Question

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

Answer 1

`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`
`\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(0)}}{2(1)}`
`\frac{1\pm\1}{2}`
`\frac{1+1}{2}=2/2=1`
`\frac{1-1}{2}=0/2=0`

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

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

Answer 2

Please see the explanation.

Explanation:

Given: `absx=x^2`

Rewrite equal zero:

`x^2 - |x| = 0" [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(x-1)=0`

`x = 0 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