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Math Help - Checking my work on this inequality

  1. #1
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    Checking my work on this inequality

    My answer disagrees with the solution given in the book. I want to make sure I'm right.

    x^2 <= y^2 +XY - 1 If X=3, find the value(s) of y that satisfy the equation.


    The given answer is -5 <= y <= 2, but that's not right, because for y=1,


    9 <= 1 + 3 - 1 is clearly not true.


    My solution is:

    9 <= y^2 +3y -1

    0 <= y^2 +3y -10

    0<= (y + 5) (y - 2)

    Y+5=0 y = -5

    y-2=0 y =2

    Are the roots and also solutions.


    For y < -5, the product of (y+5) (y-2) (try -6) is +, therefore it's part of the solution.

    For -5<y<2, the product of (y+5)(y-2) (try 1) is -, therefore it's not part of the solution.

    For y>2, the product of (y+5)(y-2) (try 3) is positive, therefore it's part of the solution.

    y <= -5
    y >= 2
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  2. #2
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    Quote Originally Posted by fcabanski View Post
    x^2 <= y^2 +XY - 1 If X=3, find the value(s) of y that satisfy the equation.
    The given answer is -5 <= y <= 2, but that's not right, because for y=1,
    My solution is:
    9 <= y^2 +3y -1
    0 <= y^2 +3y -10
    0<= (y + 5) (y - 2)
    y <= -5
    y >= 2
    This solution is correct to the problem as you have written it.
    The book’s answer is correct if the problem were x^2 {\color{red}\ge} y^2+xy-1.
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  3. #3
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    Thanks for the verification.
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