Radicals and exponents

May 2010
22
0
Hi, I am stuck on this problem, please help.

((x^2+1)^(1/2) - x^2(x^2-1)^(-1/2)) / (x^2-1)

The book says 1/(x^2+1)^3/2

While trying different things I figured out the rule that the square root of a divided by a = 1/ square root of a. There must a rule I could use here. My answer always has x^2 in the numerator.

Thanks!
 
Dec 2009
872
381
1111
Hi, I am stuck on this problem, please help.

((x^2+1)^(1/2) - x^2(x^2-1)^(-1/2)) / (x^2-1)

The book says 1/(x^2+1)^3/2

While trying different things I figured out the rule that the square root of a divided by a = 1/ square root of a. There must a rule I could use here. My answer always has x^2 in the numerator.

Thanks!
Dear dwatkins741,

So the book says that,

\(\displaystyle \frac{(x^2+1)^{\frac{1}{2}}-x^2(x^2-1)^{-\frac{1}{2}}}{x^2-1}=\frac{1}{(x^2+1)^{3/2}}\)

But notice that if x=0 then, \(\displaystyle \frac{(x^2+1)^{\frac{1}{2}}-x^2(x^2-1)^{-\frac{1}{2}}}{x^2-1}=-1\)

But \(\displaystyle \frac{1}{(x^2+1)^{3/2}}=1\)

Hence there is a mistake in this problem,

\(\displaystyle \frac{(x^2+1)^{\frac{1}{2}}-x^2(x^2-1)^{-\frac{1}{2}}}{x^2-1}\neq\frac{1}{(x^2+1)^{3/2}}\)

Hope this will help you.
 
May 2010
22
0
Correction

I sincerely apologize. I had two signs wrong when I typed the problem. Please take another look at it.

((x^2+1)^(1/2) - x^2(x^2+1)^(-1/2)) / (x^2+1)

Sorry, thanks.
 
Dec 2009
872
381
1111
I sincerely apologize. I had two signs wrong when I typed the problem. Please take another look at it.

((x^2+1)^(1/2) - x^2(x^2+1)^(-1/2)) / (x^2+1)

Sorry, thanks.
Dear dwatkins741,

Don't mention it. Now you can solve the problem. First multiply the denominatior and the numerator by, \(\displaystyle (x^2+1)^{\frac{1}{2}}\). Hope you can continue.
 
Dec 2007
3,184
558
Ottawa, Canada
((x^2+1)^(1/2) - x^2(x^2+1)^(-1/2)) / (x^2+1)
Go this way: let a = x^2 + 1 ; then:
[sqrt(a) - x^2/sqrt(a)] / a : remember that k^(-p) = 1/k^p
= [sqrt(a)sqrt(a) - x^2] / [a sqrt(a)]
= (a - x^2) / [a^1 a^(1/2)]
= (a - x^2) / a^(3/2) : remember that k^p k^q = k^(p+q)

Finish it by substituting back in...