1. ## 4x^2-1/4x^2

4x^2-1/4x^2

its supposed to equal 15/16ths somehow?Or atleast thats what my teacher told me. Could you show me how this equals 15/16ths?

2. Ok well we see that $\displaystyle 4x^2 = 16$ from the bottom so now let's solve for x

$\displaystyle 4x^{2}=16$
$\displaystyle x^{2}=4$
$\displaystyle x =\sqrt{4}$
So we get $\displaystyle x = \pm 2$
(Remember that square roots always have 2 roots)
Now we can see if it satisfies the numerator.
$\displaystyle 4(2)^2-1 = 15$
$\displaystyle 4(4)-1 = 15$
$\displaystyle 16 - 1 = 15$
and of course $\displaystyle 15 = 15$

So we found that $\displaystyle x = \pm 2$

3. Originally Posted by bilbobaggins
4x^2-1/4x^2

its supposed to equal 15/16ths somehow?Or atleast thats what my teacher told me. Could you show me how this equals 15/16ths?
SnipedYou:
You can in this particular instance assume that the numerator is 15 and the denominator is 16, but you first have to show that $\displaystyle 4x^2 - 1$ and $\displaystyle 4x^2$ are relatively prime, a problem that I doubt bilbobaggins is going to want to do.

Here's the general method:
$\displaystyle \frac{4x^2 - 1}{4x^2} = \frac{15}{16}$

$\displaystyle 16(4x^2 - 1) = 15(4x^2)$

$\displaystyle 64x^2 - 16 = 60x^2$

$\displaystyle 4x^2 = 16$

$\displaystyle x^2 = 4$

Thus
$\displaystyle x = \pm 2$

-Dan

4. I wonder if he means $\displaystyle \frac{4x^2-1}{4x^2}=\frac{15}{16}$ for all x?

Well that is false. It cannot simplify down to $\displaystyle \frac{15}{16}$.

5. Is there a case where 2 consecutive numbers aren't relatively prime?

6. Originally Posted by SnipedYou
Is there a case where 2 consecutive numbers aren't relatively prime?
No, but I suspect if bilbobaggins didn't know how to solve the problem the long way (my way) I doubt he would have even considered the concept of relatively prime in solving this.

-Dan