on both sides , we get ,

$\to \rm \left(x+\dfrac{1}{x}\right)^2=(6)^2\\\\\to \rm x^2+\dfrac{1}{x^2}+2.x.\dfrac{1}{x}=36\\\\\to \rm x^2+\dfrac{1}{x^2}+2=36\\\\\to \rm x^2+\dfrac{1}{x^2}=36-2\\\\\to \bf x^2+\dfrac{1}{x^2}=34$

Answer 2

$:\implies\sf x= 3+2\sqrt{2}$

We have to find the value of

$\sf\bigg[ x^2+\dfrac{1}{x^2}\bigg]$

Now find the value of 1/x

$\implies\sf \dfrac{1}{x}= \dfrac{1}{3+2\sqrt{2}}\times \dfrac{3-2\sqrt{2}}{3-2\sqrt{2}}\\ \\ \\ \implies\sf \dfrac{1}{x}= \dfrac{3-2\sqrt{2}}{(3)^2-(2\sqrt{2})^2}\\ \\ \\ \implies\sf \dfrac{1}{x}= \dfrac{3-2\sqrt{2}}{9-8}\\ \\ \\ \implies{\boxed{\sf{ \dfrac{1}{x}=3-2\sqrt{2}}}}$

identity used :-

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

$\sf \bigg[x+\dfrac{1}{x}\bigg]^2= x^2+\dfrac{1}{x^2}+2\\ \\ \\ \implies\sf \Big(3+\cancel{2\sqrt{2}}+3-\cancel{2\sqrt{2}})^2=x^2+ \dfrac{1}{x^2}+2\\ \\ \\\implies\sf (6)^2= x^2+\dfrac{1}{x^2}+2\\ \\ \\ \implies\sf 36-2= x^2+\dfrac{1}{x^2}\\ \\ \\ \implies{\boxed{\mathfrak{\purple{ x^2+\dfrac{1}{x^2}= 34}}}}$