# Absolute value equationx^2-4|x|+3=0

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

We have two options fo the absolute value.

lxl = x for x > 0

lxl = -x for x < 0

Then, we have two options for the equation.

x^2 - 4x + 3 = 0

OR

x^2 + 4x + 3 = 0

We will solve each.

x^2 - 4x + 3 = 0

We will factor.

==> (x-3)(x-1) = 0

==> x = { 1, 3}

x^2 + 4x + 3 = 0

==> (x+3)(x+1) = 0

==> x = { -1. -3}

**Then, the answer is: x= { 1, -1, 3, -3}**

In this kind of equation, first, we'll discuss the module.

|x| = x, for x>=0

|x| = -x, for x < 0

Now, we'll solve the equation for x>=0

x^2 - 4x + 3 =0

We'll apply quadratic formula:

x1 = [4+sqrt(16-12)]/2

x1 = (4+2)/2

x1 = 3

x2 = (4-2)/2

x2 = 1

Since both values are positive, we'll accept them as solution of the equation.

We'll solve the equation for x<0.

x^2 + 4x + 3 = 0

x1 = [-4+sqrt(16-12)]/2

x1 = (-4+2)/2

x1 = -1

x2 = -3

Since both values are negative, we'll accept them as solution of the equation.

The solutions of the equation are: {-3 ; -1 ; 1 ; 3}.