Problem:
Find all $x$ such that $|x^2+6x+6|=|x^2+4x+9|+|2x-3|$
I can't understand how to get started with this. I thought of squaring both sides of the equation to get rid of the modulus sign, but then couldn't understand how to deal with the following term:$$2|(x^2+4+9)(2x+3)|$$Any help or tips in general would be greatly appreciated. Many thanks!
On
You-know-me has already provided a good answer.
But I decided to undelete my answer in the hope that this answer might be of some help.
Since we have $x^2+4x+9=(x+2)^2+5\gt 0$, we have $$|x^2+4x+9|=x^2+4x+9.$$
Then, since we have $x^2+6x+6=0\iff x=-3\pm\sqrt{3}$ and $2x-3=0\iff x=\frac 32$, separate it into three cases :
For $x\lt -3-\sqrt 3$ or $-3+\sqrt 3\lt x\lt\frac 32$, we have $x^2+6x+6\gt 0$ and $2x-3\lt 0$. So we have $x^2+6x+6=x^2+4x+9-(2x-3)\iff x=\frac 32.$ So, in this case, there exists no such $x$.
For $-3-\sqrt 3\le x\le -3+\sqrt 3$, we have $x^2+6x+6\le 0$ and $2x-3\le 0$. So we have $-(x^2+6x+6)=x^2+4x+9-(2x-3)\iff x^2+4x+9=0.$ So, in this case, there exists no such $x$.
For $x\ge \frac 32$, we have $x^2+6x+6\ge 0$ and $2x-3\ge 0$, and we know that $x^2+6x+6=x^2+4x+9+2x-3$ always holds. So, in this case, we have $x\ge \frac 32$.
Hence, the answer is $x\ge\frac 32$.
A useful tip:
If $|a+b|=|a|+|b|$, it holds only when $a\cdot b\ge 0$.
Therefore $(x^2+4x+9)(2x-3)\ge 0$. Now since, the discriminant of the quadratic is negative, it us always greater than $0$. So, you get your answer as $x \geq\frac{3}{2}$.