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?

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- Oct 10th 2007, 03:53 PM #1

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- Oct 10th 2007, 04:33 PM #2

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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 $

- Oct 11th 2007, 04:24 AM #3
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

- Oct 11th 2007, 05:43 AM #4

- Oct 11th 2007, 07:42 PM #5

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- Oct 11th 2007, 07:51 PM #6