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Math Help - Quadratic Application

  1. #1
    Raj
    Raj is offline
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    Quadratic Application

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

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

    Now i got the equation:

     <br />
\frac{x}{1}+\frac{1}{x}=\frac{29}{10}<br />

    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 y=9x^2+bx+c has no x-intercepts but the zeros of the function are \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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    Red face

    Quote Originally Posted by Raj View Post
    1.) The sum of a number and its reciprocal is \frac{29}{10}. Determine the integers.

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

    Now i got the equation:

     <br />
\frac{x}{1}+\frac{1}{x}=\frac{29}{10}<br />

    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 x and rearrange:

    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 y=9x^2+bx+c has no x-intercepts but the zeros of the function are \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:

    <br />
c/9= \frac{7 + \sqrt{-6}}{3}\ \times \ \frac{7 - \sqrt{-6}}{3}<br />

    and:

    <br />
b/9 = \frac{7 + \sqrt{-6}}{3}\ + \ \frac{7 - \sqrt{-6}}{3}<br />

    RonL
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