How do you solve #｜x-2｜ =｜x^2-4｜#?

Question

Genesis Allen · Accepted Answer

#x = -1#, #x = -3# or #x = 2#

Given:

#abs(x-2) = abs(x^2-4)#

We must have one of the following:

a) #color(white)(0)(x-2) = (x^2-4)#

b) #color(white)(0)(x-2) = -(x^2-4)#

#color(white)()# Case a)

#x-2=x^2-4#

Subtract #(x-2)# from both sides to get:

#0 = x^2-x-2 = (x-2)(x+1)#

So #x=-1# or #x=2#

#color(white)()# Case b)

#x-2=-(x^2-4)#

Add #(x^2-4)# to both sides and transpose to get:

#0 = x^2+x-6 = (x+3)(x-2)#

So #x=-3# or #x=2#

#color(white)()# Check potential solutions

Trying each of these values of #x# as possible solutions of the original equation:

#abs((-1)-2) = abs(-3) = abs((-1)^2-4)#

#abs((2)-2) = 0 = abs((2)^2-4)#

#abs((-3)-2) = 5 = abs((-3)^2-4)#

So all the possible solutions are solutions of the original equation.

Savannah Anderson · Answer

undefined To solve the equation |x-2| = |x^2-4|, you need to consider two cases:

1. When $ x - 2 $ is positive or zero, $ |x-2| = x - 2 $.
2. When $ x - 2 $ is negative, $ |x-2| = -(x - 2) = 2 - x $.

For case 1:
$ |x-2| = x - 2 $
$ |x^2-4| = x^2 - 4 $

For case 2:
$ |x-2| = 2 - x $
$ |x^2-4| = -(x^2 - 4) = 4 - x^2 $

Solve each case separately and then check if the solutions satisfy the original equation.