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Thread: Quadratic Application

  1. #1
    Raj
    Raj is offline
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    Quadratic Application

    1.) The sum of a number and its reciprocal is $\displaystyle \frac{29}{10}$. Determine the integers.

    So i figure, let $\displaystyle x$ be a number,
    let $\displaystyle \frac{1}{x}$ be the reciprocal.

    Now i got the equation:

    $\displaystyle
    \frac{x}{1}+\frac{1}{x}=\frac{29}{10}
    $

    From here i'm not sure what the LCD is in order to put it into a quad. equation. Then ill just be able to solve for x.


    2.) The quadratic function $\displaystyle y=9x^2+bx+c$ has no x-intercepts but the zeros of the function are $\displaystyle \frac{7 \pm \sqrt{-6}}{3}$. Determine the values of b and c.

    What are "the zeros"? I dont know where to start with this one so if your in the mood to help explain im all ears
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Raj View Post
    1.) The sum of a number and its reciprocal is $\displaystyle \frac{29}{10}$. Determine the integers.

    So i figure, let $\displaystyle x$ be a number,
    let $\displaystyle \frac{1}{x}$ be the reciprocal.

    Now i got the equation:

    $\displaystyle
    \frac{x}{1}+\frac{1}{x}=\frac{29}{10}
    $

    From here i'm not sure what the LCD is in order to put it into a quad. equation. Then ill just be able to solve for x.
    Multily through by $\displaystyle x$ and rearrange:

    $\displaystyle x^2-\frac{29}{10}x+1=0$

    Solve this using the quadratic formula.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Raj View Post

    2.) The quadratic function $\displaystyle y=9x^2+bx+c$ has no x-intercepts but the zeros of the function are $\displaystyle \frac{7 \pm \sqrt{-6}}{3}$. Determine the values of b and c.

    What are "the zeros"? I dont know where to start with this one so if your in the mood to help explain im all ears
    These are the complex roots and so:

    $\displaystyle
    c/9= \frac{7 + \sqrt{-6}}{3}\ \times \ \frac{7 - \sqrt{-6}}{3}
    $

    and:

    $\displaystyle
    b/9 = \frac{7 + \sqrt{-6}}{3}\ + \ \frac{7 - \sqrt{-6}}{3}
    $

    RonL
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