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Math Help - Solving quadratic equations

  1. #1
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    Solving quadratic equations

    Hi all,

    I am unsure if this is the really right topic to do this but since it sort of involves algebra so I'm trying my luck.

    I have not touched these for at 5 years (had to serve in the army) and now I am needed to touch on maths again.

    I would prefer not to state what/how I think because I think it'll waste your time and I am really poor in maths. Even the videos on quadratic equations on youtube couldn't help me solve these problems. These questions should be of entry-level so I was hoping I could get solutions because my tutorial classes have not started and I am trying out the questions on my own.


    Q1) Find the range of values of the constants of m for which the equation has distinct real roots. 2x^2 - 3x + m = 0

    Answer (highlight to see) : m < 9/8

    Q2) The graph y = 2x + c - x^2 lies entirely below the x-axis. Find the range of values of c.

    Answer (highlight to see) : c < -1

    Q3) In the diagram, the line y = mx is tangetial to the curve y^2 = x - c. Show that c = 1 / 4m^2.



    Thank you
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  2. #2
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    Re: Solving quadratic equations

    Quote Originally Posted by transcendentx View Post
    Hi all,

    I am unsure if this is the really right topic to do this but since it sort of involves algebra so I'm trying my luck.

    I have not touched these for at 5 years (had to serve in the army) and now I am needed to touch on maths again.

    I would prefer not to state what/how I think because I think it'll waste your time and I am really poor in maths. Even the videos on quadratic equations on youtube couldn't help me solve these problems. These questions should be of entry-level so I was hoping I could get solutions because my tutorial classes have not started and I am trying out the questions on my own.


    Q1) Find the range of values of the constants of m for which the equation has distinct real roots. 2x^2 - 3x + m = 0

    Answer (highlight to see) : m < 9/8
    Use the quadratic formula:

    x = \dfrac{3 \pm \sqrt{9-4 \cdot 2 \cdot m}}4

    To get real roots the discriminant must be equal or greater than zero:

    9-4 \cdot 2 \cdot m \ge 0

    solve for m.

    Q2) The graph y = 2x + c - x^2 lies entirely below the x-axis. Find the range of values of c.

    Answer (highlight to see) : c < -1
    The graph of the function is a parabola opening downward. Choose c such that the vertex lies below the x-axis.

    y = -x^2+2x+c
    y = -(x^2-2x+1)+1+c
    y = -(x-1)^2+1+c

    That means the vertex is at V(1, 1+c)

    According to the question the y-coordinate of the vertex must be smaller than zero:

    1+c<0

    solve for c.

    Q3) In the diagram, the line y = mx is tangetial to the curve y^2 = x - c. Show that c = 1 / 4m^2.



    Thank you
    Calculate the coordinates of the point of intersection between the straight line and the curve:

    (mx)^2 = x-c

    Solve for x.

    You only get one single point of intersection (that's the tangent point!) if the discriminant equals zero. Solve for c.
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  3. #3
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    [SOLVED] Solving quadratic equations

    thank you. on a side note, for the first question, can i assume x = 0?
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  4. #4
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    Re: [SOLVED] Solving quadratic equations

    Quote Originally Posted by transcendentx View Post
    thank you. on a side note, for the first question, can i assume x = 0? <--- in general: no
    Why that? The value m = 0 is possible but it is one special value among an unlimited number of possible values.
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  5. #5
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    Re: Solving quadratic equations

    Well.. Actually even after you gave me a headstart, I'm still stuck. I read up on the basics of algebra and quadratic equations once again to refresh my memory on the laws but still could't get the solution. The video tutorials i found online weren't that complex as these questions. They were quite straight forward and can be solved easily. I think I'm missing something more basic which I needed to understand before tackling these problems but I just don't know what.



    x = \dfrac{3 \pm \sqrt{9-4 \cdot 2 \cdot m}}4

    To get real roots the discriminant must be equal or greater than zero:

    I don't understand how if i used the formula, where did x on the left side of the equation go? and how did the \ge symbol came out of now where?

    9-4 \cdot 2 \cdot m \ge 0

    solve for m.

    9-8 \cdot m \ge 0
    -8 \cdot m \ge -9
    m \le \dfrac{-9}{8}

    I'm always stuck with an wrong answer.



    The graph of the function is a parabola opening downward. Choose c such that the vertex lies below the x-axis.

    y = -x^2+2x+c
    y = -(x^2-2x+1)+1+c <- i understand the part where you factorize the minus sign out. but how did you manage to add this .. +1)+1 .. in?
    y = -(x-1)^2+1+c [COLOR="red"]

    That means the vertex is at V(1, 1+c)

    According to the question the y-coordinate of the vertex must be smaller than zero:

    1+c<0

    solve for c.



    Calculate the coordinates of the point of intersection between the straight line and the curve:

    (mx)^2 = x-c

    Solve for x.

    You only get one single point of intersection (that's the tangent point!) if the discriminant equals zero. Solve for c.

    okay i'm totally lost on this question. i think i'll need to wait for my tutorial class for an detailed explanation.
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  6. #6
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    Re: Solving quadratic equations

    Quote Originally Posted by transcendentx View Post
    Well.. Actually even after you gave me a headstart, I'm still stuck. I read up on the basics of algebra and quadratic equations once again to refresh my memory on the laws but still could't get the solution. The video tutorials i found online weren't that complex as these questions. They were quite straight forward and can be solved easily. I think I'm missing something more basic which I needed to understand before tackling these problems but I just don't know what.



    x = \dfrac{3 \pm \sqrt{9-4 \cdot 2 \cdot m}}4

    To get real roots the discriminant must be equal or greater than zero:

    [COLOR="red"]I don't understand how if i used the formula, where did x on the left side of the equation go?
    It doesn't go anywhere. You wanted to determine the value of x. This is called solving an equation.
    and how did the \ge symbol came out of now where?
    You only get real solutions for x if the value under the root-sign is positive or zero. Greater or equal to zero is translated as " \ge"

    9-4 \cdot 2 \cdot m \ge 0

    solve for m.

    9-8 \cdot m \ge 0
    -8 \cdot m \ge -9
    m \le \dfrac{-9}{-8}\le \frac98 <--- sign error

    I'm always stuck with an wrong answer.



    The graph of the function is a parabola opening downward. Choose c such that the vertex lies below the x-axis.

    y = -x^2+2x+c
    y = -(x^2-2x+1)+1+c <- i understand the part where you factorize the minus sign out. but how did you manage to add this .. +1)+1 .. in? <-- actually I added zero, so no damage has been done
    y = -(x-1)^2+1+c [COLOR="red"]

    The whole procedure is called "Completing the square". Google for it
    That means the vertex is at V(1, 1+c)

    According to the question the y-coordinate of the vertex must be smaller than zero:

    1+c<0

    solve for c.



    Calculate the coordinates of the point of intersection between the straight line and the curve:

    (mx)^2 = x-c

    Solve for x. <--- That means apply the quadratic formula and go back to the first question. It's just the same procedure.

    You only get one single point of intersection (that's the tangent point!) if the discriminant equals zero. Solve for c.

    okay i'm totally lost on this question. i think i'll need to wait for my tutorial class for an detailed explanation.
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