Find range of $m$ such that $|x^2-3x+2 | = mx$ has 4 distinct solutions

Question

$|x^2-3x+2 | = mx$

I think that $(x^2-3x+2) = mx$ or $-(x^2-3x+2) = mx$

For $-(x^2-3x+2) = mx$, $D > 0$, $m = 3 \pm \sqrt{8}$

For $(x^2-3x+2) = mx$, $D > 0$, $m = -3 \pm \sqrt{8}$

Now. What do you think it's the range? Note that $ |x^2-3x+2 | = mx$ which means $mx > 0$ isn't it?

Answer 1

Solving $|x^2-3x+2|=mx$ is same as finding the intersection of the graph of $y=|x^2-3x+2|$ and the lines $y=mx$ passing through the origin. Here is a figure to better illustrate the idea.

By changing the value of $m$ we are changing the slope of the line passing through the origin. Unless this line is tangent or passes through the middle cusp in the graph there cannot be $4$ solutions. This means the first instance the line will intersect the graph at $4$ points (tangent point is counted as double point) is when this line is tangent to the graph.

Ques: what is the point of tangency?

Let $(h,k)$ be the point on (the middle cusp) the curve such that the line $y=mx$ is tangent. Then using the fact that slope of the line $\frac{k}{h}$ should be the same as the slope of the curve (in particular the middle cusp which is given by $y=-x^2+3x-2$), we get $$m=\frac{k}{h}=\frac{dy}{dx}\Big|_{(h,k)}=-2h+3.$$ But $k=-h^2+3h-2$, so this gives $h=\pm \sqrt{2}$. Note that the negative value is not possible. So $\color{red}{h=\sqrt{2}}$.

Observe that the slope $m=-2h+3=3-2\sqrt{2}$.

Since you have asked for $4$ distinct solutions, so $$\color{red}{0 <m < 3-2\sqrt{2}}.$$

Answer 2

If $f(x) = x^2-(3+m)x+2$, the discriminant is

$D = (3+m)^2-4\cdot 2 = (m+3)^2-8.$

This gives you a quadratic polynomial in $m$. If you have two solutions for $m$, you get in total four solutions of $f(x)$.

Answer 3

Hint:

$$x^2-3x+2=\pm mx$$ has two distinct real solutions when the discriminant is strictly positive, i.e.

$$(\pm m-3)^2-8=(\pm m-3-\sqrt8)(\pm m-3+\sqrt8)>0.$$

The condition must hold for the two signs, and the general condition is

$$((m-3-\sqrt8>0\land m-3+\sqrt8>0)\lor(m-3-\sqrt8<0\land m-3+\sqrt8<0))\land\\ ((-m-3-\sqrt8>0\land-m-3+\sqrt8>0)\lor(-m-3-\sqrt8<0\land-m-3+\sqrt8<0)).$$

This simplifies to

$$(m-3-\sqrt8>0\lor m-3+\sqrt8<0)\land\\ (-m-3-\sqrt8>0\lor-m-3+\sqrt8<0).$$

The rest is yours.

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The graphical approach (as by Anurag) is interesting. We can consider the equation

$$\left|x-3+\frac2x\right|=m$$ and plot the LHS. From the plot, we see that the solution will consist of two intervals, $(0,a)$ and $(b,\infty)$, where $a,b$ are extremal values of the LHS.