How do you solve 5x-10x^2<0 using a sign chart?

Jan 22, 2017

Either $x < 0$ or $x > \frac{1}{2}$

Explanation:

$5 x - 10 {x}^{2} < 0 \Leftrightarrow 5 x \left(1 - 2 x\right) < 0$

Hence sign of $x$ changes around $x = 0$ and $1 - 2 x = 0$ i.e. $x = \frac{1}{2}$ and these two points divide real number line in three parts.

Note that on real number line, below $x = 0$, while $x < 0$, $1 - 2 x > 0$ and therefore $5 x - 10 {x}^{2} < 0$

Between $x = 0$ and $x = \frac{1}{2}$,

$x > 0$ and $1 - 2 x > 0$ and therefore $5 x - 10 {x}^{2} > 0$

and beyond $x = \frac{1}{2}$, while $x > 0$, $1 - 2 x < 0$ and hence $5 x - 10 {x}^{2} > 0$

Hence, solution is that either $x < 0$ or $x > \frac{1}{2}$

In terms of sign chart this can be expressed as

$\textcolor{w h i t e}{X X X X X X X X X X X} 0 \textcolor{w h i t e}{X X X X X X X} \frac{1}{2}$

$x \textcolor{w h i t e}{X X X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X X X} + i v e$

$\left(1 - 2 x\right) \textcolor{w h i t e}{X X X} + i v e \textcolor{w h i t e}{X X X} + i v e \textcolor{w h i t e}{X X X X} - i v e$

$5 x - 10 {x}^{2} \textcolor{w h i t e}{\times X} - i v e \textcolor{w h i t e}{X X X} + i v e \textcolor{w h i t e}{X X X X} - i v e$

and as we need $5 x - 10 {x}^{2}$ to be negative i.e. less than $0$, solution is either $x < 0$ or $x > \frac{1}{2}$.