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I am solving the following inequality, please look at it and tell me am i write or not because this is an example in Howard Anton's book, i solved it on my own as given below but in the book it is solved differently, please look at my solution and tell me am i right or not and if not then explain.

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
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4 Answers

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
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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
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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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also have to look at the case where both are negative.

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This is really a comment, not an answer to the question. You can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. –  tomasz Aug 17 '12 at 13:58
    
It looks like an answer to me. Not a very eloquent one, or even a complete one (since there are other problems than the one pointed out here), but it does seem to be an honest attempt answer the question asked. Downvoting Is Good Enough For This One. –  Henning Makholm Aug 31 '12 at 16:54
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