Answer
$$[2-\sqrt{2},2+\sqrt{2}]$$
Work Step by Step
Let $x^2 \leq 4x -2 $ be a function $f$.
Express the inequality in the form $f(x) \lt 0$. Begin by the rewriting the inequality so that $0$ is on the right side.
$x^2 \leq 4x -2 $
$x^2 -4x+2\leq 4x -2 -4x+2$
$x^2 -4x+2 \lt 0 $
Find the $x$-intercepts by solving $x^2 -4x+2=0$
Use the quadratic formula: $x = \frac{-b±\sqrt{b^2-4ac}}{2a}$ where $a=1$, $b=-4$, $c=2$
$x = \frac{-(-4)±\sqrt{-4^2-4⋅1⋅2}}{2⋅1}$
$x = 2+ \sqrt{2}$ or $2- \sqrt{2}$
These $x$-intercepts serve as boundary points that separate the number line into intervals.
The boundary points of this equation therefore divide the number line into three intervals:
$(-∞, 2- \sqrt{2})(2- \sqrt{2},2+ \sqrt{2})(2+ \sqrt{2}, +∞)$
Choose one test value within each interval and evaluate $f$ at that number.
$(-∞, 2- \sqrt{2})$
Test Value = $-1$
$f=x^2-4x+2$
$f=-1^2-4(-1)+2$
$f=5$
Conclusion: $f (x) > 0$ for all $x$ in $(-∞,2-\sqrt{2})$.
$(2-sqrt{2}, 2+sqrt{2})$
Test Value = $1$
$f=x^2-4x+2$
$f=1^2-4(1)+2$
$f=-1$
Conclusion: $f (x) < 0$ for all $x$ in $(2-\sqrt{2}, 2+\sqrt{2})$.
$(2+sqrt{2},+∞)$
Test Value = $4$
$f=x^2-4x+2$
$f=4^2-4(4)+2$
$f=2$
Conclusion: $f (x) > 0$ for all $x$ in $(2+\sqrt{2},+∞)$.
Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \leq 0$, where $f (x) = x^2-4x+2$. Based on the solution above, $f(x)\lt0$ for all $x$ in $(2-\sqrt{2},2+\sqrt{2})$. However, because the inequality involves $\geq$ (greater than or equal to), we must also include the solutions of $x^2-4x+2$, namely $2-\sqrt{2}$ and $2+\sqrt{2}$, in the solution set.
Thus, the solution set of the given inequality is:
$$[2-\sqrt{2},2+\sqrt{2}]$$
