How do you solve #｜x-2｜ =｜x^2-4｜#?

Question

1237 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-10T06:31:09

#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.