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Mathematics 8 Online
OpenStudy (anonymous):

(2+3sq) (3-3sq)

OpenStudy (anonymous):

3-3sq

OpenStudy (anonymous):

I think you missed something when you foiled. I have: \[-9s ^{2}q ^{2}-6sq+6\]

OpenStudy (anonymous):

Is it \[(2+3^2)(3-3^2)\]

OpenStudy (anonymous):

Or are s and q variables?

OpenStudy (anonymous):

Now that you mention it, I think you're right mkuehn

OpenStudy (anonymous):

Used to looking for complicated problems ;)

OpenStudy (anonymous):

It could reallly be either, but sq seems a bit too "convenient" :)

OpenStudy (anonymous):

the 3 have this sign\[\sqrt{3}\]

OpenStudy (anonymous):

waht is that

OpenStudy (anonymous):

That changes things :)

OpenStudy (anonymous):

That is the square root. If you take the square root of a number, the answer you get is a number times itself that equals that number. For example, the square root of 4 is equal to 2 because 2 x 2 = 4

OpenStudy (anonymous):

\[(2+\sqrt{3})(3-\sqrt{3})\]

OpenStudy (anonymous):

\[\sqrt{3}+3\]

OpenStudy (anonymous):

yea like that

OpenStudy (anonymous):

When you foil you end up with: \[6-2\sqrt{3}+3\sqrt{3}-3\]

OpenStudy (anonymous):

idk how you got 6

OpenStudy (anonymous):

It is 2 times 3.

OpenStudy (anonymous):

The -3 came from: \[-\sqrt{3}*\sqrt{3}=-3\]

OpenStudy (anonymous):

ok goy u

OpenStudy (anonymous):

got it .plz keep going

OpenStudy (anonymous):

You just combine like terms. In this case combine the (6 - 3) and the 3sqrt(3) - 2sqrt(3)

OpenStudy (anonymous):

So you are left with 3 + 3sqrt(3)

OpenStudy (anonymous):

The answer is above the expansion if you scroll up. I combined like terms. Think of \[\sqrt{3}\] as an x when combining like terms.

OpenStudy (anonymous):

3\[\sqrt{27xysqz}\]

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