Answer
Solution set: $ [2-\sqrt{2},2+\sqrt{2}]$
Work Step by Step
Follow the "Procedure for Solving Polynomial lnequalities",\ p.412:
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a polynomial function.
$x^{2} \leq 4x-2$
$x^{2}-4x+2 \leq 0$
... no integer factors of 2 that add up to $-4$...
... no factoring, then ...
... use the quadratic formula in the next step...
$f(x)=x^{2}-4x+2 \leq 0$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$x^{2}-4x+2 = 0$
$x=\displaystyle \frac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(2)}}{2(1)}$
$=\displaystyle \frac{4\pm\sqrt{8}}{2}=\frac{2(2\pm\sqrt{2})}{2}=2\pm\sqrt{2}$
$x=2\pm\sqrt{2}$
3. Locate these boundary points on a number line, thereby dividing the number line into intervals.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & (-\infty, 2-\sqrt{2}) & (2-\sqrt{2},2+\sqrt{2}) & (2+\sqrt{2},\infty)\\
a=test.val. & 0 & 2 & 10\\
f(a) & 0-0+2 & 4-8+2 & 100-40+2\\
f(a) \leq 0 ? & F & T & F
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $ [2-\sqrt{2},2+\sqrt{2}]$