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Math Help - Finding the Quadratic Equation

  1. #1
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    Post Finding the Quadratic Equation

    Hello people ,

    How do i solve these?

    1) Find the quadratic equation whose roots are reciprocals of the roots of x^2-7x+12=0

    2) Find the quadratic equation whose roots are double the roots of 3x^2-4x+7=0
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  2. #2
    Super Member Matt Westwood's Avatar
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    1) Either by:

    a) Solving the equation, taking the reciprocals of the roots (a and b, say), and forming the quadratic equation
    \left({x - \frac 1 a}\right)\left({x - \frac 1 b}\right)
    but that's long-winded.

    Alternatively note that for a q.e. x^2 - px + q, if the roots are a and b, we have:
    a + b = p, ab = q.

    Divide a + b = p by ab = q to get:

    \frac {a+b}{ab} = \frac 1 b + \frac 1 a = \frac p q

    and then note that \frac 1 {ab} = \frac 1 q

    So your equation is then:
    x^2 - \frac p q x + \frac 1 q
    and then multiply by q to get the answer.

    Plug in your numbers and you're done.
    Last edited by Matt Westwood; August 31st 2009 at 08:24 AM. Reason: corrected latex
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  3. #3
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    Quote Originally Posted by saberteeth View Post
    Hello people ,

    How do i solve these?

    1) Find the quadratic equation whose roots are reciprocals of the roots of x^2-7x+12=0

    2) Find the quadratic equation whose roots are double the roots of 3x^2-4x+7=0
    for (1) let a and b be the roots of the equation. Then a+b = 7 and ab = 12. The general procedure is:
    The quadratic with reciprocals of the roots will have sum of roots as \frac1{a} + \frac1{b} = \frac{a+b}{ab} = \frac7{12} and product of roots as \frac1{ab} = \frac1{12}

    You could have also observed 4 and 3 are roots. Thus the required quadratic is obtained by expanding \left(x - \frac1{3}\right)\left(x - \frac14 \right)
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by Matt Westwood View Post
    and then note that \frac 1 {ab} = \frac 1 q

    So your equation is then:
    x^2 - \frac p q x + \frac 1 {pq}
    I think the equation should read x^2 - \frac p q x + \frac 1 {q}
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  5. #5
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by Isomorphism View Post
    I think the equation should read x^2 - \frac p q x + \frac 1 {q}
    Yes it should - I just corrected it. Slip of the brain. Sorry.
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  6. #6
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    Post solution to quadratic equation

    1) Find the quadratic equation whose roots are reciprocals of the roots of

    we find the roots by factorisation

    12 = (-4) *(-3) -7 = -4 - 3
    we split the middle term

    x^2 -7x +12 = 0
    x^2 -4x -3x +12 = 0
    x(x-4) -3 (x-4) = 0
    (x-3) (x-4) = 0

    so the roots are x-3 = 0 gives x =3
    and x-4 = 0 gives x =4

    Roots are 3,4

    Reciprocal of the roots are 1/3,1/4

    Quadratic equation: x^2 - (sum of roots) x + (product of roots) = 0
    Sum of roots =1/3 +1/4 = 4/12 +3/12 = 7/12
    product of roots = 1/3 * 1/4 = 1/12

    so the quadratic equation is x^2 - 7/12 x + 1/12 = 0
    12 x^2 -7x + 1 = 0

    2) Find the quadratic equation whose roots are double the roots of

    3x^2 -4x +7 = 0

    a = 3 , b =-4 and c =7

    sum of roots = - b / a
    product of roots = c/a

    sum of roots = 4/3
    let one root is alpha and the other root is beta

    that is alpha + beta = 4/3
    alpha * beta = 7/3

    for the quadratic equation,
    now the roots are doubled means 2* alpha and 2 * beta
    are the roots

    sum of roots = (2 * alpha) + (2 * beta) = 2(alpha + beta) = 2 * 4/3 = 8/3

    product of roots = (2*alpha)(2*beta) = 4 (alpha * beta) = 4 * 7/3 = 28/3

    so the quadratic equation is x^2 - 8/3 x + 28/3 =0
    3 x^2 - 8x + 28 = 0
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