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    Someone help me with this...

    The equation kx^2 - (1 + k)x + (3k + 2) = 0 is such that the sum of its roots is twice their product. Find k and the two roots.
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    All quadratic equations can be written in the form x^2 + bx + c = 0 right? And let's say that quadratic has 2 solutions, p and q, then

    (x - p)(x - q) = 0 = x^2 + bx + c

    Expand the left hand side and compare it with the right hand side. What does b equal? What does c equal? So what can you say about the sum and product of the roots of a quadratic?
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    (Original post by Swayum)
    All quadratic equations can be written in the form x^2 + bx + c = 0 right? And let's say that quadratic has 2 solutions, p and q, then

    (x - p)(x - q) = 0 = x^2 + bx + c

    Expand the left hand side and compare it with the right hand side. What does b equal? What does c equal? So what can you say about the sum and product of the roots of a quadratic?
    I see that b = p + q and a = pq, but cannot understand how to relate this to the initial equation to find the roots.
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    b = -(p+q) and c = pq I think you mean.

    So you know that, for quadratics, the coefficient of x is the negative sum of the roots and the constant term is the product (if the coefficient of x^2 is 1).

    And you also know that "sum of its roots is twice their product".

    What equation can you write down?
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    Ok, thanks a lot, found a now

    a = -(p+q) = 2 (a-2) = 2pq

    a = 2a - 4

    4 = a
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    (Original post by Swayum)
    b = -(p+q) and c = pq I think you mean.

    So you know that, for quadratics, the coefficient of x is the negative sum of the roots and the constant term is the product (if the coefficient of x^2 is 1).

    And you also know that "sum of its roots is twice their product".

    What equation can you write down?
    actually i still do not get it...in my previous answer i assumed that the coefficient of x^2 (being a) will be equal to -(p + q).

    However, if b = -6 = -(p+q) and is twice that of the product: Pq = a -2

    then -12 = a -2
    and - 10 = a

    The answer to a should be 4 though! Please help!
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    This looks like an MEI FP1 question. MEI expects you to know 'sum of roots = -b/a' and 'product of roots = c/a' for ax^2 + bx + c = 0. So you construct an equation in k which is easy to solve.

    Plug that value for k into the original equation and you get a quadratic in x which factorises nicely. Clue - coefficients happen to sum to zero by chance. Solve for x and check that indeed sum = twice product.

    Swayum is quite right in what he says. The non-MEI way would be to divide through by the coefficient of x^2 first and work with a quadratic with just one x^2.
 
 
 
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