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Thread: Simple questions regarding roots and equations

  1. #1
    Junior Member
    Oct 2007

    Smile Simple questions regarding roots and equations

    Okay, so polynomials, quadratic equations, roots, and all that fancy stuff has got to be my weakest point in math. I have a many questions that may seem stupid of me to ask but please bear with me. Help is oh so greatly appreciated. I am not looking for direct answers, but instead I'm looking for hints and guides as I would be glad to solve these myself.

    Firstly, my teacher gave me questions such as $\displaystyle y^2+{9\over4}=4y$ and $\displaystyle x^2+2x+1=0$. He has told me to find the number of real roots each equation has by not solving the equation. How do I do this exactly?

    Secondly, my teacher gave me this equation: $\displaystyle f(x)=x^2-6x+9$. He wants me to find the x and y intercepts, and which point they occur at. How do I do this?

    Thirdly, how to I find the maximum and minimum value for this equation:$\displaystyle f(x)=-18.8x^2+7.92x+6.18$?

    Fourthly, how do I construct an equation if this information is given: A quadratic function has (-1, 0) as one of its intercepts and (3, -5) as its vertex?

    Lastly, how do I show my steps for this: what is the maximum product of two numbers that add to -10? What numbers yield this product? I got -5 and -5 with a maximum of 25 through the process of guess and check, but I'm not so sure how to construct an equation based off of that.

    Thanks for the help and thanks for putting up with me. I don't really understand my teacher's explanations for things and as a student, I can't really pay for a private tutor. MHF is a great source of help. Thank you all.

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  2. #2
    MHF Contributor red_dog's Avatar
    Jun 2007
    Medgidia, Romania
    1) Let $\displaystyle ax^2+bx+c=0$

    The discriminant of the equation is $\displaystyle \Delta =b^2-4ac$

    If $\displaystyle \Delta>0$ then the equation has two distinct real roots.

    If $\displaystyle \Delta=0$ then the equation has two equal real roots.

    If $\displaystyle \Delta<0$ then the equation has no real roots.

    2) For x-intercept you have to solve the equation $\displaystyle f(x)=0$

    For y-intercept you have to calculate f(0).

    3) Let $\displaystyle f(x)=ax^2+bx+c$

    If $\displaystyle a>0$ then the function has a minimum equal to $\displaystyle -\frac{\Delta}{4a}$ for $\displaystyle x=-\frac{b}{2a}$

    If $\displaystyle a<0$ then the function has a maximum equal to $\displaystyle -\frac{\Delta}{4a}$ for $\displaystyle x=-\frac{b}{2a}$

    In this case $\displaystyle a<0$, so find the maximum.

    4) We have $\displaystyle f(-1)=0\Rightarrow a-b+c=0$

    The vertex has the coordinates $\displaystyle V\left(-\frac{b}{2a},-\frac{\Delta}{4a}\right)$

    Then $\displaystyle -\frac{b}{2a}=3$

    and $\displaystyle f(3)=-5\Rightarrow 9a-3b+c=-5$

    Now, solve the system $\displaystyle \left\{\begin{array}{lll}a-b+c=0\\b=-6a\\9a+3b+c=-5\end{array}\right.$

    5) Let $\displaystyle x+y=-10$ and $\displaystyle P=x\cdot y$

    Then $\displaystyle P=x(-x-10)=-x^2-10x$

    The maximum of P is the maximum of the function $\displaystyle f(x)=-x^2-10x$ (see 3)
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  3. #3
    Feb 2009
    completing the square also tells you a whole load of info on the quadratic.

    $\displaystyle ax^2+bx+c\equiv p\left(x+q\right)^2+r$

    p tells you shape of parabola if > 0 it has mimima $\displaystyle \cup$ < 0 its maxima $\displaystyle \cap$

    x=-q is the line of symmetry

    (-q,r) are the coordinates of the vertex.

    you can also use $\displaystyle p\left(x+q\right)^2+r=0$ to solve the root(s) if the discriminant >= 0

    and setting $\displaystyle p\left(x+q\right)^2=0$ solves x for the vertex.

    look at what happens when you complete the square of

    $\displaystyle ax^2+bx+c$

    $\displaystyle a\left[x^2+\frac{b}{a}x+\frac{c}{a}\right]$

    $\displaystyle a\left[\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}{a}\right]\equiv a\left[\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2}\right]$

    $\displaystyle a\left(x+\frac{b}{2a}\right)^2 = \frac{a\left(b^2-4ac\right)}{4a^2}\equiv\frac{b^2-4ac}{4a}$

    $\displaystyle \left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$

    $\displaystyle x+\frac{b}{2a}= \pm\frac{\sqrt{b^2-4ac}}{2a}$

    $\displaystyle x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

    i just learned all this neat stuff in school!
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  4. #4
    Junior Member
    Oct 2007
    Wow, big, big thanks to the both of you for clearing things up! I understand a lot better now.

    I GREATLY, GREATLY appreciate it! (If I could press the thank you button more than once, I would!)
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