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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 $0<m<3-2\sqrt{2}$.

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

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closed as off-topic by YuiTo Cheng, TheSimpliFire, воитель, Cesareo, Paul Frost Jul 5 at 9:36

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enter image description here

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:

enter image description here

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  • $\begingroup$ 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? $\endgroup$ – Trey Anupong Jun 17 at 12:16
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    $\begingroup$ The discriminant of quadratic equation must be $0$ since the line touches parabola. $\endgroup$ – Aqua Jun 17 at 12:17
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    $\begingroup$ 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. :) $\endgroup$ – Trey Anupong Jun 17 at 12:34
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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: $$0<m<a.$$

Draw it!

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  • $\begingroup$ Can down-voter explain us, why did you do it? $\endgroup$ – Michael Rozenberg Jun 17 at 11:54
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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 $0<m$, not allowing equality.

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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?

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