here the answer would be basically:

i) AB: 2<x<1

ii) C: 6

iii) D: 8

could you please explain to me how to solve it and understand it properly? Thanks!

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- Mar 28th 2013, 10:57 PMRenrierange of values and conditions
here the answer would be basically:

i) AB: 2<x<1

ii) C: 6

iii) D: 8

could you please explain to me how to solve it and understand it properly? Thanks!

Attachment 27729 - Mar 29th 2013, 04:54 AMRuunRe: range of values and conditions
You are given a polynomial of degree two in $\displaystyle x$, $\displaystyle x^2-3x-10$. A convenient way of writing it is in the form $\displaystyle (x-x_1)(x-x_2)$ where $\displaystyle x_1$ and $\displaystyle x_2$ are its roots, this is, the numbers such that the polynomial is zero.

**Now the reasoning**: Because the polynomial is a continuous function, these two numbers $\displaystyle x_1,x_2$ set a change of sign in the expression $\displaystyle x^2-3x-10$. So you should check whether is positive or negative if $\displaystyle x$ is greater or smaller than $\displaystyle x_1$ and $\displaystyle x_2$.

Also for the absolute value condiction $\displaystyle |x-2|<a$ this means $\displaystyle -a<x-2<a$ - Mar 29th 2013, 06:53 AMHallsofIvyRe: range of values and conditions
Essentially, you are asked to solve an inequality: $\displaystyle x^2- 3x- 10< 0$. And the simplest way to do that is to solve the associated

**equation**. As Ruun suggests, we can factor $\displaystyle x^2- 3x- 10= (x- 5)(x+ 2)= 0$. So where is that**equal**to 0? Those points separate "> 0" from "< 0". Just pick one value in each interval to see whether it is "> 0" or "< 0".