What is the simplest way to solve this problem?
$\displaystyle |x+1|^2+3|x+1|-4=0$
Hello, Mike9182!
We have a quadratic . . . Solve it accordingly.What is the simplest way to solve this equation?
. . $\displaystyle |x+1|^2+3|x+1|-4\:=\:0$
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: .$\displaystyle x \:=\:0$
Hello Mike9182Let $\displaystyle y = |x+1|$. 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 $\displaystyle |x+1| \ge 0$
$\displaystyle \Rightarrow y =1= |x+1|$
$\displaystyle \Rightarrow x+1 = \pm 1$
$\displaystyle \Rightarrow x = 0,\, -2$
Grandad