Exponents and square roots (once more)

**t-dot** · October 20th 2007, 07:21 PM

how is x^2 the answer to this question??

$(\sqrt{x^3})(\sqrt{x})$

**Jhevon** · October 20th 2007, 07:29 PM

Originally Posted by t-dot

how is x^2 the answer to this question??

$(\sqrt{x^3})(\sqrt{x})$

see post #3 here.

hence we have: $\sqrt {x^3} \sqrt{x} = x^{3/2}x^{1/2} = x^{3/2 + 1/2} = x^2$

**SnipedYou** · October 20th 2007, 07:41 PM

I like to look at it this way

$\sqrt{x^{3}} \cdot \sqrt{x}$

Then we get

$x \sqrt{x} \cdot \sqrt{x}$

Then we have: $x^{2}$

or you could do

$\sqrt{x^{3}} \cdot \sqrt{x}$

and get

$\sqrt{x^{4}}$

and simplify to get $x^{2}$

**Soroban** · October 21st 2007, 03:46 AM

Hello, t-dot

Yet another approach . . .

How is $x^2$ the answer to this question? . $(\sqrt{x^3})(\sqrt{x})$

We have: . $\sqrt{x^3}\cdot\sqrt{x} \;=\;\sqrt{x^3\!\cdot\! x} \;=\;\sqrt{x^4} \;=\;x^2$

**topsquark** · October 21st 2007, 04:54 AM

Originally Posted by Soroban

$\sqrt{x^3}\cdot\sqrt{x} \;=\;\sqrt{x^3\!\cdot\! x} \;=\;\sqrt{x^4} \;=\;x^2$

Hmmmm... This is true so long as x is not negative. If it is, then
$\sqrt{x^3}\sqrt{x} = (i\sqrt{-x^3})(i\sqrt{-x}) = i^2\sqrt{x^4} = -x^2$

-Dan

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