# How do you solve $2|x+1|^2 = 3|x+1| + 2$?

Do you just change the absolute value signs with parentheses?

Let $|x + 1| = A$. Then we have the quadratic equation $2A^2 = 3A + 2$, or $2A^2 - 3A - 2 = 0$. Factoring, we have that $(2A + 1)(A - 2) = 0$. This yields $A =-1/2$ and $A = 2$. $A = -1 / 2$ is not possible, so$A = |x + 1| = 2$, solving, we get $x + 1 = -2$ or $2$, and $x = -3$ or $1$.

• Thank you for formatting my solution. Commented Jan 7, 2016 at 2:24

Hint: What if, instead of $\lvert x+1\rvert$, the equation had $y$? $$2y^2=3y+2$$

How would you solve it then?

Once you've found values for $y$ that satisfy this equation, "remember" that $y=\lvert x+1\rvert$.

• Thank you! Should've noticed that. I was too focused on the absolute value being squared.
– user303287
Commented Jan 7, 2016 at 2:55

Hint:

Set $t=\lvert x+1\rvert$, solve the quadratic equation for $t$, then solve $\lvert x+1\rvert=t\;$ (this last equation implies $t$ is a non-negative solution of the quadratic equation).