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Mathematics 17 Online
OpenStudy (insa):

x^2+8xy+y^2 I just need the answer but really quick

OpenStudy (lukecrayonz):

What do you need..

OpenStudy (anonymous):

do you want us to solve for x or y?...

OpenStudy (lukecrayonz):

Perhaps factor.

OpenStudy (insa):

Using perfect square identity u just need to factorize it

OpenStudy (insa):

Quick plz

OpenStudy (anonymous):

you aren't just meant to give answers on here. Sorry to be a party pooper but that's not the point of the site.

OpenStudy (anonymous):

true. no matter how stupid it sounds it will just lead to you falling behind in class. get some help from your teacher! :) @sarahusher @insa

OpenStudy (jdoe0001):

http://www.youtube.com/watch?v=xGOQYTo9AKY

OpenStudy (insa):

i know the method i dont have the answer correct as given @sarahusher @shoplovepc

OpenStudy (insa):

@jdoe0001 youtube doesnt work in my country :( i wish u could explain me?

OpenStudy (insa):

@Lukecrayonz can u help?

OpenStudy (anonymous):

okay why don't you show us your answer and we can help find where you went wrong? (I know the feeling not being able to use youtube!)

OpenStudy (insa):

i have (x+y)^2 is it correct?? btw i use youtube by proxy :) but the links actually dont work

OpenStudy (jdoe0001):

hmmm

OpenStudy (anonymous):

not quite because: \[(x+y)^{2}=x ^{2}+y ^{2}+2xy\]

OpenStudy (insa):

btw thx @jdoe0001

OpenStudy (jdoe0001):

\(\bf x^2+8xy+\square^2\) <--- neverminding the "y squared" for now, what value do you think we need there to get a "perfect square trinomial"

OpenStudy (insa):

@sarahusher but when we rearrange that would be fine right?? so no problem??

OpenStudy (insa):

@jdoe0001 y^2??

OpenStudy (jdoe0001):

well... ahemm.... well.... not quite.... do you know what the middle term is supposed to be?

OpenStudy (jdoe0001):

the middle term in a perfect square trinomial that is

OpenStudy (insa):

when we apply perfect square identity wont it be 2(4)(xy) as when multiplied it would make 8xy?

OpenStudy (jdoe0001):

yeap... so that will uncover stark naked the next term

OpenStudy (anonymous):

Sorry I had to do something...looks like @jdoe0001 has got this!

OpenStudy (insa):

anyways @jdoe0001 i gtg well that video was helpful bye:)

OpenStudy (jdoe0001):

\(\large \begin{array}{llll} x^2+&8xy+&\square^2\\ \downarrow&&\downarrow\\ x&2x\square &\square\\ \end{array}\)

OpenStudy (jdoe0001):

so we can say that \(\bf 8xy = 2x\square \implies \cfrac{8xy}{2x}=\square \)

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