# 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!

• If the number of distinct roots is three, one case has both roots same Aug 8, 2015 at 16:50
• Yes Sir, I had realised that. But I could not get around toc omputing the roots. Aug 8, 2015 at 16:51
• I'm not sure that I understand you correctly. Is $a=6$ a solution of your problem? Aug 8, 2015 at 16:52
• Sir, we have to fund values of $x$ which satisfy the equation. I suppose they will have to be in terms of $a$. Aug 8, 2015 at 16:59
• @BetterWorld: For $a=8$, there exist only two solutions. For $a=5$, there exist four distinct solutions. Are you asking for which $a$ it has exactly three distinct solutions and its solutions? Aug 8, 2015 at 17:14

Case when $x \geq 0$: $x^2+(a-8)x+1=0$

Then $x=\frac{(8-a)\pm \sqrt{a^2-16a+60}}{2}$

Case when $x < 0$: $x^2+(a-2)x+1=0$

Then $x=\frac{(2-a)\pm \sqrt{a^2-4a}}{2}$

As we're only working with real solutions: for the determinants to be non-negative, we need $a \leq 6$ or $a \geq 10$ in the former, and $a \leq 0$ or $a \geq 4$ in the latter.

Since we want 3 roots, this means one of the determinants has to be zero while the other is positive. Only $a=6$ satisfies this: for the first case, $x=1$ (only root: the determinant is zero). In the second case, $x=-2 \pm \sqrt 3$. You must also remember to check that the proposed solutions satisfy the domain of $x$ which you have fixed; indeed $1>0$ and $-2 \pm \sqrt 3<0$.

So the answer necessitates $a=6$, from which it follows that $x=1$, $x=-2 \pm \sqrt 3$ are your three distinct solutions.

EDIT: Sorry, there is another solution in $a=4$, whereupon your roots are $x=2 \pm \sqrt 3$ and $x = -1$.

• I was right! You read my thoughts Aug 8, 2015 at 17:25
• Thanks a lot Sir! Really nice solution! Aug 8, 2015 at 17:28
• For $a=4$, it has exactly three solutions $x=-1,2\pm\sqrt 3$. Aug 8, 2015 at 17:33

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 $$x^2+(a-5)x+1=+3x$$