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sammy12

  • 4 years ago

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

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  1. ChrisV
    • 4 years ago
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    X+y=5

  2. sammy12
    • 4 years ago
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    please can you explain

  3. ChrisV
    • 4 years ago
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    im trying to remember how to do this one sec

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

  5. ChrisV
    • 4 years ago
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    dunno how you came up with that but no way that works

  6. sammy12
    • 4 years ago
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    Thanks

  7. ChrisV
    • 4 years ago
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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

  8. ChrisV
    • 4 years ago
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    so x+y=5

  9. myininaya
    • 4 years ago
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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 \[(x+y)^2=28 => x+y=\pm \sqrt{28}\]

  10. myininaya
    • 4 years ago
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    so i agree with lazy :)

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