What is the Square of the Square Root of a Polynomial

**TaylorM0192** · January 21st 2010, 09:34 AM

Example: $(\sqrt{1-x^2})^2$

Inuitively, I would say it's simply $1-x^2$ . Alternatively $(\sqrt{1-x^2})(\sqrt{1-x^2})$ appears to give $\sqrt{1-2x^2+4x^2}$ which also does not seem correct.

The question arises from this problem:

$\sqrt{1+(\frac{x}{\sqrt{1-x^2}})^2}$ where I constantly get 1 by stating $\sqrt{1-x^2} = 1-x^2$ . The correct answer, according to my text book anyway, is $\frac{1}{\sqrt{1-x^2}}$ .

Thanks -

**TaylorM0192** · January 21st 2010, 09:52 AM

Nevermind, I completely forgot that I needed to multiply the denominator by $\sqrt{1-x^2}$ to keep the expression equal. The numerator as I simplified earlier, is 1 and there is nothing to simplify in the denominator $=> \frac{1}{\sqrt{1-x^2}}$

**Masterthief1324** · January 21st 2010, 05:38 PM

Originally Posted by TaylorM0192

Example: $(\sqrt{1-x^2})^2$

Inuitively, I would say it's simply $1-x^2$ . Alternatively $(\sqrt{1-x^2})(\sqrt{1-x^2})$ appears to give $\sqrt{1-2x^2+4x^2}$ which also does not seem correct.

The question arises from this problem:

$\sqrt{1+(\frac{x}{\sqrt{1-x^2}})^2}$ where I constantly get 1 by stating $\sqrt{1-x^2} = 1-x^2$ . The correct answer, according to my text book anyway, is $\frac{1}{\sqrt{1-x^2}}$ .

Thanks -

Help me out:
So you distributed the exponents to $\frac {x}{\sqrt{1-x^2}}$ , how would you simplify ... ohh I see.

So it simplifies to ..
$\sqrt {\frac {1}{1-x^2} + \frac {x^2}{1-x^2}}$

right? If so, then where did the $x^2$ in the numerator go?

**Pulock2009** · January 22nd 2010, 10:34 PM

√(1+((x/(√(1-x²))))²)
=√(1+(((x²)/(1-x²))))
=√(((1-x²+x²)/(1-x²)))
=√((1/(1-x²)))
=(1/(√(1-x²)))

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