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    guys sorry it's me again i was factorising
    3x2 + 10x + 7 = 0

    in this case, would it be appropriate to use the quadratic formula, or would it not work because 3x2 isn't exactly the same as ax
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    (Original post by entertainmyfaith)
    guys sorry it's me again i was factorising
    3x2 + 10x + 7 = 0

    in this case, would it be appropriate to use the quadratic formula, or would it not work because 3x2 isn't exactly the same as ax
    How can a quadratic formula "not work"?

    Completing the square, factorising, and using the quadratic formula (which is a general result from completing the square) are all valid methods of solving a quadratic, so they all work.

    You can use the quadratic formula if you want, but you can also easily factorise this quadratic too.
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    no you can use it you would just make a= three
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    (Original post by RDKGames)
    How can a quadratic formula "not work"?

    Completing the square, factorising, and using the quadratic formula (which is a general result from completing the square) are all valid methods of solving a quadratic, so they all work.

    You can use the quadratic formula if you want, but you can also easily factorise this quadratic too.
    i didn't mean the formula not working, more if it was appropriate for the method because i was just unsure if 3x2 + 10x +7 = 0 was equivalent to ax + bx + c = 0- does the squared notation affect it?
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    (Original post by entertainmyfaith)
    i didn't mean the formula not working, more if it was appropriate for the method because i was just unsure if 3x2 + 10x +7 = 0 was equivalent to ax + bx + c = 0- does the squared notation affect it?
    Your "ax+bx+c=0" is not a quadratic expression, so no they are no equivalent.

    If you meant ax^2+bx+c=0 then yes of course they are equivalent since we just have a=3, b=10, c=7
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    Your gonna use the quadratic formula to FACTORISE.......
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    (Original post by Ex-pert)
    Your gonna use the quadratic formula to FACTORISE.......
    Don't see how that's a crime. The formula would give her solutions x_1,x_2 to the equation so all that is needed to be done is plug them in the form (x-x_1)(x-x_2)=0 which would give the factorised form.
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    (Original post by entertainmyfaith)
    guys sorry it's me again i was factorising
    3x2 + 10x + 7 = 0

    in this case, would it be appropriate to use the quadratic formula, or would it not work because 3x2 isn't exactly the same as ax
    the answer is;

    [solution removed]
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    (Original post by RDKGames)
    Don't see how that's a crime. The formula would give her solutions x_1,x_2 to the equation so all that is needed to be done is plug them in the form (x-x_1)(x-x_2)=0 which would give the factorised form.
    But there is another easier way
    I dont see the reason why he is making it look so complicated
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    (Original post by Ex-pert)
    But there is another easier way
    I dont see the reason why he is making it look so complicated
    I agree, but if that's the way she wants to try first then she can go ahead and settle on it for the time being.
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    i tried factorising it but i couldn't do it so i just used the quadratic formula. it got me the right answer
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    (Original post by entertainmyfaith)
    i tried factorising it but i couldn't do it so i just used the quadratic formula. it got me the right answer
    ok
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    (Original post by entertainmyfaith)
    i tried factorising it but i couldn't do it so i just used the quadratic formula. it got me the right answer
    Thats what i want to see
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    (Original post by entertainmyfaith)
    i tried factorising it but i couldn't do it so i just used the quadratic formula. it got me the right answer
    To factorise something of the form ax^2+bx+c=0 you want to look for two numbers \lambda, \mu with the following properties: \lambda + \mu = b and \lambda \mu =ac

    Then you take the form \displaystyle \left( x+\frac{\lambda}{a} \right) \left( x+\frac{ \mu}{a} \right)=0 and plug those in there for factorisation.

    In your example, you have a=3, b=10, c=7 with \lambda + \mu = 10 and \lambda \mu = 21. Shouldn't take too long to notice that \lambda = 3 and \mu = 7 are valid conditions which make those properties true.

    So then you have (x+\frac{3}{3})(x+\frac{7}{3})=0 which can be simplified further.
 
 
 
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