Question

If x=√5+2 find the value of x ⁴+1/x⁴​

Answer 1

322

Explanation:

Given equation:

`x=√(5)+2`

Subtract 2 from both sides of the equation so that only the radical is on the right side of the equation:

`\begin{aligned}x-2&=√(5)+2-2\\x-2&=√(5)\end{aligned}`

Square both sides:

`\begin{aligned}(x-2)^2&=(√(5))^2\\(x-2)(x-2)&=5\\x^2-2x-2x+4&=5\\x^2-4x+4&=5\end{aligned}`

Subtract 4 from both sides of the equation so that the x-terms are on the left side, and the constant is on the right side:

`\begin{aligned}x^2-4x+4-4&=5-4\\x^2-4x&=1\end{aligned}`

Divide both sides by x:

`\begin{aligned}(x^2)/(x)-(4x)/(x)&=(1)/(x)\\\\x-4&=(1)/(x)\end{aligned}`

Rearrange so that all the terms in x are on the left side and the constant is on the right side:

`\begin{aligned}x-4-x&=(1)/(x)-x\\\\-4&=(1)/(x)-x\\\\(1)/(x)-x}&=-4\end{aligned}`

Square both sides of the equation:

`\begin{aligned}\left((1)/(x)-x\right)^2&=(-4)^2\\\\\left((1)/(x)-x\right)\left((1)/(x)-x\right)&=16\\\\(1)/(x)\cdot(1)/(x)-(1)/(x)\cdot x-(1)/(x)\cdot x+(-x)\cdot(-x)&=16\\\\(1)/(x^2)-2+x^2&=16\\\\(1)/(x^2)-2+x^2+2&=16+2\\\\(1)/(x^2)+x^2&=18\end{aligned}`

Square both sides again:

`\begin{aligned}\left((1)/(x^2)+x^2\right)^2&=(18)^2\\\\\left((1)/(x^2)+x^2\right)\left((1)/(x^2)+x^2\right)&=324\\\\(1)/(x^2)\cdot(1)/(x^2)+(1)/(x^2)\cdot x^2+(1)/(x^2)\cdot x^2+(x^2)\cdot(x^2)&=324\\\\(1)/(x^4)+2+x^4&=324\\\\(1)/(x^4)+2+x^4-2&=324-2\\\\(1)/(x^4)+x^4&=322\end{aligned}`

Therefore, the value of x⁴ + 1/x⁴ when x = √5 + 2 is 322.

`\large\boxed{x=√(5)+2\implies(1)/(x^4)+x^4&=\boxed{322}}`