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Math Help - Algebra with pq-formula

  1. #1
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    Algebra with pq-formula

    Hello! I am currently having a homework for tomorrow with pretty advanced tasks, for me atleast, that I should solve. I have though got stuck on some of them but there is one in perticular that i have no idea how to solve and I was hoping that you might help me out with this. I start with writing the task.
    (It is in Swedish so I am going to try to translate it the best i can)

    a) Show that if an Quadratic equation has the roots x_{1}=a  x_{2}=b then the equation is x^2-(a+b)x+ab=0

    b) Solve the Equation above and show that is has the solutions x_{1}=a  x_{2}=b


    I am quite sure that you need to use the pq-formula for this one but then you get x=\frac{(a+b)}{2} \pm \sqrt{-(\frac{(a+b)}{2})^2}-ab

    This in my eyes does not say much.
    If someone is willing to help me out here I'm very glad!
    Thanks in advance!
    //Rickard Liljeros

    PS. I didn't put this in urgent help section since I maybe after this problem will ask for some other algebra help and thought it would be unnecessary to start a new thread for each one.
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  2. #2
    A riddle wrapped in an enigma
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    If x_1=a is a root, then x-a is a factor.

    If x_2=b is a root, then x-b is a factor.

    Therefore,

    (x-a)(x-b)=x(x-b)-a(x-b)=x^2-bx-ax+ab=0

    Simplifying (factoring out x in 2nd and 3rd terms):

    x^2-(a+b)x+ab=0
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  3. #3
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    Quote Originally Posted by masters View Post
    If x_1=a is a root, then x-a is a factor.

    If x_2=b is a root, then x-b is a factor.

    Therefore,

    (x-a)(x-b)=x(x-b)-a(x-b)=x^2-bx-ax+ab=0

    Simplifying (factoring out x in 2nd and 3rd terms):

    x^2-(a+b)x+ab=0

    "If x_1=a is a root, then x-a is a factor.

    If x_2=b is a root, then x-b is a factor."

    How come it is like this? Is it a rule of some kind?
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  4. #4
    A riddle wrapped in an enigma
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    You have probably learned that a zero of a function f(x) is any value c such that f(c)=0. When the function is graphed, the real zeros of the function are the x-intercepts of the graph.

    If f(x)=a_nx^n+...+a_1x+a_0 is a polynomial function, then

    c is a zero of the polynomial function f(x),

    x-c is a factor of the polynomial function f(x), and

    c is a root or solution of the polynomial equation f(x)=0.
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  5. #5
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    Quote Originally Posted by masters View Post
    You have probably learned that a zero of a function f(x) is any value c such that f(c)=0. When the function is graphed, the real zeros of the function are the x-intercepts of the graph.

    If f(x)=a_nx^n+...+a_1x+a_0 is a polynomial function, then

    c is a zero of the polynomial function f(x),

    x-c is a factor of the polynomial function f(x), and

    c is a root or solution of the polynomial equation f(x)=0.

    I have not learnd this yet, going on the first year in high school. But I think i maybe understand this.
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  6. #6
    A riddle wrapped in an enigma
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    Quote Originally Posted by liljeros View Post
    a) Show that if an Quadratic equation has the roots x_{1}=a x_{2}=b then the equation is x^2-(a+b)x+ab=0

    b) Solve the Equation above and show that is has the solutions x_{1}=a x_{2}=b

    I am quite sure that you need to use the pq-formula for this one but then you get x=\frac{(a+b)}{2} \pm \sqrt{-(\frac{(a+b)}{2})^2}-ab
    If you continue with your "pq" formula, you will arrive at the same conclusion; that x_1=a \ and \ x_2=b

    This is a much longer process, but if that's the way you have to do it, then just continue with what you already started.
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  7. #7
    A riddle wrapped in an enigma
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    x=\frac{a+b}{2}\pm\sqrt{(\frac{a+b}{2})^2-ab}
    ---------------------------------------------------------------
    x_1=\frac{a+b}{2} + \sqrt{\frac{a^2+2ab+b^2}{4}-ab}

    x_1=\frac{a+b}{2}+\sqrt{\frac{a^2+2ab+b^2}{4}-\frac{4ab}{4}}

    x_1=\frac{a+b}{2}+\sqrt{\frac{a^2+2ab+b^2-4ab}{4}}

    x_1=\frac{a+b}{2}+\sqrt{\frac{a^2-2ab+b^2}{4}}

    x_1=\frac{a+b}{2}+\frac{a-b}{2}

    x_1=\frac{a+b+a-b}{2}

    x_1=\frac{2a}{2}

    x_1=a

    Now, perform a similar task to find x_2
    Last edited by masters; May 22nd 2008 at 11:15 AM.
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  8. #8
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    Quote Originally Posted by masters View Post
    x=\frac{a+b}{2}\pm\sqrt{(\frac{a+b}{2})^2-ab}
    ---------------------------------------------------------------
    x_1=\frac{a+b}{2} + \sqrt{\frac{a^2+2ab+b^2}{4}}-ab

    x_1=\frac{a+b}{2}+\sqrt{\frac{a^2+2ab+b^2}{4}}-\frac{4ab}{4}

    x_1=\frac{a+b}{2}+\sqrt{\frac{a^2+2ab+b^2-4ab}{4}}

    x_1=\frac{a+b}{2}+\sqrt{\frac{a^2-2ab+b^2}{4}}

    x_1=\frac{a+b}{2}+\frac{a-b}{2}

    x_1=\frac{a+b+a-b}{2}

    x_1=\frac{2a}{2}

    x_1=a

    Now, perform a similar task to find x_2
    You really are the master! A friend to me explained the thing you were talking about before and it seems like I have learnd that and it helped me alot! Thank you *2

    Very kind of you to help me out!
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  9. #9
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    Quote Originally Posted by masters View Post
    x=\frac{a+b}{2}\pm\sqrt{(\frac{a+b}{2})^2-ab}
    ---------------------------------------------------------------
    x_1=\frac{a+b}{2} + \sqrt{\frac{a^2+2ab+b^2}{4}}-ab

    x_1=\frac{a+b}{2}+\sqrt{\frac{a^2+2ab+b^2}{4}}-\frac{4ab}{4}
    Just to be clear there is a typo in the last two of the first three lines of masters' post. (Probably because of the same typo in the original poster's statement.) They should read:
    x=\frac{a+b}{2}\pm\sqrt{(\frac{a+b}{2})^2-ab}

    x_1=\frac{a+b}{2} + \sqrt{\frac{a^2+2ab+b^2}{4}-ab}

    x_1=\frac{a+b}{2}+\sqrt{\frac{a^2+2ab+b^2}{4}-\frac{4ab}{4}}

    -Dan
    Last edited by topsquark; May 22nd 2008 at 11:25 AM.
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  10. #10
    A riddle wrapped in an enigma
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    Thanks, Dan. I fixed it.
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