Chapter A, Problem 10E

To determine

To rewrite: The expression without the absolute value symbol.

Answer to Problem 10E

The equivalent form of the given expression is $\{\begin{array}{l}\text{1}-\text{2}{x}^2\text{ if }-\frac{1}{\sqrt{2}}\le x\le \frac{1}{\sqrt{2}}\\ \text{2}{x}^2-1\text{ if }x<-\frac{1}{\sqrt{2}}\text{ or }x>\frac{1}{\sqrt{2}}\end{array}$.

Explanation of Solution

Given:

The expression is $\left|1-2x^2\right|$.

Definition used:

The absolute value of a number a, denoted by $\left|a\right|$, is the distance from a to 0 on the real

number line. Distances are always positive or 0, so $\left|a\right|\ge 0$ and $\{\begin{array}{l}\left|a\right|=a\text{ if }a\ge 0\\ \left|a\right|=-a\text{ if }a<0\end{array}$.

Calculation:

Solve the above expression as,

$\begin{array}{l}1-2x^2<0\\ 1<2x^2\\ \frac{1}{2}<x^2\\ x^2>\frac{1}{2}\end{array}$

Take square root on both sides,

$\begin{array}{c}\sqrt{x^2}>\sqrt{\frac{1}{2}}\\ x>\frac{1}{\sqrt{2}}\\ \left|x\right|>\frac{1}{\sqrt{2}}\end{array}$

The expression can be re written as,

$\begin{array}{l}x>\frac{1}{\sqrt{2}}\text{ or}\\ x<-\frac{1}{\sqrt{2}}\end{array}$

Thus, the expression becomes,

$\left|1-2x^2\right|=\{\begin{array}{l}\text{1}-\text{2}{x}^2\text{ if }-\frac{1}{\sqrt{2}}\le x\le \frac{1}{\sqrt{2}}\\ \text{2}{x}^2-1\text{ if }x<-\frac{1}{\sqrt{2}}\text{ or }x>\frac{1}{\sqrt{2}}\end{array}$

Hence, the equivalent form of the given expression is $\{\begin{array}{l}\text{1}-\text{2}{x}^2\text{ if }-\frac{1}{\sqrt{2}}\le x\le \frac{1}{\sqrt{2}}\\ \text{2}{x}^2-1\text{ if }x<-\frac{1}{\sqrt{2}}\text{ or }x>\frac{1}{\sqrt{2}}\end{array}$.