Distinct Roots of $x^2+(a-5)x+1=3|x|$ Problem:

$$x^2+(a-5)x+1=3|x|$$ Find 3 distinct solutions to the above problem.

A friend of mine at my coaching center came up with this problem which nobody was able to solve. Unfortunately, I have been unable to contact my professor and understand how to solve this problem. Despite thinking for a long time, I could not come up with anything.
The only things that struck me was that I should open up the modulus sign (first by taking $x\ge0\Rightarrow |x|=x$ and then by taking $x<0\Rightarrow |x|=-x$). 
Also, the question could perhaps then be tackled by using relations between the roots of the quadratic equations (I know only that the sum of both roots of a quadratic equation $ax^2+bx+c$ is $\dfrac{-b}{a},$ and that their product is $\dfrac{c}{a}$).
Unfortunately I could not proceed any further. I would be truly grateful if somebody would kindly show me how to solve this problem. Many, many thanks in advance!

HINT...If you want to find solutions which comprise precisely three values of $x$, you will need to find the values of $a$ for which the parabola has either of the lines $y=\pm3x$ as tangents.
So, for example, if $y=-3x$ is tangent, then the equation $$x^2+(a-5)x+1=-3x$$ must have double roots, and this leads to the possible values $a=4, 0$
However, the $y$ value at the point of tangency must be non-negative, which means only $a=4$
then you can then write down the $x$ value at the tangent point, and for that chosen value of $a$ go ahead and find the other two values of $x$ by solving $$x^2+(a-5)x+1=+3x$$
