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Math Help - Graph Help

  1. #1
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    Graph Help

    1)Solve the following quadratic equations

    a)x2 + 4x + 3 =

    b) x2 - 5x + 6

    c) x2 + 2x – 3

    d)x2 – 2x – 8



    2)Give the values of m and c

    a)y = 2x + 3

    b) y = 4x – 2


    c)y = 2x – 5

    d)y = 1/2x – 8

    e) 3y = 6x – 9

    f) 1/2y = 2x + 3

    These are the answers I came up for question no #2. Comparing the given equations with the standard format of
    y = mx + c

    ------------------------------------------------------------
    a)
    y = 2x + 3

    m = 2 and c = 3

    --------------------------------------------------------------
    b)
    y = 4x - 2

    m = 4 and c = -2

    ---------------------------------------------------------------
    c)
    y = 2x - 5

    m = 2 and c = -5

    --------------------------------------------------------------
    d)
    y = 1/2x - 8

    m = 1/2 and c = -8

    ---------------------------------------------------------------
    e)
    3y = 6x - 9

    rearranging y = 2x -3

    m = 2 and c = -3

    ---------------------------------------------------------------
    f)
    1/2y = 2x + 3

    rearranging y = 4x + 6

    m = 4 and c = 6

    -----------------------------------------------------------------
    Last edited by Soul; July 29th 2009 at 07:26 PM.
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  2. #2
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    Quote Originally Posted by Soul View Post
    1)Solve the following quadratic equations

    a)x2 + 4x + 3

    b) x2 - 5x + 6

    c) x2 + 2x – 3

    d)x2 – 2x – 8

    2)Give the values of m and c

    a)y = 2x + 3

    b) y = 4x – 2

    c)y = 2x – 5

    d)y = 1/2x – 8

    e) 3y = 6x – 9

    f) 1/2y = 2x + 3
    By "solve" the first set do you mean FACTOR?

    And I'm slightly confused on how you have your examples set up. Is it set up the way I have you quoted?
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  3. #3
    Member SENTINEL4's Avatar
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    Quote Originally Posted by Soul View Post
    1)Solve the following quadratic equations

    a)x2 + 4x + 3
    x^2+4x+3=0 It's in this form ax^2+bx+c=0 in your problem it's a=1 , b=4 , c=3.
    First you find b^2-4*a*c which here is D = 4^2-4*1*3=16-12=4 and then you use \frac{-b+\sqrt{D}}{2a} to find the first solution and \frac{-b-\sqrt{D}}{2a} to find the second.
    Here the solutions are : \frac{-4+2}{2}=-1 and \frac{-4-2}{2}=-3

    Try the rest...
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  4. #4
    Member SENTINEL4's Avatar
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    Quote Originally Posted by Soul View Post
    2)Give the values of m and c

    a)y = 2x + 3b) y = 4x – 2c)y = 2x – 5

    a)y = 1/2x – 8e) 3y = 6x – 9f) 1/2y = 2x + 3
    What do you mean to give the values of m and c ???
    What is m and c ?
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  5. #5
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    Quote Originally Posted by SENTINEL4 View Post
    What do you mean to give the values of m and c ???
    What is m and c ?
    I suspect the OP means to compare them to the formula for a straight line - often denoted as y = mx + c

    In relation to post 3 if the discriminant is a perfect square that means there are rational roots and you should be able to factor that equation. For example x^2+4x+3 = (x+3)(x+1)
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  6. #6
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    Quote Originally Posted by Soul View Post
    1)Solve the following quadratic equations

    a)x2 + 4x + 3
    b) x2 - 5x + 6
    c) x2 + 2x – 3
    b)x2 – 2x – 8
    These are expressions; having no "equals" signs in them, they cannot be "equations". Please reply with clarification.

    Quote Originally Posted by Soul View Post
    2)Give the values of m and c

    a)y = 2x + 3
    b) y = 4x – 2
    c)y = 2x – 5
    a)y = 1/2x – 8
    e) 3y = 6x – 9
    f) 1/2y = 2x + 3
    For the first four, the equations are already written in the form "y = mx + c". Copy off the values of "m" and "c".

    For the last two, solve for "y=", and then copy off the values.
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  7. #7
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    Oh that makes sense. Well if these equations are set to zero than an easier method (if you are comfortable factoring) is to do so:

    x^2+4x+3 factors to (x+3)(x+1). And thus the resultant equation:

    (x+3)(x+1)=0 we can solve by taking any of the two terms and setting it individually equal to zero:

    (x+3)=0\\(x+1)=0

    Using the quadratic equation also works, but might be a bit of a waste on equations that factor nicely.

    On the second set, "Give the values of M and C" I'm assuming refers to:

    y=mx+b;

    where M is the slope of our equation and b is our y-intercept and the equation is the slope-intercept form of a line.

    Thusly, arranging each of the equations a-f and putting them in slope-intercept form will yield our M and C.
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