Fully simplify
$\displaystyle \frac{x^2+x-12}{x^2-9x+18}$
$\displaystyle \frac{x-12}{-9x+18}=0$?
$\displaystyle x-12=0*-9x+18=0$
$\displaystyle x=12$
Yes or no?
Definitely No!
$\displaystyle \frac{x^2+x-12}{x^2-9x+18} = \frac{(x+4)(x-3)}{(x-6)(x-3)}$
1. Cancel the common factor.
2. Determine the domain of the term at the LHS.
3. Set the numerator equal to zero, solve for x.
By the way: The simplification isn't necessary here: Multiply both sides of the equation by the denominator after you've determined the domain of the complete term.
Sorry, but this is completely wrong. By your reasoning, you would simplify something like $\displaystyle \frac{x^2 - 4}{x^2 - 1}$ as $\displaystyle \frac{-4}{-1} = 4$.
You are expected to factorise the numerator and the denominator and then cancel the common factor.
Can you factorise $\displaystyle x^2 + x - 12$? Can you factorise $\displaystyle x^2 - 9x + 18$?
The domain is which input values give real output values. Essentially this boils down to a few simple rules of which you should use the first
The denominator can never equal 0
Values inside logarithms must be greater than 0
Values inside even roots (such as square root) must be greater than 0
In your case $\displaystyle \frac{(x+4)}{(x-6)}$ which means any value that makes $\displaystyle x-6 = 0$ is not part of the domain
So... it's safe to assume the following; $\displaystyle x+4=0$ , $\displaystyle x+6\neq{0}$ ; $\displaystyle x=-4$?
then how does this work out?
$\displaystyle \frac{x^2+x-12}{x^2-9x+18}$Originally Posted by earboth
$\displaystyle x^2-9x+18\neq{0}$ ; $\displaystyle x^2+x-12=0$
does this not grant two answers for $\displaystyle x$?
Edit: O wait, I think I got it now. The domain also includes $\displaystyle x-3\neq{0}$ which in the end means you do have to factorise both nominator and denominator of the original equation.