Squares and Inverses!

If we square $x^2 + \frac {1}{x^2}$ , then we get $x^4 + \frac {1}{x^4}$ + 2, which is equal to 47 squared, or 2209. All we do now is subtract 2 to get $\boxed{2207}$ .

Solution 1:

$x^2 + \dfrac{1}{x^2} = 47$

$\Rightarrow \large \left(x^2 + \dfrac{1}{x^2} \right)^2 = (47)^2$

$\Rightarrow x^4 + 2+ \dfrac{1}{x^4} = 2209$

$\Rightarrow x^4 + \dfrac{1}{x^4} = \boxed{2207}$

Solution 2:

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

$= \large \left(x^2\right)^2 + \large \left(\dfrac{1}{x^2}\right)^2$

$= \large \left(x^2 + \dfrac{1}{x^2}\right)^2 - 2 \cdot x^2 \cdot \dfrac{1}{x^2}$

$= \large \left(47\right)^2 -2$

$= \boxed{2207}$

$\begin{aligned} x^2+\frac 1{x^2} & = 47 \\ \left(x^2+\frac 1{x^2} \right)^2 & = 47^2 \\ x^4+2+\frac 1{x^4} & = 2209 \\ \implies x^4+\frac 1{x^4} & = 2207 \end{aligned}$

$x^2$ + $\dfrac{1}{x^2}=47$

$(x^2+\dfrac{1}{x^2})^2$ $=47^2$

$x^4+2+\dfrac{1}{x^4}$ $=2209$

$x^4+\dfrac{1}{x^4}$ $=2207$