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Math Help - further maths quadratics and roots again

  1. #1
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    further maths quadratics and roots again

    Getting used to these equations now just stuck on this one.

    The equation 2x - 7x + 4 = 0 has roots a and b. find the quadratic equation with integer coefficients and with roots 1/a , 1/b

    i have the sum = 7/2 and the product = 2

    pleeease help
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  2. #2
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    Hello, djmccabie!

    The equation 2x^2 - 7x + 4 \:= \:0 has roots a and b.
    Find the quadratic equation with integer coefficients and with roots: . \frac{1}{a^2},\;\;\frac{1}{b^2}

    i have the sum = 7/2 and the product = 2 . . . . Good!
    Looking ahead, we want the equation with: . \begin{array}{ccccc}\text{Sum} &=& \dfrac{1}{a^2} +\dfrac{1}{b^2} &=& \dfrac{a^2+b^2}{a^2b^2}\\ \\[-3mm] \text{Product} &=& \dfrac{1}{a^2}\cdot\dfrac{1}{b^2} &=&\dfrac{1}{a^2b^2} \end{array}


    We have: . \begin{array}{cccc}a + b &=&\frac{7}{2} & {\color{blue}[1]} \\ ab &=& 2 & {\color{blue}[2]} \end{array}

    Square [1]: . (a + b)^2 \:=\:\left(\frac{7}{2}\right)^2 \quad\Rightarrow\quad a^2 + 2ab + b^2 \:=\:\frac{49}{4}

    . . \text{and we have: }\;(a^2 + b^2) + 2\underbrace{(ab)}_{2} \:=\:\frac{49}{4} \quad\Rightarrow\quad \boxed{a^2+b^2 \:=\:\frac{33}{4}}

    Square [2]: . (ab)^2 \:=\:2^2 \quad\Rightarrow\quad \boxed{a^2b^2 \:=\:4}


    . . \text{Sum} \;=\;\frac{a^2+b^2}{a^2b^2} \;=\;\frac{\frac{33}{4}}{4} \;=\;\frac{33}{16}

    . . \text{Product} \;=\;\frac{1}{a^2b^2} \;=\;\frac{1}{4}


    The equation is: . x^2 - \frac{33}{16}x + \frac{1}{4} \:=\:0 \quad\Rightarrow\quad \boxed{16x^2 - 33x + 4 \:=\:0}

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  3. #3
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    Thanks a lot! i guess you enjoy going through this website lol great help
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