OpenStudy (anonymous):

(square root of 2)+x)(square root of 8)-x)

7 years ago
OpenStudy (anonymous):

Foil it out.

7 years ago
OpenStudy (anonymous):

i dont no how to foil it

7 years ago
OpenStudy (anonymous):

i dont no how to foil it

7 years ago
OpenStudy (anonymous):

is the answer 4+2 times the square root of x -x^2

7 years ago
OpenStudy (anonymous):

\[(\sqrt{2} + x)(\sqrt{8} -x)\]\[=(\sqrt{2})(\sqrt{8})\ \ + \ \ (\sqrt{8})x \ + \ \ \ (\sqrt{2})(-x)\ \ \ \ +\ \ \ (x)(-x)\] ^First terms ^Inner terms ^Outer terms ^Last terms

7 years ago
OpenStudy (anonymous):

i dont no how to foil it

7 years ago
OpenStudy (anonymous):

i no how to do that part but i dont no how multiply square roots by a normal number

7 years ago
OpenStudy (anonymous):

Then you multiply them. \[\sqrt{2}\sqrt{8} = \sqrt{2\cdot 8} = \sqrt{16} = 4\] \[\sqrt{8} = \sqrt{4\cdot 2} = 2\sqrt{2}\]\[\implies \sqrt{8}x - \sqrt{2}x = 2\sqrt{2}x - \sqrt{2}x = \sqrt{2}x\] \[x\cdot -x = -x^2\]

7 years ago
OpenStudy (anonymous):

i dont no how to foil it

7 years ago
OpenStudy (anonymous):

so wats the answer

7 years ago
OpenStudy (anonymous):

I showed you how to foil it, and gave you the solution.

7 years ago
OpenStudy (anonymous):

i dont no how to foil it

7 years ago
OpenStudy (anonymous):

im sorry i dont understand

7 years ago
OpenStudy (anonymous):

To foil it, you sum the product of the First terms, the inner terms, the outer terms, and the last terms.

7 years ago
OpenStudy (anonymous):

i dont no how to foil it

7 years ago
OpenStudy (anonymous):

\[i got up \to 4-\sqrt{2x}+2\sqrt{2x}x^{2}\]

7 years ago
OpenStudy (anonymous):

i dont no how to foil it

7 years ago
OpenStudy (anonymous):

sorry minus x^2 at the end

7 years ago
OpenStudy (anonymous):

Ok so now from the two middle terms you can factor out a \((\sqrt{2})x\)

7 years ago
OpenStudy (anonymous):

sry idk wat ur tellin me 2 do is this it \[ 4-2x \times 2-x^{2}\]

7 years ago
OpenStudy (anonymous):

Nope

7 years ago
OpenStudy (anonymous):

i dont no how to foil it

7 years ago
OpenStudy (anonymous):

sry lol

7 years ago
OpenStudy (anonymous):

Ok, lets see here.

7 years ago
OpenStudy (anonymous):

Let's ignore the first and last terms They're 4 and -x^2. You have those right

7 years ago
OpenStudy (anonymous):

So all we care about for now is this: \[(2\sqrt{2})x - (\sqrt{2})x\]

7 years ago
OpenStudy (anonymous):

Right?

7 years ago
OpenStudy (anonymous):

yep

7 years ago
OpenStudy (anonymous):

Ok lets try this:\[\]2ab - ab\[\]What is that the same as?

7 years ago
OpenStudy (anonymous):

Not sure?

7 years ago
OpenStudy (anonymous):

i dont no how to foil it

7 years ago
OpenStudy (anonymous):

idk is it \[2\sqrt{2x}-2x\]

7 years ago
OpenStudy (anonymous):

Hrm. Maybe this will work better.\[\]Let \(\sqrt{2}x = oranges\)\[\]Therefore \[2(\sqrt{2}x) - \sqrt{2}x\]\[=2(\sqrt{2}x) - 1(\sqrt{2}x)\]\[=2oranges - 1oranges\]\[= ?\]

7 years ago
OpenStudy (anonymous):

i dont no how to foil it

7 years ago
OpenStudy (anonymous):

1 orange

7 years ago
OpenStudy (anonymous):

Right!

7 years ago
OpenStudy (anonymous):

And what is oranges again (we defined it above)

7 years ago
OpenStudy (anonymous):

i dont no how to foil it

7 years ago
OpenStudy (anonymous):

square root of 2x

7 years ago
OpenStudy (anonymous):

(sqrt 2)x

7 years ago
OpenStudy (anonymous):

But yes

7 years ago
OpenStudy (anonymous):

So therefore\[2(\sqrt{2})x - 1(\sqrt{2})x = 1(\sqrt{2})x\]

7 years ago
OpenStudy (anonymous):

i dont no how to foil it

7 years ago
OpenStudy (anonymous):

omg yes it does!!!!!

7 years ago
OpenStudy (anonymous):

So that's the form of the middle term

7 years ago
OpenStudy (anonymous):

is that it were the outside terms correct

7 years ago
OpenStudy (anonymous):

Now we put back in your first and last terms, and we have: \[4 + 2(\sqrt{2})x - (\sqrt{2})x - x^2\]\[=4 + (\sqrt{2})x - x^2\]

7 years ago
OpenStudy (anonymous):

i dont no how to foil it

7 years ago
OpenStudy (anonymous):

thnx soooooooooo much ur like a genius

7 years ago
OpenStudy (anonymous):

Just takes practice. Nothing special about me ;P

7 years ago
OpenStudy (anonymous):

Happy to help though

7 years ago
OpenStudy (anonymous):

alright well thnx 4 the help but i gtg and do more math

7 years ago
OpenStudy (anonymous):

Have fun =)

7 years ago
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