# Thread: Easiest way to solve

1. ## Easiest way to solve

What is the simplest way to solve this problem?

$\displaystyle |x+1|^2+3|x+1|-4=0$

2. Hello, Mike9182!

What is the simplest way to solve this equation?

. . $\displaystyle |x+1|^2+3|x+1|-4\:=\:0$
We have a quadratic . . . Solve it accordingly.

Factor: .$\displaystyle \left(|x+1| + 4\right)\left(|x+1| - 1\right) \:=\:0$

Set each factor equal to zero and solve:

. . $\displaystyle |x+1| + 4 \:=\:0 \quad\Rightarrow\quad|x+1| \:=\:-4$ . . . no roots
. . . .
An absolute value is never negative.

. . $\displaystyle |x+1| - 1 \:=\:0 \quad\Rightarrow\quad |x+1| \:=\:1 \quad\Rightarrow\quad x+1 \:=\:\pm1 \quad\Rightarrow\quad x \:=\:0,\:2$

But $\displaystyle x=2$ is extraneous.

. . Therefore, the only solution is: .$\displaystyle x \:=\:0$

3. ## Absolute value

Hello Mike9182
Originally Posted by Mike9182
What is the simplest way to solve this problem?

$\displaystyle |x+1|^2+3|x+1|-4=0$
Let $\displaystyle y = |x+1|$. Then:

$\displaystyle y^2 +3y - 4=0$

$\displaystyle \Rightarrow (y+4)(y-1)=0$

$\displaystyle \Rightarrow y=-4, \,1$

But $\displaystyle y = -4$ is impossible, since $\displaystyle |x+1| \ge 0$

$\displaystyle \Rightarrow y =1= |x+1|$

$\displaystyle \Rightarrow x+1 = \pm 1$

$\displaystyle \Rightarrow x = 0,\, -2$

$\displaystyle x+1 \:=\:\pm1 \quad\Rightarrow\quad x \:=\:0,\:2$