x^2+8xy+y^2 I just need the answer but really quick
What do you need..
do you want us to solve for x or y?...
Perhaps factor.
Using perfect square identity u just need to factorize it
Quick plz
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.
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
i know the method i dont have the answer correct as given @sarahusher @shoplovepc
@jdoe0001 youtube doesnt work in my country :( i wish u could explain me?
@Lukecrayonz can u help?
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!)
i have (x+y)^2 is it correct?? btw i use youtube by proxy :) but the links actually dont work
hmmm
not quite because: \[(x+y)^{2}=x ^{2}+y ^{2}+2xy\]
btw thx @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"
@sarahusher but when we rearrange that would be fine right?? so no problem??
@jdoe0001 y^2??
well... ahemm.... well.... not quite.... do you know what the middle term is supposed to be?
the middle term in a perfect square trinomial that is
when we apply perfect square identity wont it be 2(4)(xy) as when multiplied it would make 8xy?
yeap... so that will uncover stark naked the next term
Sorry I had to do something...looks like @jdoe0001 has got this!
anyways @jdoe0001 i gtg well that video was helpful bye:)
\(\large \begin{array}{llll} x^2+&8xy+&\square^2\\ \downarrow&&\downarrow\\ x&2x\square &\square\\ \end{array}\)
so we can say that \(\bf 8xy = 2x\square \implies \cfrac{8xy}{2x}=\square \)
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