# 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?

• Welcome to Math SE ! Could you explain what you tried and where you are stuck ? People will help you but I don't think that anyone would accept to do your homework if you don't show your efforts. Cheers :) – Claude Leibovici Sep 14 '14 at 9:09
• @ClaudeLeibovici Thanks for showing interest buddy, actually thi isn't a homework - was just practicing some algebra question and got stuck on this one. I was confused about how to approach this one. Sometimes i guess even a simple thing doesn't click :) – Mohit Singh Sep 14 '14 at 9:36

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.

• wow! Now this is really an amazing approach Martin! Just loved it! Thanks a lot... :) – Mohit Singh Sep 14 '14 at 11:38

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?

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 $$x^2+12x+34=-c$$

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

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

3. You need that the first equation has two solutions and the second equation has one solution or conversely

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.

• that was quick buddy :) – Mohit Singh Sep 14 '14 at 9:39

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

• Thanks a lot @Alessandro , very well explained! :) – Mohit Singh Sep 14 '14 at 9:34