Question

if x=3+2root2 then find the value of x4+1/x4

Answer 1

I hope it will help you

Answer 2

$x^{4}+\dfrac{1}{x^{4}}=1154$

Step-by-step explanation:

We have,

$x=3+2\sqrt{2}$

To find, the value of $x^{4} +\dfrac{1}{x^{4}} =?$

∴ $\dfrac{1}{x} =\dfrac{1}{3+2\sqrt{2}}$

Rationalising, we get

$\dfrac{1}{x} =\dfrac{1}{3+2\sqrt{2}}\times \dfrac{3-2\sqrt{2}}{3-2\sqrt{2}}$

$=\dfrac{3-2\sqrt{2}}{3^2-(2\sqrt{2})^2}=\dfrac{3-2\sqrt{2}}{9-8}$

$\dfrac{1}{x} =3-2\sqrt{2}$

∴ $(x+\frac{1}{x})^{2}=x^{2}+(\dfrac{1}{x})^{2}+2.x.\dfrac{1}{x}$

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

⇒ $x^{2}+\dfrac{1}{x^{2}} =36-2=34$     .....(1)

Again squaring (1), we get

$(x^{2}+\dfrac{1}{x^{2}} )^{2} =34^{2}$

$x^{4}+\dfrac{1}{x^{4}} +2=1156$

⇒ $x^{4}+\dfrac{1}{x^{4}} =1156-2=1154$  

Hence, $x^{4}+\dfrac{1}{x^{4}}=1154$