An algebra problem by manish kumar singh

Case 1: $x\ge0$

When $x\ge0$ , the absolute value will do nothing, yielding the equation: $x^2+x-2=0$ . This factors to $(x+2)(x-1)=0$ . The two solutions to this equation are $x=-2$ and $x=1$ . However, $x=-2$ is an extraneous solution, because this case requires that $x\ge0$ .

Case 2: $x<0$

When $x<0$ , the absolute value will negate the $x$ , yielding the equation $x^2-x-2=0$ . This factors to $(x-2)(x+1)=0$ . The two solutions to this equation are $x=2$ and $x=-1$ . However, $x=2$ is an extraneous solution because this case requires that $x<0$ .

Solving both cases yields the solutions $\boxed{x=1}$ and $\boxed{x=-1}$ .

Did the same way. Perhaps better presented as follows:

$\begin{aligned} \text{We note that} \quad \quad |x| & = \begin{cases} \color{#D61F06}{-x} & \color{#D61F06} {\text{for }x \le 0} \\ \color{#3D99F6}{+x} & \color{#3D99F6}{\text{for } x > 0} \end{cases} \\ \Rightarrow x^2 + |x| - 2 = 0 & \Rightarrow \begin{cases} x^2 \color{#D61F06}{-x} - 2 = 0 & \color{#D61F06} {\text{for }x \le 0} & \Rightarrow x = \begin{cases} x = \color{#3D99F6}{-1} < 0 & \color{#3D99F6} {\text{accepted.}} \\ x = \color{#D61F06}{2 > 0} & \color{#D61F06} {\text{rejected.}} \end{cases} \\ x^2 \color{#3D99F6}{+x} - 2 = 0 & \color{#3D99F6}{\text{for } x > 0} & \Rightarrow x = \begin{cases} x = \color{#3D99F6}{1} > 0 & \color{#3D99F6} {\text{accepted.}} \\ x = \color{#D61F06}{-2 < 0} & \color{#D61F06} {\text{rejected.}} \end{cases} \end{cases} \end{aligned}$

Therefore, the values of $x$ satisfy the equation are $\boxed{\pm1}$ .

x^2 can be written as |x^2| , then consider |x| as t and solve simply we ger t = 1 and thus |x| = 1 and so , x = + 1, - 1 .

The graphs of y=x^2 and y=|x|-1 can be superimposed into one graph which clearly has two x-intercepts