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Math Help - Moar Quadratics!

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    Moar Quadratics!

    1. Find an equation for the straight lines which pass through the point (1, -2) and are tangent to the parabola with equation y = x2.

    Would my two equations be y = x2 and -2 = m + c? Where do I go from here? Usually I would make equation 1 equal equation 2, but y is not equal to -2.

    2. Solve the following for x by completing the square:

    2x2 - 4x + 1 = 0

    What I tried was:

    2(x2 - (4/2)x + (1/2)) =0

    (x - 1)2 + 1 + (1/2) - 1 = 0

    (x - 1)2 - (1/2) = 0

    x - 1 = sqrt(1/2) = 0

    x = 1 + sqrt(1/2)

    Thanks in advance.
    Last edited by Fratricide; April 21st 2013 at 12:24 AM.
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    Re: Moar Quadratics!

    Let's work the first problem before moving to the second. Using the point-slope formula, with unknown slope m, what is the family of lines which pass through the given point?
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    Re: Moar Quadratics!

    y - y1 = m(x - x1)

    y + 2 = m(x - 1)
    y = mx - m + 2
    y = m(x - 1) + 2
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    Re: Moar Quadratics!

    Yes, this is correct:

    y-(-2)=m(x-1)

    and so:

    y+2=m(x-1)

    This is what you correctly found,

    Now subtract 2 from both sides to solve for y...I think you just got a little careless when you did this.
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    Re: Moar Quadratics!

    Yep, I see it.

    y = m(x - 1) - 2

    Would we then do something like...?

    x2 = m(x - 1) - 2

    -x2 + m(x - 1) - 2 = 0

    If we were to solve this by equating the discriminant to 0, would the b constant of said discriminant be?
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    Re: Moar Quadratics!

    First, you want to write the quadratic in standard form ax^2+bx+c=0. I recommend distributing the m, and then multiplying through by -1. What do you get?
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    Re: Moar Quadratics!

    x2 - mx + m + 2 = 0
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    MHF Contributor MarkFL's Avatar
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    Re: Moar Quadratics!

    Yes, so equate the discriminant to zero and solve for m...what do you find?
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    Re: Moar Quadratics!

    What is the b value of the discriminant? Is it m?
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    Re: Moar Quadratics!

    It is -m actually, but this won't affect the computation of the discriminant, but it is good to recognize that it we have:

    (1)x^2+(-m)x+(m+2)=0

    This way we can easily see that:

    a=1,\,b=-m,\,c=m+2.
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    Re: Moar Quadratics!

    Ah, I see. I'll complete this equation later and post my results.
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    Re: Moar Quadratics!

    m2 - 4m - 8 = 0

    Therefore m = 2 +- 2sqrt3.

    So do we then just substitute this m value back into y = mx - m - 2?
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  13. #13
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    Re: Moar Quadratics!

    Yes, this will give you the two lines that satisfying the given conditions:

    y=x^2,y=(2+2sqrt(3))(x-1)-2,y=(2-2sqrt(3))(x-1)-2 where x=-2 to 4 - Wolfram|Alpha
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    Re: Moar Quadratics!

    Awesome.

    Where did I go wrong with the following equation?

    2(x2 - (4/2)x + (1/2)) =0

    (x - 1)2 + 1 + (1/2) - 1 = 0

    (x - 1)2 - (1/2) = 0

    x - 1 = sqrt(1/2) = 0

    x = 1 + sqrt(1/2)
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  15. #15
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    Re: Moar Quadratics!

    When you took the square root of both sides, you did not consider the negative root.

    If x^2=y then x=\pm\sqrt{y}
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