For what real number $c$, this equation has exactly three solutions? For what real number c does the equation $|x^2 + 12x + 34| = c$ has exactly three solutions?
 A: Hint:
$$ |x^2+12x+34|=c $$
$$ \sqrt{(x^2+12x+34)^2}=c $$
$$ (x^2+12x+34)^2=c^2 $$
Can you solve it from here?
A: First of all rewrite your equation as $$|(x-6)^2-2|=c$$
Then take away the modulus, which means that you have to solve $$(x-6)^2-2-c=0\lor(x-6)^2-2+c=0$$
Since they are both quadratics they'll have 2 solutions each, for a total of 4 solutions, unless you make one of them a perfect square, by picking either $c=2$ or $c=-2$.
You should immediately see that only one of these 2 values makes sense
A: You need no fancy rewriting; to begin with, the equality
$$
|A|=B
$$
is equivalent to a set of two equalities
$$
A=B\quad\text{or}\quad A=-B
$$
so you get the solutions of your equation by putting together the solutions of
$$
x^2+12x+34=c
$$
and
$$
x^2+12x+34=-c
$$


*

*Note that no solution of the first can be a solution of the second, because this would happen only for $c=0$ and…

*Either equation can have zero, one or two solutions; what's the way to decide what case we're in?

*You need that the first equation has two solutions and the second equation has one solution or conversely
A: Notice $|x^2 + 12 x +  34| = |(x+6)^2 -2|$, the equation 
$$|x^2 + 12 x  + 24| = c\tag{*1}$$
is invariant under the transform $x \mapsto -12 -x$, Unless $x = -6$ itself is a solution, the number of solutions for the equation $(*1)$ is an even number! To have an odd number of solutions, say 3 solutions, the only possible choice for $c$ is given by $|(-6+6)^2 - 2| = 2$. 
By direct substitution, one can check that when $c = 2$, $(*1)$ does have exactly 3 solution.
A: It should not be very difficult to plot the graph of $y=|x^2+12x+34|$. (You have probably learned how to draw graphs of quadratic functions such as $y=x^2+12x+34$; for example by completing the square. To add the absolute value you just "mirror" the part which is below the $x$-axis.)
Here is the plot from WolframAlpha:

If you are able to draw the graph, you should be also able to find for which $c$ the horizontal line $y=c$ intersects this graph in exactly 3 points.
