# Solve the Absolute Value Equation

Find the value of x from the absolute value equation |x² + 4x + 3| + 2x + 3 = 0

## Solution to the absolute value equation

We know when solving the absolute value equations value of |x| becomes

\left | x \right |= \begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x=0 \\ x & \text{ if } x>0 \end{cases}

Before solving the problem we can rearrange the equation into

|x² + 4x + 3| = -2x – 3

Now solving the equation

We can divide the equation into two cases, that is

Case 1: x² + 4x + 3 = -2x – 3, when x² + 4x + 3 is grater than 0

Case 2: x² + 4x + 3 = 2x + 3, when x² + 4x + 3 is less than 0

From case 1

x² + 4x + 3 = -2x – 3

⇒ x² + 6x + 6 = 0

### Apply quadratic formula to solve the equation

\begin{aligned} x&=\frac{-6\pm\sqrt{6^2-4\times1\times6}}{2\times1} \\ \\ &=\frac{-6\pm\sqrt{36-24}}{2} \\ \\ &=\frac{-6\pm\sqrt{12}}{2} \\ \\ \Rightarrow x&=-3\pm\sqrt{3}\\ \\ \end{aligned}

That is the solution to the problem is **x = -3 + √3** and **x = -3 – √3** from the case 1

Check answers. (Plug the solution and make sure they work)

When x = -3 + √3

|x² + 4x + 3| =? -2x – 3

⇒ |(-3 + √3)² + 4( -3 + √3) + 3| =? -2(-3 + √3) – 3

⇒ |9 – 6√3 + 3 – 12 + 4√3 + 3| =? 6 – 2√3 – 3

⇒ |- 2√3 + 3| =? 3 – 2√3

⇒ |- 2√3 + 3| ≠ 3 – 2√3

So, x = -3 + √3 is not a solution

When x = -3 – √3

|x² + 4x + 3| =? -2x – 3

⇒ |(-3 – √3)² + 4( -3 – √3) + 3| =? -2(-3 – √3) – 3

⇒ |9 + 6√3 + 3 – 12 – 4√3 + 3| =? 6 + 2√3 – 3

⇒ |- 2√3 + 3| =? 3 – 2√3

⇒ |- 2√3 + 3| = 3 – 2√3

So, x = -3 – √3 is a solution

From case 2

x² + 4x + 3 = 2x + 3

⇒ x² + 2x = 0

⇒ x(x + 2) = 0

so, x = 0 & x = -2

Check answers. (Plug the solution and make sure they work)

When x = 0

|x² + 4x + 3| =? -2x – 3

⇒ |0 + 0 + 3| =? 0 – 3

⇒ |3| =? -3

⇒ 3 ≠ -3

So, x = 0 is not a solution

When x = -2

|x² + 4x + 3| =? -2x – 3

⇒ |4 – 8 + 3| =? 4 – 3

⇒ |-1| =? 1

⇒ 1 = 1

So, x = -2 is a solution

That is, in general solution to the absolute value equation are **x = -2 & x = -3 – √3**