Solving a quadratic inequality $x^2-3x-10>0$

Question

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.

Answer 1

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

Answer 2

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 $ax^{2}+bx+c=a(x-x_1)(x-x_2)$

Answer 3

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