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Here's a question I've encountered in a recent high school examination.

Find the range of values of m such that the line $y=mx-3$ intersects with the graph of $ y=2-|3x - 5|$ at exactly two points.
The answer is $-3 < m < 3$.

Picture The suggested method to solve this by the exam setter is:
1. Sketch the absolute value line(s).
2. Calculate the two gradients of the absolute value line(s).
3. Use logic to determine the range of values in which the straight line intersects with both absolute value line(s).

Picture of another method

However, I tried using another method.
1. Simultaneously solving
2. Turning them into a single quadratic equation
3. Using the discriminant to find 2 intersections

But I'm not getting the correct answer. What went wrong?

I also tried splitting $y = 2-|3x - 5|$ into 2 different equations, $y=2 - 3x + 5$ and $y=2 + 3x - 5$, but that wasn't too successful either.Failed attempt

I have no idea why both methods are wrong and am confused.

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    $\begingroup$ Welcome to Mathematics Stack Exchange. Note that when you square both sides you may bring in extraneous solutions (e.g., when you square $x=2$, you get $x^2=4$, which also has solution $x=-2)$ $\endgroup$ Jul 7 '19 at 15:27
  • $\begingroup$ also, when you multiplied both sides of an inequality by $m+3$, you failed to consider the possibility that $m+3<0,$ in which case the direction of the inequality should be reversed $\endgroup$ Jul 7 '19 at 15:29
  • $\begingroup$ So trying to make things quadratic will cause extraneous solutions to occur, and therefore discriminants cannot be possibly used in this case? $\endgroup$
    – helpme
    Jul 8 '19 at 21:46
  • $\begingroup$ If you go from $mx-5=-|3x-5|$ to $(mx-5)^2=(3x-5)^2$ you get (as you showed) $(m^2-9)x^2+(30-10m)x=0.$ Assuming $m^2\ne9$ (i.e., it's a bona fide quadratic), the discriminant is $(30-10m)^2$, which is always positive; there are always two solutions: $x=0$ and $x=10/(m+3)$. But while there are always two solutions to $(mx-5)^2=(3x-5)^2$ (for $m^2\ne9$), there are not always two solutions to $(mx-5)=-|3x-5|$, as my answer shows $\endgroup$ Jul 9 '19 at 4:37
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You want to find values of $m$ such that the line $y=mx-3$ intersects

with the graph of $ y=2-|3x - 5|$ at exactly two points.

Note that for $x\le\dfrac53, y=2+3x-5=3x-3,$ so $y=mx-3$ intersects $y=3x-3$ only when $x=0\le\dfrac53$ (unless $m=3$, in which case there are infinitely many intersection points).

Thus (when $m\ne3)$ we have one intersection point when $x\le\dfrac53,$ so we want exactly one intersection point when $x>\dfrac53$. When $x>\dfrac53$, $y=2-3x+5=7-3x,$ and this intersects $y=mx-3$ when $x=\dfrac{10}{m+3}$ (unless $m+3=0$, in which case there is no intersection), as you correctly calculated. Now we want $\dfrac{10}{m+3}\gt\dfrac53;$ i.e., $\dfrac6{m+3}>1.$

This happens when $m+3>0$ and $m+3<6$; i.e., $-3<m<3$.

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  • $\begingroup$ Ok, I understand all of this if "x=10/3+m" refers to the decreasing part of the modulus function. However, why does "x=10/3+m" refer to the decreasing part of the modulus function? The "3x-5" above it shows that the gradient is positive, and that it refers to the increasing part of the modulus function. $\endgroup$
    – helpme
    Jul 8 '19 at 21:43
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    $\begingroup$ $x=\dfrac{10}{m+3}$ is the solution for when $7-3x$, which is what I think you are calling the "decreasing part of the modulus function", intersects $y=mx-3$; $7-3x$ is $2-|3x-5|$ when $x>\dfrac53$. When $x\le\dfrac53$ then $2-|3x-5|=2+3x-5=3x-3$, which I think you are calling the "increasing part of the modulus function" (it has a positive slope) $\endgroup$ Jul 9 '19 at 3:22
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Hint: You have to distinguish the cases: $x\geq \frac{5}{3}$ and you get the function $$y=2-3x+5=-3x+7$$ And in the other case: $x<\frac{5}{3}$ then we have $$y=2+3x-5=3x-3$$

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I'd first change the coordinate system doing $y\to y+2$, so the problem is to determine $m$ in such a way that the line $y=mx-5$ intersects with the graph of $y=-|3x-5|$.

This becomes finding the intersections \begin{cases} y=mx-5 \\ y^2=(3x-5)^2 \\ y\le0 \end{cases} The resolvent becomes $m^2x^2-10mx+25=9x^2-30x+25$, so $$ (m^2-9)x^2+(30-10m)x=0 $$ For $m=3$, the equation becomes $0=0$: the line $y=3x-5$ indeed includes infinitely many points of the given graph.

For $m\ne3$, there is no need to find the discriminant, because the equation becomes $(m-3)(m+3)x^2-10(m-3)x=0$; in the case $m=-3$ the only solution is $x=0$, otherwise the solutions are $x=0$ and $x=10/(m+3)$.

Note that we can so eliminate the cases $m=3$ and $m=-3$ from consideration.

For $x=0$ you find $y=-5$, which satisfies the requirement. For $m\ne\pm3$, the second intersection is at $$ x=\frac{10}{m+3} $$ and we get $$ y=\frac{10m}{m+3}-5=\frac{10m-5m-15}{m+3}=5\frac{m-3}{m+3} $$ which is $\le0$ if and only if $-3<m<3$ (remember we removed the cases $m=\pm3$ beforehand).

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  • $\begingroup$ It looks like I was too focused on trying to find the discriminant that I forgot about factoring x. Why is the discriminant a bad idea here? Also, why does y have to be equal or less than 0? (The maximum value of y is 5/3 here) $\endgroup$
    – helpme
    Jul 8 '19 at 21:50
  • $\begingroup$ @helpme The discriminant of $ax^2+bx$ is $b^2$ and it's useless to compute it as it is surely $\ge0$. With my transformation, the graph is $y=-|3x-5|$, which takes on all negative values (and no positive value). Note that $5/3$ is where $3x-5$ changes sign, but has no bearing with $y$. $\endgroup$
    – egreg
    Jul 8 '19 at 21:53
  • $\begingroup$ Ah, so that's why the transformation was used. I understand your method now. I tried using the discriminant to get the range of m, with 100m^2-600m+900>=0, but it was wrong. Why was it wrong? $\endgroup$
    – helpme
    Jul 8 '19 at 22:06
  • $\begingroup$ @helpme You know that $(10m-30)^2\ge0$ for every $m$, don't you? $\endgroup$
    – egreg
    Jul 8 '19 at 22:08
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    $\begingroup$ @J.W.Tanner The $m=-3$ case is easy to analyze, but you're right. I fixed it. $\endgroup$
    – egreg
    Jul 9 '19 at 8:37
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$$mx-3 = 2-|3x-5|\implies |3x-5|^2 = (5-mx)^2$$

So $$x^2(9-m^2)-10x(3-m)=0$$ So $x=0$ is always a soluton. Say $x\ne 0$ then we get

$$(3+m)\Big(x(3-m)-10x\Big)=0$$

If $m=-3$ we have no solution if $m=3$ every $x$ is a solution.

So if $m\neq \pm 3$ $$x ={10\over 3+m}\neq 0$$ and thus $2$ solution.

Now you have check when this value fits starting equation:

$$\Big|{3-m\over 3+m}\Big| ={3-m\over 3+m}$$

So $${3-m\over 3+m}\geq 0\implies m\in (-3,3)$$

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  • $\begingroup$ The problem with your approach is that not all solutions of the quadratic equation are solutions to the original one. E.g. when $m=3$, every $x$ is a solution to the quadratic equation, but $x=2$ is not a solution to the original equation. $\endgroup$
    – user1551
    Jul 9 '19 at 10:59
  • $\begingroup$ But $m=3$ does not fit the condition. $\endgroup$
    – Aqua
    Jul 9 '19 at 11:10
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    $\begingroup$ What I am saying is that the two equations are not equivalent. You have shown that when $m=3$, every $x$ is a solution to the quadratic equation. However, as the solutions to the quadratic equation are not necessarily solutions to the original equation (such as when $x=2$), it is not immediately clear why the original equation has infinite many solutions when $m=3$. Although the latter is indeed true, this is something that you need to prove. $\endgroup$
    – user1551
    Jul 9 '19 at 11:18
  • $\begingroup$ Yes, I see. Thanky you for pointing out. $\endgroup$
    – Aqua
    Jul 9 '19 at 11:27

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