Math Is Fun Forum

Cammock

Member

Registered: 2009-11-10

Posts: 6

Fraction binomials

Does anyone know anything about fraction binomials with X in the common numerator?

E.G.:

(X/(6/25)) × (X/(1/5)) + (X/(111/250)) = 2/3

A value can easily be ascertained for the numerator after the multiplication, but the numerator is not X squared or X.

I.E.:

((148/5125) / (6/125)) + ((148/5125) / (111/250)) = 2/3

A square root of this numerator works before the multiplication, but the numerator is not X.

I.E.:

((√148/5125) / (6/25)) × ((√148/5125) / (1/5)) + ((148/5125) / (111/250)) = 2/3

Factoring this fraction binomial seems very difficult or impossible.

Can anyone help?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Fraction binomials

Hi;

Is this what you meant?

This equals approx. .00105 not 2 / 3

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Anakin

Member

Registered: 2009-10-04

Posts: 145

Re: Fraction binomials

I think you're assuming that for any value of x, the first equation holds true.

Let's say

So that means only when x is equal to 8, does the left side equal to the right side.

When you put in a random number, it does not work out. Ex. when we substitute in 5 for x:

Obviously, that is not the case.

Cammock

Member

Registered: 2009-11-10

Posts: 6

Re: Fraction binomials

Thanks fellas, your comments really helped me understand the problem, it seems that this is bascially not a proper binomial.

If I have anymore problems I will be sure to ask.

Graham Cammock