# Thread: Exponents and square roots (once more)

1. ## Exponents and square roots (once more)

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

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

2. Originally Posted by t-dot
how is x^2 the answer to this question??

$\displaystyle (\sqrt{x^3})(\sqrt{x})$
see post #3 here.

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

3. I like to look at it this way

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

Then we get

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

Then we have: $\displaystyle x^{2}$

or you could do

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

and get

$\displaystyle \sqrt{x^{4}}$

and simplify to get $\displaystyle x^{2}$

4. Hello, t-dot

Yet another approach . . .

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

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

5. Originally Posted by Soroban
$\displaystyle \sqrt{x^3}\cdot\sqrt{x} \;=\;\sqrt{x^3\!\cdot\! x} \;=\;\sqrt{x^4} \;=\;x^2$

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

-Dan