Radical Expressions

[SARAHrawr]

(4root3)(2root6) I dont remember how to multipy radical expressions.

 

[mathandpolitics]

Multiply the insides, multiply the outsides, then simplify if possible.

[4(root3)]*[2(root6)] =

8(root18) =

8*3(root2) =

24(root2) answer

 

[BigGlenntheHeavy]

\(\displaystyle 4th \ root \ of \ 3 \ times \ the \ 2nd \ root \of \ 6 \ = \ (3^{1/4})(6^{1/2})\)

$\displaystyle = \ (3^{1/4})(3^{1/2})(2^{1/2}) \ = \ (3^{1/4})(3^{2/4})(2^{1/2}) \ = \ (3^{3/4})(2^{1/2}) \ = \ 3^{3/4}\sqrt2.$

 

[mmm4444bot]

These are square roots, yes?

Here's how we type square roots: sqrt(x).

4 sqrt(3) * 2 sqrt(6)

We can rearrange this, using The Commutative Property of Multiplication

4 * 2 * sqrt(3) * sqrt(6)

You know what 4 times 2 is.

8 * sqrt(3) * sqrt(6)

There is a property for multiplying square roots.

sqrt(x) * sqrt(y) = sqrt(xy)

In other words, we multiply the radicands together.

8 * sqrt(3*6)

8 * sqrt(18)

Now we simplify the radical sqrt(18) by looking for squared factors of 18.

18 is 3^2 times 2

8 * sqrt(9*2)

Using the property above, this is the same as 8 * sqrt(9) * sqrt(2)

8 * 3 * sqrt(2)

24 sqrt(2) is the simplified product.

 

[SARAHrawr]

omg thank you I get it now

 

[lookagain]

(4root3)(2root6) I dont remember how to multipy radical expressions.

You can split up $\displaystyle \sqrt{6}$ into the product of two other square roots in simplifying it:


$\displaystyle (4 \sqrt{3})(2 \sqrt{6}) =$

$\displaystyle (4\sqrt {3})(2 \sqrt{3} \sqrt{2}) =$

$\displaystyle 4(2)(\sqrt{3})(\sqrt{3})\sqrt{2} =$

$\displaystyle 8(3)\sqrt{2} =$

$\displaystyle \boxed{24 \sqrt{2}}$

 

[Denis]

$\displaystyle (4 \sqrt{3})(2 \sqrt{6}) =$

$\displaystyle (4\sqrt {3})(2 \sqrt{3} \sqrt{2}) =$

$\displaystyle 4(2)(\sqrt{3})(\sqrt{3})\sqrt{2} =$

1st step should be:
= 8SQRT(3)SQRT(6) ; then:
= 8SQRT(3)SQRT(3)SQRT(2)

 

[lookagain]

$\displaystyle (4 \sqrt{3})(2 \sqrt{6}) =$

$\displaystyle (4\sqrt {3})(2 \sqrt{3} \sqrt{2}) =$

$\displaystyle 4(2)(\sqrt{3})(\sqrt{3})\sqrt{2} =$

"1st step should be:
= 8SQRT(3)SQRT(6) ; then:
= 8SQRT(3)SQRT(3)SQRT(2)

Wrong. The first step may be/might be what is posted in this quote box,
but mine retains the original order/order of operations at the outset.

 

[Denis]

Disagree: you way is SUBPAR! Go remove your post :wink:
(Ted: just having fun!)

 

[lookagain]

Agree: I am SUBPAR! I should remove my post. :wink: minus :wink: $\displaystyle \ = \ \ no \ wink \ for \ you$
\(\displaystyle (\text{Ted: just} \ \\)\(\displaystyle \ \ \ \not{h} \not {a} \not{v} \not{i} \not {n} \not{g} \ \not {f}\not{u} \not{n} \ \ \text{causing trouble by instigating!)}\)

 

[Denis]

Agree: I am SUBPAR! I should remove my post. :wink: minus :wink: $\displaystyle \ = \ \ no \ wink \ for \ you$
\(\displaystyle (\text{Ted: just} \ \\)\(\displaystyle \ \ \ \not{h} \not {a} \not{v} \not{i} \not {n} \not{g} \ \not {f}\not{u} \not{n} \ \ \text{causing trouble by instigating!)}\)

Looks like you're still in kindergarTen.