$|x^2-3x+2|=mx$ with 4 distinct real solutions, find range of $m$ [closed]

I am working on my scholarship exam practice but not sure how to start here.

Find the range of $$m$$ such that the equation $$|x^2-3x+2|=mx$$ has $$4$$ distinct real solutions $$\alpha, \beta, \gamma, \delta$$.

The answer provided is $$0.

Could you please give a solution or at least a hint to this question?

closed as off-topic by YuiTo Cheng, TheSimpliFire, Shogun, Cesareo, Paul FrostJul 5 at 9:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YuiTo Cheng, TheSimpliFire, Shogun, Cesareo, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question. Let $$m_0$$ be a value when $$y=m_0x$$ touches $$y=-x^2+3x-2$$, then the answer is $$m\in (0,m_0)$$.

That is $$x^2+x(m-3)+2=0\implies (m-3)^2=8\implies m_0 = 3-\sqrt{8}$$

See why $$\color{red}{m=3+\sqrt{8}}$$ is not good: • I am still struggling at how $x^2+x(m-3)+2=0$ went to $(m-3)^2=8$. What did you do with the $x$? And how did $(m-3)^2$ become squared? – Trey Anupong Jun 17 at 12:16
• The discriminant of quadratic equation must be $0$ since the line touches parabola. – Aqua Jun 17 at 12:17
• Yes, that makes sense. Now I have two possibilities: $m=3+2\sqrt{2}$ or $m=3-2\sqrt{2}$, which are on the red and light green lines of your graph. The red line cannot be used since it does not touch 4 points. In the exam conditions, I will probably test it by substituting the $x$ in range of $(1,2)$ into $y=mx$ for both m's we have got. I got it all now, thank you Aqua and all for the help. :) – Trey Anupong Jun 17 at 12:34

Let $$y=ax$$, where $$a>0$$, be a tangent to the parabola $$y=-x^2+3x-2$$ on $$(1,2)$$.

Thus, we'll obtain the answer: $$0

Draw it!

• Can down-voter explain us, why did you do it? – Michael Rozenberg Jun 17 at 11:54

Absolute values are not easy to deal with in practice, so start by noting that any solution to $$|x^2-3x+2|=mx$$ is either a solution to $$x^2-3x+2=mx$$ or a solution to $$x^2-3x+2=-mx\rm .$$

I am sure you will be able to find the non-negative values of $$m$$ for which the first equation has two real solutions, and also the non-negative values of $$m$$ for which the second equation has two solutions. The only thing you finally need to ensure, to satisfy the question, is that 2 solutions + 2 solutions = 4 solutions. And that (as you will see) is why the inequality is written as $$0, not allowing equality.

Procedure (one of possibly many): On the region where we have $$|x^2-3x + 2| = x^2 - 3x + 2$$ we are interested in the solutions to the equation $$x^2 - 3x + 2 = mx$$ and on what region where we have $$|x^2 -3x + 2| = -x^2+3x-2\tag2$$ we are interested in the solutions to the equation $$-x^2+3x-2 = mx$$ Each of these two equations have at most two solutions. If we want four solutions, we need both of them to have two solutions. Which $$m$$ makes that happen?