Rationalize this denominator?

Question

With an explanation please.

14/ square root of 7 - square root of 5.

Answer 1

multiply by the ?cognate or somthing like that on top and bottom

14/sq.root7-sq. root5 times (sq.root7 + sq.root5)/sq.root7 + sq.root5

so its 14(sq.root7 - sq.root5)/2

then >7(sq.root7 - sq.root5)<

......................the guy above me is wrong cause when u square the bottom its gona be 156/(sqrt7-sqrt5)^2

and that ends up being 156/12-2sqrt7sqrt5

and there are still sqrts i the denominator

Answer 2

It's called a conjugate. You multiply the numerator and denominator by sqrt7+sqrt5:

(14(sqrt7+sqrt5))/(7-5) When you multiply by the conjugate, you take the sqare of the first term(sqrt7) and subtract the square of the second term(sqrt5)

So you have:

(14(sqrt7+sqrt5))/2

Or

7(sqrt7+sqrt5)

Answer 3

there are a couple of different reasons to rationalize a denominator:

one common reason is you have a complex number there

like  x / (a-2i)  but you may want to rationalize things like y /(√3 +√2)   or  x / (x-1)

so here's the trick:  given the expression a + b

when you multiply it by a - b you get    a² -ab + ab -b² = a² - b²

so 1/ (a - 2i ) =  (a+2i) / (a² + 4) since - i² = +1

or 1 /(√3 +√2) = (√3-√2) / (3 -2) = √3 - √2

or x / (x-1) = (x² +x) / (x² -1) <== this doesn't get you anywhere, does it?  In some cases it will, however.

So for any A ± B  you multiply top and bottom (numerator and denominator) by A -/+ B

Answer 4

14 /  sqrt 7-sqrt 5

square both numerator and denominator

14^2/ 7-5

156/2=78

Answer 5

http://img294.imageshack.us/img294/9361/...

Answer 6

  __14___

  7^.5 - 5^.5

times conjugate(change + to -) on numerator and denominator to be equal.

=  __14___  * 7^.5 + 5^.5    

   7^.5 - 5^.5 * 7^.5 + 5^.5

=___14(7^.5 +5^.5)__  Simplify

        7-5

=7(7^.5+5^.5)  Bam!