# Thread: Finding the values of x for a boundary on a graph

1. ## Finding the values of x for a boundary on a graph

Hi I need to find the values of x for which

$\displaystyle \frac{x^{3}+5x-12}{x-3} > 4$

Now I could - 4(x-3)/(x-3) from the RHS to the LHS, which is what the markscheme does, but why cant I do this:

$\displaystyle x^{3}+5x-12> 4(x-3)$
$\displaystyle x^{3}+5x-12> 4x-12$
$\displaystyle x^{3}+5x> 4x$

$\displaystyle x^{3}+x$

that only gives me x = 0 and i as opposed to the other way which gives me 0,3,i

Why doesn't the stupid math work?

Thanks

2. Originally Posted by thomas49th
Hi I need to find the values of x for which

$\displaystyle \frac{x^{3}+5x-12}{x-3} > 4$

Now I could - 4(x-3)/(x-3) from the RHS to the LHS, which is what the markscheme does, but why cant I do this:

$\displaystyle x^{3}+5x-12> 4(x-3)$
$\displaystyle x^{3}+5x-12> 4x-12$
$\displaystyle x^{3}+5x> 4x$

$\displaystyle x^{3}+x$

that only gives me x = 0 and i as opposed to the other way which gives me 0,3,i

Why doesn't the stupid math work?

Thanks
move the 4 to the left side:
$\displaystyle \frac{x^{3}+5x-12}{x-3} - 4 > 0$
$\displaystyle \frac{x^{3}+5x-12-4(x-3)}{x-3} > 0$
$\displaystyle \frac{x^{3}+5x-12-4x+12}{x-3} > 0$
$\displaystyle \frac{x^{3}+x}{x-3} > 0$
$\displaystyle \frac{x(x^{2}+1)}{x-3} > 0$

Can you finish the rest?

Good luck!!

3. Sorry I didn't make myself clear The method what you posted above is what works , but I understand why the method I posted earlier doesn't work either. I'm multiplying x-3 to the other side of the inequality but following it through makes me lose x=3 root. I don't see why

Thanks

4. Originally Posted by thomas49th
I'm multiplying x-3 to the other side of the inequality but following it through makes me lose x=3 root.
Don't multiply through by $\displaystyle x-3$. This could be a minus number, and you should know that if you multiply an inequality by a minus you need to swap the sign around.

The method to use is to either subtract like in the above example, or multiply through by $\displaystyle (x-3)^2$ as you know this is a positive number.