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I am solving the following inequality, please look at it and tell me whether am I correct or not. This is an example in Howard Anton's book and I solved it on my own as given below, but the book has solved it differently! I want to confirm that my solution is also valid.

alt text

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Hint: if both x-5 and x+2 have the same sign (positive or negative), their product will be greater than 0. – J. M. Aug 15 '10 at 12:35
Just so you know: "Hay" is properly spelled "Hey" and "Hey Dear" is generally only used between married couples. On a forum, you should introduce yourself using either simply "Hey" or "Hey everyone". On Stack Exchange, you don't need to "Hey" at all, you can just post your problem and your motivation – Casebash Aug 15 '10 at 12:39
oo exelent sir, thank you. – Zia ur Rahman Aug 15 '10 at 12:49
Exelent should be excellent. Next time, when you post a thread just say something regarding the question. You don't need to use hey,hi sort of things while posting the question. – anonymous Aug 15 '10 at 12:55
I don't see this is a (differential and integral) calculus question. Why was the calculus tag used?. – Américo Tavares Aug 15 '10 at 17:33
up vote 5 down vote accepted

For ab to be positive either

  • a and b are both positive
  • a and b are both negative

Here, a=x-5 and b=x+2

They are both positive if x>5. They are both negative if x<-2. Either of these will solve the problem

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Remove the "non" part in your answer and it becomes accurate; what he has is a strict inequality (">" instead of "≥") – J. M. Aug 15 '10 at 12:37
@J. You are right, I misread the problem – Casebash Aug 15 '10 at 12:42
Thank you Sir i have understood it know. – Zia ur Rahman Aug 15 '10 at 12:43

If you graph the function $y=x^2-3x-10$, you can see that the solution is $x<-2$ or $x>5$.

alt text

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image failure!! – The Chaz 2.0 Sep 15 '11 at 13:23

Casebash's answer is very good.

Here is a second answer. You can apply the following

Theorem: If the roots $x_{1},x_{2}$ of $f(x)=ax^{2}+bx+c$ are real and $x_{1}\neq x_{2}$ (with $x_{1} < x_{2}$), then, the signal of $f(x)$ is:

  • opposite to the signal of $a$ for $x\in \left[ x_{1},x_{2}\right] $,
  • the same of $a$ for $x\in \left] -\infty ,x_{1}\right[ \vee x\in \left] x_{2},-\infty \right[ $.

Since in your case $a=1>0$, $x_{1}=-2<5=x_{2}$, you have $x^{2}-3x-10>0$ for $x\in \left] -\infty ,-2\right[ \vee x\in \left] 5,\infty \right[ $.

Addendum: A possible proof of this theorem is to use the explanation of Casebash, taking into consideration that $ax^{2}+bx+c=a(x-x_1)(x-x_2)$

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