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Math Help - Intersections of 2 lines

  1. #1
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    Intersections of 2 lines

    Ok in this question, I can't draw graphs or use a calculator (I'm really wishing I could) but I need to find the intersection(s) of y=\sqrt{x} and y=-x+2

    I keep getting (1,4) or (1,2) but the answer is actually (1,1) so I'm not sure how to get to it?
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  2. #2
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    In the problem, the equations are
    y = sqrt(x) Which can be written as
    y^2 = x.
    Substitute this value in the second equation. You get
    y = -y^2 + 2
    Or
    y^2 + y - 2 = 0
    If you factorize yoy get
    ( y - 1 )( y + 2 ) = 0
    Or y = -2 or y = 1.
    Since y = sqrt(x), y cannot be negative.
    So y = 1 and x = 1.
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  3. #3
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    How did you get "(1, 4) or (1, 2)"? If x= 1 then y= -x+ 2= -1+ 2= 1, not either 4 or 2. Similarly, if x= 1, y= \sqrt{x}= \sqrt{1}= 1.

    I would have done that a slightly different way. Since y= \sqrt{x} and y= -x+ 2, \sqrt{x}= -x+ 2 (because they are both equal to y). Now square both sides. x= (-x+ 2)^2= x^2- 4x+ 4 so that x^2- 5x+ 4= 0. Notice that that is an equation for x where sa-ri-ga-ma got an equation for y.

    x^2- 5x+ 4= (x- 4)(x- 1)= 0 gives x= 4 and x= 1. If x= 4, y= -x+ 2= -4+ 2= -2. But y= \sqrt{x} cannot be negative so that is NOT a solution. If x= 1, y= -1+ 2= 1. The graphs intersect at (1, 1).
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  4. #4
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    :D

    HallsofIvy had prove the answer right.
    Just wanna show you the graph of it for ADDITIONAL PROVING.

    Note the the line of function y=sqrt(x) OR y= \sqrt{x} is "cut-off" at the y=0, it's because y can't be less than zero due to y=\sqrt{x}.

    The graph shows that it really only have 1 point of intersection.

    Hope it helps.

    Intersections of 2 lines-prove.jpg
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