Question

Put the following radical expression into simplified form. Assume the variable represents a positive number.

Answer 1

\[\frac{ 4\sqrt{12x^4} }{ \sqrt{3x} }\]

Answer 2

I don't know if I did it right...
I factored \[\sqrt{12x^4}=\sqrt{4}\sqrt{3}=2\sqrt{3}\]
Does the x^2 remain in the radical or outside?

Answer 3

Outside

Answer 4

Then I multiplied it by the 4 & got...
\[4*2\sqrt{3}=8\sqrt{3}\]

Answer 5

There are several ways in which you could carry out this simplification.

You're off to a good start. To answer your question, that x^2, the square root of x^4, would be written outside.

Answer 6

so far, so good!

Answer 7

Do i multiply the top & bottom by
\[\sqrt{3x}\] to get rid of the bottom square root?

Answer 8

Here's a slightly different approach:\[\frac{ 4\sqrt{12x^4} }{ \sqrt{3x} }=\frac{ 4\sqrt{3}\sqrt{4} \sqrt{x^4}}{ \sqrt{3}\sqrt{x}}\]

Answer 9

Clearly you can cancel that Sqrt(3) in numerator and denominator, and reduce Sqrt(4) to 2. What do you think would be the next step?

Answer 10

How would you reduce\[\frac{ 4\sqrt{3}\sqrt{4} \sqrt{x^4}}{ \sqrt{3}\sqrt{x}}=\frac{ 8\sqrt{3}x^2 }{ \sqrt{3}x ^{1/2} }\]

Answer 11

\[\frac{ 8x^2\sqrt{3} }{ 3x }\]
Or does the 3 in the denominator remain in a radical?

Answer 12

Cookie: Cancel the Sqrt(3) that appears in both numerator and denominator:\[\frac{ 8\sqrt{3}x^2 }{ \sqrt{3}x ^{1/2} }\] and then use the rule \[\frac{ x^a }{ x^b }=x ^{a-b}\]

Answer 13

How do i cancel the top?

Answer 14

@Miracrown

Answer 15

\[\frac{ 8\sqrt{3}x^2 }{ \sqrt{3}x ^{1/2} }\] becomes \[\frac{ 8x^2 }{ x ^{1/2} }\]after we've cancelled that Sqrt(3). Then,\[\frac{ x^2 }{ x ^{1/2} }\]

Answer 16

may be re-written as \[x ^{2-\frac{ 1 }{ 2 }}=?\]

Answer 17

and so the final answer is\[8x ^{2-\frac{ 1 }{ 2 }}=?\]

Answer 18

Great working with you! But i need to get off the 'Net now. Good night!

Answer 19

Goodnight

Answer 20

8x^3/2