Common Algebra Problem

Given $x + \frac{1}{x} = 5$ , $x$ is a solution to $x^2 - 5x + 1 = 0$ . $(1)$

Thus $x^2 - 4x + 4 = x + 3$ , i.e. $(x-2)^2 = (x+3)$ . $(2)$

We can use Equation $(2)$ to substitute in the expression:

$(x-2)^2 + \frac{25}{(x-2)^2} = (x+3) + \frac{25}{x+3}.$

Let t be the value of this expression: $t = (x+3) + \frac{25}{x+3}$ .

Then $t(x+3) = (x+3)^2 + 25 = x^2 + 6x + 34$ . $(3)$

Use Equation $(1)$ to get the identity $x^2 = 5x -1$ . $(4)$ Use $(4)$ to substitute for $x^2$ in $(3)$ .

Then $t(x+3) = 5x - 1 + 6x + 34 = 11x + 33 = 11(x+3)$ . The solution is $t=11.$

Given that $x+\dfrac 1x = 5 \implies x^2 - 5x + 1 = 0$ . Then

$\begin{aligned} (x-2)^2 + \frac {25}{(x-2)^2} & = x^2 - 4x + 4 + \frac {25}{x^2 - 4x + 4} \\ & = \blue{x^2 - 5x + 1} +x + 3 + \frac {25}{\blue{x^2 - 5x + 1} +x + 3} & \small \blue{\text{Note that }x^2 - 5x + 1 = 0} \\ & = x + 3 + \frac {25}{x + 3} \\ & = \frac {(x+3)^2 + 25}{x + 3} \\ & = \frac {x^2 + 6x + 9 + 25}{x + 3} \\ & = \frac {\blue{x^2 - 5x + 1} + 11x + 8 + 25}{x + 3} \\ & = \frac {11x+33}{x + 3} \\ & = \boxed {11} \end{aligned}$

$x +\frac1x = 5 \quad\Leftrightarrow \quad x^2 -5x + 1 = 0 \quad\Leftrightarrow \quad (x-2)^2 - x - 3 = 0 \quad\Leftrightarrow \quad (x-2)^2 = x + 3$

Let $(x-2)^2 + \frac{25}{(x-2)^2} = k$ , then

$x + 3 + \frac{25}{x+3} = k \quad\Leftrightarrow \quad (x+3)^2 - k (x+3)^2 + 25 = 0 \quad\Leftrightarrow \quad x^2 + x(6-k) + (34-3k) = 0$

We also know from earlier that $x^2 - 5x + 1 = 0$ . By comparing coefficients, $-5 = 6-k$ and $1 = 34-3k$ must both be satisfied. Both of these equations produce $k =\boxed{11}$ .