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.
Solving a quadratic inequality
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0I don't see this is a (differential and integral) calculus question. Why was the calculus tag used?. – 2010-08-15
3 Answers
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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0@J. You are right, I misread the problem – 2010-08-15
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 $ax^{2}+bx+c=a(x-x_1)(x-x_2)$
If you graph the function $y=x^2-3x-10$, you can see that the solution is $x<-2$ or $x>5$.
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0image failure!! – 2011-09-15