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    Just looked through my book and it's all back in my head.

    Thanks!
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    (Original post by youngeN)
    Could someone please do the working out step by step to these two questions as I forgot everything I learnt in class.

    Calculate the value of the Discrimiant and hence state the number of real roots

    A).  x^2 +7x + 3 =0

    B).  3x^2 - 2x - 1 = 0
    No, that's not what we are here for.

    The discriminant is the part of the quadratic equation that is:

    b^2-4ac

    where a, b and c are the coefficients of the terms in the quadratic equation  ax^2+bx+c = 0 respectively.

    If the discriminant is bigger than zero, then there exists two real roots.

    If it is equal to zero, then there exists only one real root.

    Lastly if it is less than zero then no real roots exist, i.e they are imaginary.
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    (Original post by youngeN)
    Could someone please do the working out step by step to these two questions as I forgot everything I learnt in class.

    Calculate the value of the Discrimiant and hence state the number of real roots

    A).  x^2 +7x + 3 =0

    B).  3x^2 - 2x - 1 = 0
    All quadratic equations can be expressed in the form

    ax^2 + bx + c = 0

    So for A), a=1, b=7, c=3.

    The discriminant is equal to b^2 - 4ac.

    If the discriminant is 0, there is one real root (a repeated root). If it's greater than 0, there are two real roots. If it's less than 0, there are no real roots. Why? Check the quadratic formula. The discriminant is square rooted.

    You can't square root a negative number, so if the discriminant is a negative number, the whole thing doesn't work and there's no roots. (Unless you do Further Maths - then you'll discover soon that it's a little more complicated than that.)

    If the discriminant is 0, there's only one square root of 0, which is 0. If it's positive, there will be two square roots (e.g. \sqrt 1 = 1 AND \sqrt 1 = -1 - both work!).

    If you want more explanation just ask
 
 
 

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