1. ## Re: Square root fractions.....

Let a = 1+x = 1+sqrt(3)/2 and b = 1-x = 1-sqrt(3)/2 ; then expression becomes:

[a(1 - sqrt(b)) + b(1 + sqrt(a))] / [(1 + sqrt(a))(1 - sqrt(b))]
= [a(1 - sqrt(b)) + b(1 + sqrt(a))] / [1 + sqrt(a) - sqrt(b) - sqrt(ab)]

Since sqrt(ab) = 1/2, above simplifies to:
2[a(1 - sqrt(b)) + b(1 + sqrt(a))] / [1 + 2(sqrt(a) - sqrt(b))]

Substitute a = 1+sqrt(3)/2 and b = 1-sqrt(3)/2 and you'll get a nice 1 as answer

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

Substituting (√3)/2 into √(1±x), for x, may not look like it has much potential to be simplified, but it can be simplified quite nicely.

$\displaystyle \sqrt{1\pm\frac{\sqrt{3}\,}{2}\,}=\sqrt{\frac{4\pm 2\sqrt{3}\,}{4}\,}$
$\displaystyle =\frac{\sqrt{4\pm2\sqrt{3}\,}}{2}}$

$\displaystyle =\frac{\sqrt{3+1\pm2\sqrt{3}\,}}{2}}$

$\displaystyle =\frac{\sqrt{(\sqrt{3})^2\pm2\sqrt{3}\,+1}}{2}}$

$\displaystyle =\frac{\sqrt{\left(\sqrt{3}\pm1\right)^2}}{2}}$

$\displaystyle =\frac{\sqrt{3}}{2}\pm1$
This should make plugging in $\displaystyle \frac{\sqrt{3}}{2}$ for x much easier.

3. ## Re: Square root fractions.....

Originally Posted by annam911
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!
Are you sure that the answer is 1??

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

Yes, you can check it with wolphram alpha if you want.

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