Question

if x is equals to 2 + root 3 then find x power 4 + 1 by x power 4

Answer 1

$x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3} \\ \\ \frac{1}{x} = 2 - \sqrt{3}$


Now,

x + 1/x = 2 + root3 + 2 - root3

=> x + 1/x = 4

On squaring both sides, we get

x^2 + 1/x^2 + 2(x)(1/x) = 16

=> x^2 + 1/x^2 + 2 = 16

=> x^2 + 1/x^2 = 14

Again, on squaring both sides, we get

x^4 + 1/x^4 + 2 (x^2) (1/x^2) = 196

=> x^4 + 1/x^4 + 2 = 196

=> x^4 + 1/x^4 = 194

Answer 2

Answer:

Value of given expression is 194.

Step-by-step explanation:

Given: x = 2 + √3

To find: $x^4+\frac{1}{x^4}$

Consider,

x² = ( 2 + √3 )² = 2² + (√3)² + 2(2)(√3) = 4 + 3 + 4√3 = 7 + 4√3

$x^4=(7+4\sqrt{3})^2=7^2+(4\sqrt{3})^2+2(7)(4\sqrt{3})=49+48+56\sqrt{3}=97+56\sqrt{3}$

$\frac{1}{x^4}=\frac{1}{97+56\sqrt{3}}=\frac{1}{97+56\sqrt{3}}\times\frac{97-56\sqrt{3}}{97-56\sqrt{3}}=\frac{97-56\sqrt{3}}{(97)^2-(56\sqrt{3})^2}=\frac{97-56\sqrt{3}}{9409-9408}=97-56\sqrt{3}$

So,  $x^4+\frac{1}{x^4}=97+56\sqrt{3}+97-56\sqrt{3}=97+97=194$

Therefore, Value of given expression is 194.