Mathematics
42 Online
OpenStudy (anonymous):
(square root of 2)+x)(square root of 8)-x)
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
Foil it out.
OpenStudy (anonymous):
i dont no how to foil it
OpenStudy (anonymous):
i dont no how to foil it
OpenStudy (anonymous):
is the answer 4+2 times the square root of x -x^2
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
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
i dont no how to foil it
OpenStudy (anonymous):
i no how to do that part but i dont no how multiply square roots by a normal number
OpenStudy (anonymous):
Then you multiply them.
\[\sqrt{2}\sqrt{8} = \sqrt{2\cdot 8} = \sqrt{16} = 4\]
\[\sqrt{8} = \sqrt{4\cdot 2} = 2\sqrt{2}\]\[\implies \sqrt{8}x - \sqrt{2}x = 2\sqrt{2}x - \sqrt{2}x = \sqrt{2}x\]
\[x\cdot -x = -x^2\]
OpenStudy (anonymous):
i dont no how to foil it
OpenStudy (anonymous):
so wats the answer
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
I showed you how to foil it, and gave you the solution.
OpenStudy (anonymous):
i dont no how to foil it
OpenStudy (anonymous):
im sorry i dont understand
OpenStudy (anonymous):
To foil it, you sum the product of the First terms, the inner terms, the outer terms, and the last terms.
OpenStudy (anonymous):
i dont no how to foil it
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
\[i got up \to 4-\sqrt{2x}+2\sqrt{2x}x^{2}\]
OpenStudy (anonymous):
i dont no how to foil it
OpenStudy (anonymous):
sorry minus x^2 at the end
OpenStudy (anonymous):
Ok so now from the two middle terms you can factor out a \((\sqrt{2})x\)
OpenStudy (anonymous):
sry idk wat ur tellin me 2 do is this it
\[ 4-2x \times 2-x^{2}\]
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
Nope
OpenStudy (anonymous):
i dont no how to foil it
OpenStudy (anonymous):
sry lol
OpenStudy (anonymous):
Ok, lets see here.
OpenStudy (anonymous):
Let's ignore the first and last terms They're 4 and -x^2. You have those right
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
So all we care about for now is this:
\[(2\sqrt{2})x - (\sqrt{2})x\]
OpenStudy (anonymous):
Right?
OpenStudy (anonymous):
yep
OpenStudy (anonymous):
Ok lets try this:\[\]2ab - ab\[\]What is that the same as?
OpenStudy (anonymous):
Not sure?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
i dont no how to foil it
OpenStudy (anonymous):
idk is it \[2\sqrt{2x}-2x\]
OpenStudy (anonymous):
Hrm. Maybe this will work better.\[\]Let \(\sqrt{2}x = oranges\)\[\]Therefore \[2(\sqrt{2}x) - \sqrt{2}x\]\[=2(\sqrt{2}x) - 1(\sqrt{2}x)\]\[=2oranges - 1oranges\]\[= ?\]
OpenStudy (anonymous):
i dont no how to foil it
OpenStudy (anonymous):
1 orange
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
Right!
OpenStudy (anonymous):
And what is oranges again (we defined it above)
OpenStudy (anonymous):
i dont no how to foil it
OpenStudy (anonymous):
square root of 2x
OpenStudy (anonymous):
(sqrt 2)x
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
But yes
OpenStudy (anonymous):
So therefore\[2(\sqrt{2})x - 1(\sqrt{2})x = 1(\sqrt{2})x\]
OpenStudy (anonymous):
i dont no how to foil it
OpenStudy (anonymous):
omg yes it does!!!!!
OpenStudy (anonymous):
So that's the form of the middle term
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
is that it were the outside terms correct
OpenStudy (anonymous):
Now we put back in your first and last terms, and we have:
\[4 + 2(\sqrt{2})x - (\sqrt{2})x - x^2\]\[=4 + (\sqrt{2})x - x^2\]
OpenStudy (anonymous):
i dont no how to foil it
OpenStudy (anonymous):
thnx soooooooooo much ur like a genius
OpenStudy (anonymous):
Just takes practice. Nothing special about me ;P
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
Happy to help though
OpenStudy (anonymous):
alright well thnx 4 the help but i gtg and do more math
OpenStudy (anonymous):
Have fun =)