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Thread: 4x^2-1/4x^2

  1. #1
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    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?
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  2. #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 $
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bilbobaggins View Post
    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
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  4. #4
    Senior Member DivideBy0's Avatar
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    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 $\displaystyle \frac{15}{16}$.
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  5. #5
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    Is there a case where 2 consecutive numbers aren't relatively prime?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SnipedYou View Post
    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
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