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

2 Answers
flamespirit919
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
\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

Douglas K.
Apr 27, 2017

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

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