If $|x^2-x-5|+(3x+6)=0$, then how many values of $x$ exist?

Question

If $|x^2-x-5|+(3x+6)=0$, then how many values of $x$ exist?

What I did: when $x^2-x-5\geq0$ $x^2-x-5=-(3x+6)$ On solving I got $x=-1$. Then I took intersection of $x=-1$ and domain that is zero. On intersection there was no such value of $x$.

When $x^2-x-5<0$, on solving I got my answer in root. Now, I don't know what to do next and I don't know if I am doing it in a right way.

Please guys let me know if my way of solving it is right.

Answer 1

Observe that when we have $|a|=b$, then $b \geq 0$. You are given

$$|x^2-x-5|=-(3x+6).$$

So here if there is a solution then it can only occur when $3x+6 \leq 0$, same as saying $x \leq -2$.

Now for your second case, when $x^2-x-5 < 0$, the quadratic you get is $x^2-4x-11=0$, it has two roots $x=2 \pm \sqrt{15}$. None of the roots satisfy the condition $x \leq -2$, hence NO solution.

Graphically things are as shown below.

Answer 2

Hint:

What you have done is correct. Your equation is equivalent to the two systems: $$ \begin{cases} x^2-x-5\ge0\\ x^2-x-5+3x+6=0 \end{cases} \quad \lor \quad \begin{cases} x^2-x-5<0\\ -x^2+x+5+3x+6=0 \end{cases} $$ and you have correctly found that the first system has no solutions.

Now, the equation in the second system has some solution ($ x=2\pm\sqrt{15}$) in the interval that solve the inequality $(\frac{1-\sqrt{21}}{2},\frac{1+\sqrt{21}}{2})$? Can you evaluate this? (you can use approximation of the roots).