# How do you solve the equation absx=x^2?

Apr 27, 2017

$x = - 1 , 0 , 1$

#### Explanation:

We can rewrite the absolute value as
$x = {x}^{2}$
$- x = {x}^{2}$

Solving for the first one we get
${x}^{2} - x = 0$
$\setminus \frac{- \left(- 1\right) \setminus \pm \setminus \sqrt{{\left(- 1\right)}^{2} - 4 \left(1\right) \left(0\right)}}{2 \left(1\right)}$
$\setminus \frac{1 \setminus \pm \setminus 1}{2}$
$\setminus \frac{1 + 1}{2} = \frac{2}{2} = 1$
$\setminus \frac{1 - 1}{2} = \frac{0}{2} = 0$

Solving for the second equation we get
${x}^{2} + x = 0$
$\setminus \frac{- \left(1\right) \setminus \pm \setminus \sqrt{{\left(1\right)}^{2} - 4 \left(1\right) \left(0\right)}}{2 \left(1\right)}$
$\setminus \frac{- 1 \setminus \pm \setminus 1}{2}$
$\setminus \frac{- 1 + 1}{2} = \frac{0}{2} = 0$
$\setminus \frac{- 1 - 1}{2} = - \frac{2}{2} = - 1$

So our answers are $x = - 1 , 0 , 1$

Apr 27, 2017

#### Explanation:

Given: $\left\mid x \right\mid = {x}^{2}$

Rewrite equal zero:

${x}^{2} - | x | = 0 \text{ [1]}$

Because the definition of the absolute value function is

|x|={(x;x>=0),(-x;x<0):}

Equation [1] breaks into two equations:

x^2-x= 0;x >= 0" [2]"

x^2+x = 0; x<0" [3]"

We can factor equation [2] and obtain two roots:

$x \left(x - 1\right) = 0$

$x = 0 \mathmr{and} x = 1$

We can remove a factor of x from equation [3] but we must discard it because its root is not within the domain

$x + 1 = 0$

$x = - 1$

We have 3 roots, -1, 0 and 1