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Math Help - Two equations and 3 variables

  1. #1
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    Two equations and 3 variables

    Another problem from the GRE program. Determine which option is greater:

    x = v + 2
    y = 2v

    Option A: x^2 + y^2
    Option B: (x+y)^2

    I decided that B is greater because it results in: x^2 + y^2 + 2xy. So I guess that this 2xy makes the possitive difference. However, I have tried it out with v=1 and v=-1 and both of them are ok with my result, B is greater. However, in the corrections it says that it canīt be determined which one is greater. Could anyone help me to know why??

    Thank you so much!
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  2. #2
    Super Member Deadstar's Avatar
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    Check your working again.

    If v = -1.

    x = 1,
    y = -2

    So 2xy = -4

    Hence option A gives
    1^2 + (-2)^2 = 1 + 4 = 5
    Option B gives
    1^2 + (-2)^2 + (-4) = 5 - 4 = 1

    What happens when -2 < v < 0?
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  3. #3
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    Hello, artabro!

    Determine which option is greater: . \begin{array}{ccc}x \;=\;v+2 \\ y \:=\:2v \end{array}

    . . \begin{array}{cccc}\text{Option A:} & x^2+y^2 \\ \text{Option B:} & (x+y)^2 \end{array}

    I decided that B is greater because it results in: x^2 + y^2 + 2xy
    So I guess that this 2xy makes the positive difference.
    How do we know this is a positive difference?

    Option A: . x^2 + y^2 \;=\;(v+2)^2 + (2v)^2 \;=\;5v^2 + 4v + 4

    Option B: . (x+y)^2 \;=\;(v+2+2v)^2 \-=\;(3v+2)^2 \;=\;9v^2 + 12v + 4


    How do they compare?

    . . . . . \text{Option B} \;\;\begin{array}{c}> \\[-2mm] < \end{array}\;\;\text{Option A}

    . . 9v^2 + 12v + 4 \;\;\begin{array}{c}> \\[-2mm]<\end{array}\;\;5v^2 + 4v + 4

    . . . . . 4v^2 + 8v \;\;\begin{array}{c}> \\[-2mm]<\end{array}\;\;0

    . . . . . 4v(v+2)\;\;\begin{array}{c}>\\[-2mm] <\end{array}\;\;0


    \text{Option B }\: > \:\text{ Option A }\:\text{ if }\,v < \text{-}2\,\text{ or }\,v > 0

    \text{Option B }\:<\:\text{ Option A }\:\text{ if }\,\text{-}2 < v < 0


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