sammy12
If xy = 6 and x^2 + y^2 = 16, then what is the value of (x + y)?



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ChrisV
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X+y=5

sammy12
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please can you explain

ChrisV
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im trying to remember how to do this one sec

lazypig
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\[x ^{2}+y ^{2}+2xy =16+6\times2\]
\[\left( x +y \right)^{2}=28\]
\[x +y =\pm \sqrt{28}\]

ChrisV
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dunno how you came up with that but no way that works

sammy12
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Thanks

ChrisV
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its obvious the values of x and y are 2 and 3 but i cannot for the life of me remember how to explain it

ChrisV
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so x+y=5

myininaya
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\[(x+y)^2=x^2+2xy+y^2=x^2+y^2+2(xy)=16+2(6)=16+12=28\]
so we have
\[(x+y)^2=28 => x+y=\pm \sqrt{28}\]

myininaya
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so i agree with lazy :)