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Thread: quadratic equation problem

  1. #1
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    quadratic equation problem

    Given that one of the root of the quadratic equation 2x(4x-9)=4-3p is half the other root.find:

    a)value of p

    b) root of the quadratic equation..


    anyone sir...please help!
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  2. #2
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    Quote Originally Posted by mastermin346 View Post
    Given that one of the root of the quadratic equation 2x(4x-9)=4-3p is half the other root.find:

    a)value of p

    b) root of the quadratic equation..


    anyone sir...please help!
    If α and β are the roots of a quadratic equation ax^2 + bx + c = 0, then
    the product of the roots α*β = c/a. and sum of the roots (α + β) = -(b/a)
    Using sum of the roots equation, find α. Using the product of roots equation, find p.
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  3. #3
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    Quote Originally Posted by mastermin346 View Post
    Given that one of the root of the quadratic equation 2x(4x-9)=4-3p is half the other root.find:

    a)value of p

    b) root of the quadratic equation..


    anyone sir...please help!
    Hi mastermin346,

    sa-ri-ga-ma's method is a fast way...

    the long way to do it is

    the equation in $\displaystyle ax^2+bx+c$ form is $\displaystyle 8x^2-18x+(3p-4)=0$

    One root is $\displaystyle \frac{18+\sqrt{18^2-32(3p-4)}}{16}$

    The other root is $\displaystyle \frac{18-\sqrt{18^2-32(3p-4)}}{16}$

    The larger root must be the first one if the roots are real.

    Then twice the second root is the first.

    $\displaystyle 2\left(18-\sqrt{324-32(3p-4)}\right)=18+\sqrt{324-32(3p-4)}$

    $\displaystyle 18=3\sqrt{324-32(3p-4)}$

    $\displaystyle \sqrt{324-32(3p-4)}=6$

    $\displaystyle 324-32(3p-4)=36$

    $\displaystyle 32(3p-4)=324-36=288$

    $\displaystyle 4(3p-4)=36$

    $\displaystyle 3p-4=9$

    $\displaystyle p=\frac{13}{3}$

    The equation is

    $\displaystyle 8x^2-18x+13-4=0$

    $\displaystyle 8x^2-18x+9=0$

    Either find the roots of this or substitute "p" into

    $\displaystyle \frac{18\pm\sqrt{324-32(9)}}{16}$
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