How do you solve $∣ ∣ x^{2} + 3 x - 2 ∣ ∣ = 2$ $| x^2+3x-2 | =2$ ?

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Answer 1 · 2016-08-29T21:05:46

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

Case 1: The absolute value is positive

$x^{2} + 3 x - 2 = 2$ $x^2 + 3x - 2 = 2$

$x^{2} + 3 x - 2 - 2 = 0$ $x^2 + 3x - 2 - 2 = 0$

$x^{2} + 3 x - 4 = 0$ $x^2 + 3x - 4 = 0$

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

$x = - 4 and 1$ $x = -4 and 1$

Case 2: The absolute value is negative

$- (x^{2} + 3 x - 2) = 2$ $-(x^2 + 3x - 2) = 2$

$- x^{2} - 3 x + 2 = 2$ $-x^2 - 3x + 2 = 2$

$- x^{2} - 3 x + 2 - 2 = 0$ $-x^2 - 3x + 2 - 2 = 0$

$- x^{2} - 3 x = 0$ $-x^2 - 3x = 0$

$- x (x + 3) = 0$ $-x(x + 3) = 0$

$x = 0 and - 3$ $x = 0 and -3$

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

Hopefully this helps!

How do you solve ∣∣x2+3x−2∣∣=2| x^2+3x-2 | =2?