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From Kaplan Diag. test equation

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isisalaska Community Manager Default Avatar
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From Kaplan Diag. test equation

Post Wed Mar 21, 2007 6:45 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    For the equation x^2 + 2x + m = 5, where m is a constant, 3 is one solution for x. What is the other solution?

    a) -5
    b) -2
    c) -1
    d0 3
    E) 5

    What do you do? do you subtiutute 3 where x is? I thought it was 5...

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    Neo2000 Legendary Member
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    Post Wed Mar 21, 2007 7:50 am
    Yup, you substitute 3. See, a solution for X implies that X=3 will satisfy the given equation.

    You get 9 +6 +m = 5
    or m = -10

    so you finally get x^2 +2x -10 = 5
    x^2 +2x-15 = 0
    (X+5)(X-3) = 0

    so the solution is X = -5 or X = 3

    Option A

    Cybermusings Legendary Member Default Avatar
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    Post Tue Mar 27, 2007 9:55 am
    For the equation x^2 + 2x + m = 5, where m is a constant, 3 is one solution for x. What is the other solution?

    a) -5
    b) -2
    c) -1
    d0 3
    E) 5

    x^2 + 2x + m = 5

    3 is one solution for x

    Thus the quadratic equation has to look like

    x^2 + 2x + m - 5 = 0

    For 3 to be one root it has to look like this

    x^2 - 3x + x + m - 5 = 0

    Therefore even m - 5 = -3

    Therefore m = -3 + 5 = 2

    Thus x^2 - 3x + x + 2 - 5

    = x ^2 - 3x + x - 3

    = x(x-3) + 1(x-3)

    Thus the other root of the equation is -1

    I aint too good at explaining but hope this helps

    Post Fri Sep 29, 2017 2:30 pm
    since 3 is one of the solutions of the quadratic equation, we will represent the whole solution with a letter say 'y'.
    x=3 or x=y
    and x-3=0 or x-y=0
    (x-3)(x-y)=0
    expanding the equation above, we have
    x(x-y) - 3(x-y) = 0
    x^2 - xy-3x+ 3y =0
    x^2 - (y+3)x + 3y =0
    Comparing this newly derived form of the equation with the original question, we observe that:
    -(y+3)=2 ------------(1)
    -y-3=2
    -y=2+3=5
    -y=5
    y=-5
    Hence, we obtained the other solution of the equation as required. option A is correct

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