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):

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 - abWhat 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