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

Answer 1

Kennedy Adams

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

Example 1: There is a positive absolute value

#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#

Situation 2: There is a negative absolute value

#-(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#

Verify that each and every solution to the original equation satisfies the equation by checking all of the solutions.

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Answer 2

Avery Alexander

To solve the equation |x^2 + 3x - 2| = 2, you need to consider two cases:

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

Solve each case separately:

Case 1: x^2 + 3x - 2 = 2 x^2 + 3x - 4 = 0 (x + 4)(x - 1) = 0 x + 4 = 0 or x - 1 = 0 x = -4 or x = 1

Case 2: x^2 + 3x - 2 = -2 x^2 + 3x = 0 x(x + 3) = 0 x = 0 or x + 3 = 0 x = 0 or x = -3

So, the solutions to the equation |x^2 + 3x - 2| = 2 are x = -4, x = 1, x = 0, and x = -3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.