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March 5th, 2017, 09:45 PM   #1
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Simplifying Nested Fractions

Can anyone direct me where I went wrong with this example:

$\displaystyle \frac{\sqrt{4-x^{2}}-\frac{1}{2\sqrt{4-x^{2}}}}{4-x^{2}}$

My first inkling was to simplify the top section by multiplying by the reciprocal of the top right fraction - by $\displaystyle {2\sqrt{4-x^{2}}}$ to give like denominators.

So left side should look like this:
$\displaystyle {2\sqrt{4-x^{2}}}*{\sqrt{4-x^{2}}}$

Which should resolve to 2*$\displaystyle 4-x^{2}$

The right side of the top of the original fraction should now resolve to -1 after we multiply with the reciprocal.

We now have :
$\displaystyle \frac{2*(4-x^{2})-1}{4-x^{2}}$

I see the opportunity to cancel out like terms, removing the $\displaystyle 4-x^{2}$ term from both sides of the fraction.
That leaves $\displaystyle \frac{2*1-1}{1}$
Or simply 1.

Too bad after all that work and 'logic' the answer is wrong and I really don't understand what I did wrong. I am trying to shore up my deficiencies in algebra so I can return to school. Calculus I was long ago for me, I took a B, but I am sure that I need to retake it at this point. Apparently algebra too!!

Any assistance is greatly appreciated. No answers please, but directions on what went wrong and perhaps why.

Thank you.
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March 6th, 2017, 06:33 AM   #2
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You have to divide (or multiply) the numerator and denominator by the same thing. If the numerator (or denominator) contains more than one term, your division (or multiplication) must be applied to each term in the numerator (or denominator).

You didn't apply the above rule consistently.
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March 6th, 2017, 07:07 AM   #3
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!!!
Did not cross multiply to make like terms for my top side of the equation! Thank you for the guidance, let's see what I make of it now:

$\displaystyle \frac{\sqrt{4-x^{2}}-\frac{1}{2\sqrt{4-x^{2}}}}{4-x^{2}}$

Cross multiply only the top side:
$\displaystyle \frac{\sqrt{4-x^{2}}}{1}*\frac{2\sqrt{4-x^{2}}}{2\sqrt{4-x^{2}}}-\frac{1}{2\sqrt{4-x^{2}}}*\frac{1}{1}$

We are only cross multiplying the denominators to make equivalent denominators so that we can simplify the top fraction.

From this I get:
$\displaystyle \frac{\frac{\sqrt{4-x^{2}}*2\sqrt{4-x^{2}}-1}{2\sqrt{4-x^{2}}}}{2\sqrt{4-x^{2}}}$

We can take out our like terms and get:
$\displaystyle \frac{\sqrt{4-x^{2}}-1}{2\sqrt{4-x^{2}}}$

Have I got it now?

Last edited by skipjack; March 6th, 2017 at 07:27 AM.
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March 6th, 2017, 07:22 AM   #4
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You may find it easier this way:

let u = 4 - x^2
expression becomes:
[sqrt(u) - 1/(2sqrt(u))] / u

simplify:
you should get (2u - 1) / (2u sqrt(u))

Substitute back in.

Makes the step by step process easier...

Note:
you can assign a value to x (like x=1)
and then evaluate the original expression
and your results....get my drift?
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Last edited by Denis; March 6th, 2017 at 07:34 AM. Reason: added note
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March 6th, 2017, 07:35 AM   #5
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As chopnhack requested that we don't supply answers, I will limit my advice to saying that "take out our like terms" is a vague description that shouldn't be used as an alternative to giving the working in full (which would enable identification of any mistakes).
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March 6th, 2017, 07:48 AM   #6
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Taking Denis's approach even farther

$Let\ u = \sqrt{4 - x^2} \implies \dfrac{\sqrt{4 - x^2} - \dfrac{1}{2\sqrt{4 - x^2}}}{4 - x^2} = \dfrac{u - \dfrac{1}{2u}}{u^2}.$
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March 6th, 2017, 07:54 AM   #7
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Sorry for the vagueness. What I meant by remove the like terms is to divide out the $\displaystyle 2\sqrt{4-x^{2}}$ from the third equation in my last post. It appeared in both the numerator and denominator of the top part of the nested fraction.

I will also explore the substitution methods you fine gents are posting in a moment. I am still curious to work this out and see what I am doing wrong. Sometimes these complicated fractions have a way of getting me spun around and I need to nail this skill down for deriving future equations.

Thanks!

Last edited by chopnhack; March 6th, 2017 at 08:14 AM.
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March 6th, 2017, 08:13 AM   #8
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On the u front:

$\displaystyle \frac{\sqrt{4 - x^2} - \frac{1}{2\sqrt{4 - x^2}}}{4 - x^2} = \frac{u - \frac{1}{2u}}{u^2}$

so this can further be simplified by creating the same denominator for the top half of the nested fraction:

$\displaystyle \frac{u - \frac{1}{2u}}{u^2}$

becomes:

$\displaystyle \frac{\frac{2u*u}{2u}-\frac{1}{2u}}{u^{2}}$

becomes:

$\displaystyle \frac{\frac{2u*u-1}{2u}}{u^{2}}$

using the property of (a/b)/c is equivalent to a/bc gives:

$\displaystyle \frac{2u*u-1}{2u*u^{2}}$

from here I can see two paths to further resolve the equation, but one makes the equation zero, which I will omit since it doesn't make sense:

$\displaystyle \frac{2u^{2}-1}{2u^{3}}$

I see the ability to reduce down our u's:

$\displaystyle \frac{2(1)-1}{2u}$

or $\displaystyle \frac{1}{2u}$ = $\displaystyle \frac{1}{2\sqrt{4 - x^2}}$

Unfortunately I know that this is not the correct answer. Please assist me in finding the flaw in the process, thanks!

Last edited by chopnhack; March 6th, 2017 at 08:15 AM.
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March 6th, 2017, 09:51 AM   #9
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Quote:
Originally Posted by chopnhack View Post
$\displaystyle \frac{2u^{2}-1}{2u^{3}}$

I see the ability to reduce down our u's:

$\displaystyle \frac{2(1)-1}{2u}$
Nooooooooooo.....
Try same procedure with (10 - 1)/15 ; you'll see why
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March 6th, 2017, 10:02 AM   #10
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The wording "reduce down our u's" is just as vague as "take out our like terms".
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