How do you solve #| x^2+3x-2 | =2#?

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Answer 1 · 2016-08-29T21:05:46

The solution set is #{1, 0, -3, -4}#

Case 1: The absolute value is positive

#x^2 + 3x - 2 = 2#

#x^2 + 3x - 2 - 2 = 0#

#x^2 + 3x - 4 = 0#

#(x + 4)(x - 1) = 0#

#x = -4 and 1#

Case 2: The absolute value is negative

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

#-x^2 - 3x + 2 = 2#

#-x^2 - 3x + 2 - 2 = 0#

#-x^2 - 3x = 0#

#-x(x + 3) = 0#

#x = 0 and -3#

Check all the solutions in the original equation. You will find that all solutions satisfy the equation.

Hopefully this helps!