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    I am stuck on this equation and would like some help please. Apparently it can be solved without using the quadratic formula:

    30x^+49x+20=0
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    The equation factorises into 2 linear factors. From there values of x can be easily determined. As the discriminant, b^2 -4ac > 0 we know that there are 2 distinct real roots to this equation.
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    You need to split 30x into it's factors such as 3 and 10, 6 and 5, etc. and also split 20 into it's factors such as 2 and 10, 5 and 4. Then try combinations of those until you get the answer. E.g. (3x+5)(10x+4) clearly doesn't work.
    Be assured that it does fully factorise and can be easily solved once you've factorised it no need to worry about negative numbers since the entire equation is positive.
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    (Original post by mollyxrose)
    You need to split 30x into it's factors such as 3 and 10, 6 and 5, etc. and also split 20 into it's factors such as 2 and 10, 5 and 4. Then try combinations of those until you get the answer. E.g. (3x+5)(10x+4) clearly doesn't work.
    Be assured that it does fully factorise and can be easily solved once you've factorised it no need to worry about negative numbers since the entire equation is positive.
    Thanks for taking the trouble to respond. I was hoping for a method that might make the process of factorising more efficient. However I followed your advice and eventually came up with the factors and solved the equation.
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    (Original post by Ihol)
    Thanks for taking the trouble to respond. I was hoping for a method that might make the process of factorising more efficient. However I followed your advice and eventually came up with the factors and solved the equation.
    You're welcome. Unfortunately I don't have a more efficient method - I tend just to use trial and error myself
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    there IS a way you can factorise ANY quadratic (even when b^2-4ac<0) - but standard factorisation questions are so easy nowadays no one uses it.

    if you want to know it, its:

    \displaystyle ax^{2}+bx+c \equiv \frac{1}{4a}((2ax+b)+S)((2ax+b)-S) where S = \sqrt{b^{2}-4ac}}

    there are also some standard tips like:

    in (assuming "a" is +ve) ax^2+bx+c:

    positives throughout mean +in each bracket

    negatives throughout mean one -ve, one+ve

    negative x term, +ve number means 2 negatives

    +ve x term, -ve number means one bracket +ve, one -ve.

    also, if the coefficient of x (the b) is odd, the brackets will have an even and an odd number in them
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    (Original post by Ihol)
    Thanks for taking the trouble to respond. I was hoping for a method that might make the process of factorising more efficient. However I followed your advice and eventually came up with the factors and solved the equation.
    The method of splitting the middle term would work.
    see attachment


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  1. File Type: doc C1 Factorise tricky quadratics.doc (28.0 KB, 58 views)
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    (Original post by gdunne42)
    The method of splitting the middle term would work.
    see attachment


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    What would the middle terms be though?
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    (Original post by Jack_devin)
    What would the middle terms be though?
    Factors of 600 that add to 49

    Not that hard to find
    6 x 100
    12 x 50
    24 x 25


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    (Original post by gdunne42)
    Factors of 600 that add to 49

    Not that hard to find
    6 x 100
    12 x 50
    24 x 25


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    Oh, wow. I was being extremely stupid, I feel embarassed, lmao.

    Thank you for your time.
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    (Original post by Jack_devin)
    Oh, wow. I was being extremely stupid, I feel embarassed, lmao.

    Thank you for your time.
    No prob, you are welcome


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