Question

simplifying this radical √12x3y8

Answer 1

\[\huge \sqrt{\color{#3366CF}{12}\color{#3399AA}{x^3}\color{#662FFF}{y^8}}=\sqrt{\color{#3366CF}{12}}\cdot \sqrt{\color{#3399AA}{x^3}}\cdot \sqrt{\color{#662FFF}{y^8}}\]Hmm so if we break it up, we can deal with each piece separately.
Let's look at the 12 first.

Answer 2

6X2= 12

Answer 3

We want a FACTOR of 12 that is a perfect square. Perfects squares are as such, 4, 9, 16, 25, 36, ...\[\large \sqrt4=2\]See how we can take the square root and get a nice even number?
Hmm Those are factors of 12 yes, but neither 6 nor 2 is a perfect square.

Answer 4

Can you think of some other factors of 12? c:

Answer 5

2X5 / 4X3

Answer 6

4x3, yah let's try that c:
\[\huge \sqrt{\color{#3366CF}{12}}\qquad =\qquad \sqrt{\color{#3366CF}{4}}\cdot \sqrt{\color{#3366CF}{3}}\qquad =\qquad \color{#3366CF}{2}\sqrt{\color{#3366CF}{3}}\]

Answer 7

Understand what we did there? :o

Answer 8

mhmm 2X2=4 and nothing goes into 3 ....so yeah i undertstand

Answer 9

That simplifies the 12, that's as far as we can break it down.
So we're 1/3 done c:

Answer 10

For the X term, you'll want to remember how to rewrite a radical using FRACTIONS in the exponent.

Here is how our X term would change,\[\huge \sqrt{\color{#3399AA}{x^3}} \qquad = \qquad (\color{#3399AA}{x^3})^{1/2}\]

Answer 11

my teacher was talking about that but i didnt get it

Answer 12

\[\huge \sqrt{\color{#3399AA}{x^3}} \qquad = \qquad (\color{#3399AA}{x^3})^{1/2}\qquad = \qquad \color{#3399AA}{x^{3/2}}\]

Answer 13

Hmm ok lemme see if I can think of a simple example :O