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Mathematics 22 Online
OpenStudy (anonymous):

Multiply (sqrt2)(-2sqrt2 +3)

OpenStudy (anonymous):

\[(\sqrt2)(-2\sqrt2 +3)\]

OpenStudy (anonymous):

-4+sqrt(2)*3

OpenStudy (anonymous):

approx equivalent to 0.243

OpenStudy (anonymous):

\[so \it should look like -4+\sqrt(2)*3 ?\]

OpenStudy (anonymous):

yeah, would you like some more detail?

OpenStudy (anonymous):

If you could please =)

OpenStudy (anonymous):

& should the sqrt go all the way across, or just across the (2)?

OpenStudy (anonymous):

Just distribute it out: \[\sqrt{2}(-2\sqrt{2} + 3)\]\[=-2(\sqrt{2})(\sqrt{2}) + 3\sqrt{2}\] \[=-2(\sqrt{2(2)}) + 3 \sqrt{2}\]\[=-2(\sqrt{4}) + 3\sqrt{2}\]\[=-2(2) + 3\sqrt{2}\]\[= -4 + 3\sqrt{2}\]

OpenStudy (anonymous):

cool, step one is distribution so you've got sqrt(2) * (-2) * sqrt(2)

OpenStudy (anonymous):

sqrt(2) * sqrt(2) = 2 so the first distribution is is equal to -4, agreed?

OpenStudy (anonymous):

second distribution is sqrt(2) * 3, this unfortunately can't be simplified any further :-(

OpenStudy (anonymous):

agreed!

OpenStudy (anonymous):

so final answer is -4 + sort(2)*3

OpenStudy (anonymous):

why are yours and Polpak's answers different?

OpenStudy (anonymous):

Because he put the 3 on the end, rather than on the beginning.

OpenStudy (anonymous):

\(3\sqrt{2} = 3*\sqrt{2} = \sqrt{2}* 3\)

OpenStudy (anonymous):

multiplication is commutative.

OpenStudy (anonymous):

I think the law is called commutative? A*B = B*A

OpenStudy (anonymous):

Nice Polpak, good call!

OpenStudy (anonymous):

so it's suppose to be -4+3sqrt2 ?

OpenStudy (anonymous):

'suppose to be' is sorta whatever you want. It's the same thing.

OpenStudy (anonymous):

it can be either, good luck. give me a shout if you have another question :-)

OpenStudy (curry):

stion polpak i need your help take a look at my question

OpenStudy (anonymous):

you can write it a lot of different ways. If it were me, I'd write it as: \[3\sqrt{2} - 4\] Since I prefer to have a positive leading coefficient whenever possible.

OpenStudy (anonymous):

But it's all the same number.

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