1. ## Square root fractions.....

Hi guys, I've been stuck on this question for days!
If x= √3/2
Find the value of the expression
1+x/1+√(1+x) + 1-x/1-√(1-x)
Thanks!

2. ## Re: Square root fractions.....

Just to confirm, its $x = \sqrt{\frac{3}{2}}$

And $\displaystyle \frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}$

?

Yes

4. ## Re: Square root fractions.....

Actually no sorry x = √(3)/2

5. ## Re: Square root fractions.....

Maybe start by rationalising the denominator

$\displaystyle \frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}$

$\displaystyle \frac{1+x}{1+\sqrt{1+x}}\times \frac{1-\sqrt{1-x}}{1-\sqrt{1-x}}+\frac{1-x}{1-\sqrt{1-x}}\times \frac{1+\sqrt{1+x}}{1+\sqrt{1+x}}$

What do you get?

6. ## Re: Square root fractions.....

I've tried everything! The answer is ment to be 1, maybe ive gone wrong in my calculation somewhere

7. ## Re: Square root fractions.....

Originally Posted by annam911
Actually no sorry x = √(3)/2

8. ## Re: Square root fractions.....

Originally Posted by annam911
The answer is ment to be 1, maybe ive gone wrong in my calculation somewhere
Show me your attempts. Maybe I can pick up a mistake.

9. ## Re: Square root fractions.....

following on from what u had....
(1+x)(1-√1-x)/)1+√1+x)(1-√1-x) + (1-x)(1+√1+x)/(1-√1-x)1+√1+x)

= (1+x)(1-√1-x) + (1-x)(1+√1+x)/1+√1+x)(1-√1-x)

=2-√1-x - x√1-x + √1+x -x1+x/(1+√1+x)(1-√1-x)

=2-√1-x(1+x) + √1+x(1-x)/(1+√1+x)(1-√1-x)

Thats as far as i could get :s

10. ## Re: Square root fractions.....

In the denominator apply the difference of two squares. $(a+b)(a-b) = a^2-b^2$

11. ## Re: Square root fractions.....

Originally Posted by pickslides
Maybe start by rationalising the denominator

$\displaystyle \frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}$

$\displaystyle \frac{1+x}{1+\sqrt{1+x}}\times \frac{1-\sqrt{1-x}}{1-\sqrt{1-x}}+\frac{1-x}{1-\sqrt{1-x}}\times \frac{1+\sqrt{1+x}}{1+\sqrt{1+x}}$

What do you get?
Opps, I meant to say

$\displaystyle \frac{1+x}{1+\sqrt{1+x}}\times \frac{1-\sqrt{1+x}}{1-\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}\times \frac{1+\sqrt{1-x}}{1+\sqrt{1-x}}$

go from there,

12. ## Re: Square root fractions.....

would you be able to solve it and ill work from there haha im losing my patience with it

14. ## Re: Square root fractions.....

hmmm i dont understand how that working out results in 1

15. ## Re: Square root fractions.....

You'll get the most simplified form of it by using the hint pickslides has given you:
$\frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}=\frac{(1+x)(1-\sqrt{1-x})+(1-x)(1+\sqrt{1+x})}{(1+\sqrt{1+x})(1-\sqrt{1-x})}$
$=\frac{1+x-\sqrt{1-x}-x\sqrt{1-x}+1-x+\sqrt{1+x}-x\sqrt{1+x}}{(1+\sqrt{1x})(1-\sqrt{1-x})}$
$=\frac{2-\sqrt{1-x}-x\sqrt{1-x}+\sqrt{1+x}-x\sqrt{1+x}}{(1+\sqrt{1x})(1-\sqrt{1-x})}$

Now you just have to substitute the given $x$ value in this expression which is not that easy.

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