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

Nov 16, 2016

Move every term to one side, factor the expression, and determine when each factor is$+$or$-$. Multiply these$+$and$-$intervals from the factors to get the$+$and$-$intervals for the product (and thus the original inequality).

Explanation:

${x}^{2} \le 4 x$
${x}^{2} - 4 x \le 0$ Move everything to one side
$x \left(x - 4\right) \le 0$ Factor the expression

This is now written as the product of two factors, $x$ and $x - 4$. The values of $x$ that make the factors $0$ are the "points of interest", because it is only at these points where the value of ${x}^{2} - 4 x$ could change from being$\le 0$ to not being$\le 0$. We set each factor equal to $0$ and solve for $x$:

$x = 0 \mathmr{and} x - 4 = 0$
$\implies x = 0 \mathmr{and} x = 4$

So we have three intervals to examine: below 0, between 0 and 4, and above 4.

A sign chart can be made like this:
. . . . . . . . . .---------0----------4-----------
$x$
$x - 4$
. . . . . . . . . .----------------------------------
$x \left(x - 4\right)$

Fill in the sign chart with + and - signs to reflect where each factor is$\le 0$. For example: the factor $x - 4$ is negative when $x < 4$; it's 0 when $x = 4$; and it's positive when $x > 4$.

. . . . . . . . . .---------0---------4-----------
$x$ . . . . . . . . $- -$ $0 + + + + +$
$x - 4$. . . . . $- - - -$ $0 + + +$
. . . . . . . . . .----------------------------------
$x \left(x - 4\right)$

To complete the bottom row of the sign chart, each interval is filled with the product of all the signs above it. For example, when $x < 0$, $\left(-\right)$ x $\left(-\right)$ = $\left(+\right)$.

. . . . . . . . . .---------0--------4-----------
$x$ . . . . . . . . $- -$ $0 + + + + +$
$x - 4$ . . . . $- - - -$ $0 + + +$
. . . . . . . . . .----------------------------------
$x \left(x - 4\right)$ . $+ + 0 - - 0 + + +$

This tells you that $x \left(x - 4\right)$ (in other words, ${x}^{2} - 4 x$) will be negative only when $x$ is between 0 and 4, and it will be 0 at those endpoints, i.e. ${x}^{2} - 4 x \le 0$ when $0 \le x \le 4$. And since ${x}^{2} - 4 x \le 0$ is equivalent to our original inequality ${x}^{2} \le 4 x$, we are done.

The solution is $0 \le x \le 4$.
Or $x \in \left[0 , 4\right]$ if you prefer interval notation.

Go ahead and try some values for yourself: say $x = 3$. We should expect the inequality to be true. Well, is it?

${x}^{2} \le 4 x$
$\iff {3}^{2} \le 4 \left(3\right)$ (when $x = 3$)
$\iff 9 \le 12$
Which is true!