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    1. The roots of the equation mx2 + nx + 1 = 0 are -1 and 3. Find the values of m and n, substitute these values into the equation, and arrange the equation tidily, i.e. with integer coefficients.

    Can someone explain how m = -1/3 and n = 2/3?

    (x+1)(x-3) = x^2 - 2x - 3 = 0
    I get m = 1, n = -2 ??????
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    (Original post by ckfeister)
    1. The roots of the equation mx2 + nx + 1 = 0 are -1 and 3. Find the values of m and n, substitute these values into the equation, and arrange the equation tidily, i.e. with integer coefficients.

    Can someone explain how m = -1/3 and n = 2/3?

    (x+1)(x-3) = x^2 - 2x - 3 = 0
    I get m = 1, n = -2 ??????
    Divide your answer by -3 to match the constant then compare coefficients.
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    (Original post by RDKGames)
    Divide your answer by -3 to match the constant then compare coefficients.
    Do you always divide by the biggest number or?
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    (Original post by ckfeister)
    Do you always divide by the biggest number or?
    What? No...

    The reason for division by -3 is that then your constant is the same in both mx^2+nx+1=0 and -\frac{1}{3}x^2+\frac{2}{3}x+1=0 so then you can compare the coefficients directly. If you were to compare them before the division then your constant would tell you that -3=1 which is clearly non sense.

    Easier way you could've done this question in simply plug in x=-1 and x=3 into mx^2+nx+1=0 then solve simultaneous equations.
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    (Original post by RDKGames)
    What? No...

    The reason for division by -3 is that then your constant is the same in both mx^2+nx+1=0 and -\frac{1}{3}x^2+\frac{2}{3}x+1=0 so then you can compare the coefficients directly. If you were to compare them before the division then your constant would tell you that -3=1 which is clearly non sense.

    Easier way you could've done this question in simply plug in x=-1 and x=3 into mx^2+nx+1=0 then solve simultaneous equations.
    What stupid mistake am I doing this time..

    2. The equation 3x2 + 8x - 1 = 0 has roots α and β. Without solving this equation, find a quadratic equation whose roots are 2α and 2β, and arrange the equation with integer coefficients.

    αβ= - 1/3
    α + β= -8/3

    2(αβ) = -1/3 * 2 = -2/3
    2(α + β) 2(-8/3) = -`16/3
    x2 - 16/3x - 2/3 = 0
    3x2 - 16x - 2 = 0

    Answer is: 3x2 + 16x - 4 = 0 ????????/
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    (Original post by ckfeister)
    What stupid mistake am I doing this time..

    2. The equation 3x2 + 8x - 1 = 0 has roots α and β. Without solving this equation, find a quadratic equation whose roots are 2α and 2β, and arrange the equation with integer coefficients.

    αβ= - 1/3
    α + β= -8/3

    2(αβ) = -1/3 * 2 = -2/3
    2(α + β) 2(-8/3) = -`16/3
    x2 - 16/3x - 2/3 = 0
    3x2 - 16x - 2 = 0

    Answer is: 3x2 + 16x - 4 = 0 ????????/
    Two mistakes. 2\alpha \times 2 \beta \neq 2 \alpha \beta and the coefficient of x should be -\frac{b}{a}.
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    (Original post by Mr M)
    Two mistakes. 2\alpha \times 2 \beta \neq 2 \alpha \beta and the coefficient of x should be -\frac{b}{a}.
    Nevermind, I get it now thanks.
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    (Original post by ckfeister)
    How do I find alpha and beta? The course on cloudlearn just leaves you in the dark...
    You don't need to. This is the point. You can find associated roots without solving the equations.

    You know \alpha+\beta = -\frac{8}{3} and \alpha \beta = -\frac{1}{3}.

    The sum of the roots of the new quadratic is twice the original sum but the product is four times the original product.

    Now use your rules for sum and product to form the new quadratic. Note the sign changes for the coefficient of x. Multiply by 3 to obtain integers.
 
 
 
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