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$$y=mx^2+(m+7)x+m+22$$

Steps I took:

$$mx^2+(m+7)x+m+22=7$$

$$mx^{ 2 }+(m+7)x+m+15=0$$

Now this equation will only have real solutions if and only if $b^2-4ac>0$

$$(m+7)^ 2-4(m)(m+15)>0$$

$$m^{ 2 }+14m+49-4m^2-60m>0$$

$$-3m^{ 2 }-46m+49>0$$

$$-3m^{ 2 }-46m+49>0\Rightarrow 3m^{ 2 }+46m-49<0$$

$$(m-1)(m+\frac { 49 }{ 3 } )<0$$

$$(m-1)(m+\frac { 49 }{ 3 } )<0\Rightarrow m<1,m>-\frac { 49 }{ 3 } ;m<-\frac { 49 }{ 3 } ,m>1,$$

I am now stuck and I don't know what to do in order to figure out the actual solution set of $m$ to fit the conditions given in the problem. As a matter of fact, I realized that figuring out where the solution set lies is the most difficult part for me in these type of problems. I would like to fix that. I am sure that I just need to learn what to do at this point in order to get the right solution in the end.

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  • $\begingroup$ $m=(\frac { -49 }{ 3 } , 1)$ $\endgroup$ – Pratyush Oct 8 '15 at 9:05
  • $\begingroup$ @Pratyush Do you mean $m \in (\frac{-49}{3}, 1)$ $\endgroup$ – Shailesh Oct 8 '15 at 9:11
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    $\begingroup$ @Shailesh right $\endgroup$ – Pratyush Oct 8 '15 at 9:12
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Your error

Assuming what you have done is OK, only the final step you made an error. The product of two numbers is negative if one of them is positive and the other is negative. So the final line should read like.

$$(m-1)(m+\frac { 49 }{ 3 } )<0\Rightarrow m \lt 1,m \color{blue}{\gt} -\frac { 49 }{ 3 } ; m \gt 1, m \color{blue}{\lt} -\frac { 49 }{ 3 } $$.

The second one is impossible so the solution set is $\color{blue}{-\frac { 49 }{ 3 } \lt m \lt 1}$.

However, you have to $\color{red}{exclude}$ $m = 0$ otherwise the equation will reduce to $y = 22$ which is parallel to $y= 7$.

Hence we split the interval as

$\color{blue}{-\frac { 49 }{ 3 } \lt m \lt 0}$ and $\color{blue}{0 \lt m \lt 1}$ which you can also write as $\color{blue}{m \in (-\frac{49}{3}, 0) \cup (0,1)}$

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  • $\begingroup$ I understand that in order for $(m-1)(m+\frac { 49 }{ 3 } )<0$, one has to be negative and the other positive; I guess I got lazy and didn't write it properly. However, what I don't understand is how the solution to this problem in the booklet I am using saids: $(0,1)$ or $(-\frac { 49 }{ 3 } ,0)$ $\endgroup$ – Cherry_Developer Oct 8 '15 at 9:24

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