prove that x²+2xy+3y²≥0 for all x, y E R

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prove that x²+2xy+3y²≥0 for all x, y E R

Mathematics
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hint: Expand \((x+y)^2\)
i'm sorry i'm still lost
what can you say about the number \((x+y)^2\)? can it ever be negative?

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yes
how?
remember x, y E R
well i thought anything squared would be true bit i can't get the two brackets the same as i can't get 3y squared
try and first expand \((x+y)^2\) and then see what else you need to add to this in order to get your original expression of \(x^2+2xy+3y^2\)
so the answere would be (x+y)^2 + y^2
not quite - what do you get when you expand \((x+y)^2\) ?
(x+y)^2+2y^2
x^2 + 2xy + y ^2
correct - now what do you need to add to this in order to get to your original expression?
2y to be squared
perfect!
so we end up with:\[x^2+2xy+3y^2=(x+y)^2+2y^2\]now - can you see that this can never be negative?
ok thanks while i have you i have to do this using the same axsis and scales graph the functions f(x)=|x| and g(x)= |2x - 3|
yw :) could you please post that as a separate question. thanks.
using the same axsis and scales graph the functions f(x)=|x| and g(x)= |2x - 3|
<--- I meant post in the list to the left
oh ok sorry
np :) it just helps posting each question separately as others can then see each question more easily and also learn from them. :)

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