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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