# How do you solve x^2<=4x-2 using a sign chart?

Apr 4, 2017

$2 - \sqrt{2} \le x \le 2 + \sqrt{2}$

#### Explanation:

We have ${x}^{2} \le 4 x - 2$ i.e. ${x}^{2} - 4 x + 2 \le 0$ and using the quadratic formula is

$x = \frac{4 \pm \sqrt{{4}^{2} - 4 \times 1 \times 2}}{2} = 2 \pm \sqrt{2}$

and our inequality is therefore

$\left(x - 2 + \sqrt{2}\right) \left(x - 2 - \sqrt{2}\right) \le 0$

From this, we know that the product $\left(x - 2 + \sqrt{2}\right) \left(x - 2 - \sqrt{2}\right)$ is negative or equal to $0$. It is apparent that sign of binomials $\left(x - 2 + \sqrt{2}\right)$ and $x - 2 - \sqrt{2}$ will change around the values $2 - \sqrt{2}$ and $2 + \sqrt{2}$ respectively. In a sign chart, we divide the real number line using these values, i.e. below $2 - \sqrt{2}$, between $2 - \sqrt{2}$ and $2 + \sqrt{2}$ and above $2 + \sqrt{2}$ and see how the sign of ${x}^{2} - 4 x + 2$ changes.

Sign Chart

$\textcolor{w h i t e}{X X X X X X X X X X X} 2 - \sqrt{2} \textcolor{w h i t e}{X X X X X} 2 + \sqrt{2}$

$\left(x - 2 + \sqrt{2}\right) \textcolor{w h i t e}{X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X X X} + i v e$

$\left(x - 2 - \sqrt{2}\right) \textcolor{w h i t e}{X} - i v e \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e$

$\left({x}^{2} - 4 x + 2\right) \textcolor{w h i t e}{X X} + i v e \textcolor{w h i t e}{X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e$

It is observed that ${x}^{2} - 4 x + 2 \le 0$ when either $x \ge 2 - \sqrt{2}$ or $x \le 2 + \sqrt{2}$ i.e. $x$ lies between $2 - \sqrt{2}$ and $2 + \sqrt{2}$ including these numbers or $2 - \sqrt{2} \le x \le 2 + \sqrt{2}$, which is the solution for the inequality.

In interval form solution is $\left[2 - \sqrt{2} , 2 + \sqrt{2}\right]$

graph{x^2-4x+2 [-2.976, 7.024, -2.5, 2.5]}