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    Hello

    Is there a certain rule when you can and cannot use factorisation to solve a quadratric equation?

    I have just started algebra so my knowledge is very limited, however I came across this equation which I am struggling to use factorisation with.

    2x^2 + 9x - 7

    My understanding is to multiply the co-efficent (or whatever) with the constant, so 2 x -7 = -14.

    Then to find 2 numbers that can be multiplied together to make -14 and that add together to make 9. I cannot think of any that would work? 2 x 7 = 14 so can't be that as its negative, -2 x 7 = -14 but -2 + 7 = 5.

    Its probably really obvious that I am missing so can anyone politely give me a nudge in the right direction, and if there are rules I am missing then please inform me

    Thank you
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    there might be something I'm missing too but you can use the quadratic formula for things that don't factorise:

    using ax^2+bx+c = 0

    http://upload.wikimedia.org/math/3/c...3ccad7990f.png
    (didn't know how to type out the formula so here's the link to an image of it)

    if its still confusing you just ask any questions


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    if the image does not open just type quadratic formula into google images and the one that has b2-4ac in the square root is the formula you need


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    Yeah in the end I used the quadratric equation, but I find using that is just long, I thought I was getting the hang of using the factorization method until now. There must be something I am missing. I dont want to waste the page on an exam using the factorisation method when theres a 'rule' when it can't be used, and waste time and marks. This was the question:

    2x^2+ 9x – 7 = 0
    Give your solutions correct to 3 significant figures.

    Hmm, maybe when theres a zero the factorisation method can't be used and the quadratric equation to be used instead. Just looked up the page and there was another equation with no '=0' on the end and the factorisation method worked for that. Can anyone confirm?

    Thank you
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    x=\dfrac{-b \pm \sqrt{b^2-4ac} } {2a}

    Yeah, it doesn't look like that quadratic factorizes into rational roots. It would if the first term was negative or the last term was positive, though.

    A hint is that when it asks you for 3 s.f., it probably won't be a rational root and you will instead be expected to graph it on a calculator or something.

    Of course the quadratic formula always works, but most methods to determine whether it has rational roots or not are based on finding whether the determinant b^2-4ac is a perfect square, which is basically doing the quadratic formula anyways.

    After enough practice, you will be able to guess rational roots (if any) more quickly than any identifying method.
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    All quadratic equations should really be arranged so they're =0 before you solve them - it doesn't affect whether it can be factorised. The clue that this one probably can't be factorised is that it asks for the answer to 3 s.f. If it could be factorised, the solutions for this equation would be either integers or end in .5. It should be solved using the quadratic formula or completing the square.
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    I thought you could only use the completing square method when there is a '-' sign and both are square numbers. I may be wrong though as I haven't looked into that part yet
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    Completing the square is just the long way of doing the quadratic formula. By going through the motions of completing the square, you are doing exactly what you would do to the coefficients as you would with the quadratic formula.

    Anyways, the =0 thing is a misconception on your part.

    If your function is:

    f(x)=k(x-r_1)(x-r_1)

    then the roots (x-intercepts) are the x values where the following equation holds true:

    0=k(x-r_1)(x-r_1)

    Namely:

    x=r_1,r_2

    Can you see why this is so?
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    (Original post by Tinkletea)
    I thought you could only use the completing square method when there is a '-' sign and both are square numbers. I may be wrong though as I haven't looked into that part yet
    That's difference of two squares.
    You probably haven't come across completing the square yet, so don't worry. It's basically where the formula comes from, so you don't need to memorise the formula to solve equations that don't factorise.
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    (Original post by aznkid66)
    Completing the square is just the long way of doing the quadratic formula. By going through the motions of completing the square, you are doing exactly what you would do to the coefficients as you would with the quadratic formula.

    Anyways, the =0 thing is a misconception on your part.

    If your function is:

    f(x)=k(x-r_1)(x-r_1)

    then the roots (x-intercepts) are the x values where the following equation holds true:

    0=k(x-r_1)(x-r_1)

    Namely:

    x=r_1,r_2

    Can you see why this is so?

    I'll be honest, I dont understand a word of that lol.

    To other poster, nope I haven't come across that part yet then, got the difference of 2 squares mixed up
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    Well, basically, I'm saying that the roots of a quadratic are its x-intercepts, and x-intercepts are where y=0, so the roots of the quadratic are where it equals 0.

    Thus, these two are equivalent:

    "Find where x is a root of the quadratic."

    "Find the x where the quadratic is equal to 0."
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    (Original post by Tinkletea)
    Hello

    Is there a certain rule when you can and cannot use factorisation to solve a quadratric equation?

    I have just started algebra so my knowledge is very limited, however I came across this equation which I am struggling to use factorisation with.

    2x^2 + 9x - 7

    My understanding is to multiply the co-efficent (or whatever) with the constant, so 2 x -7 = -14.

    Then to find 2 numbers that can be multiplied together to make -14 and that add together to make 9. I cannot think of any that would work? 2 x 7 = 14 so can't be that as its negative, -2 x 7 = -14 but -2 + 7 = 5.

    Thank you
    That would work with some modification, but to find that two numbers
    only for real integer (maybe rational) roots will be easy.
    You always can arrange a quadratic to form of
    ax^2+bx+c=0
    Assuming two real roots x_1 and x_2
    then from equation follows
    a(x-x_1)(x-x_2)=0 which is the original equation in another
    form (or factorized by roots)
    Expanding it
    ax^2-a(x_1+x_2)x+ax_1x_2=0
    Looking to the first equation and dividing by a \neq 0
    you will get
    x_1+x_2=-\frac{b}{a}
    x_1\cdot x_2=\frac{c}{a}
    Where x_1 and x_2 are two numbers you wrote.
    THe formula above is known as Viéte formula.
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    If the equation is in the form ax2+bx+c=0 then basically there is only 1 rule which you have already explained Basically two numbers that add to make b and multiply to make ac. if not then you must ust the quadratic equation or complete the square
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    full walk through.

    2x^2 + 9x - 7

    Step.1 multiply 2 and -7 to give -14

    step2. find a number that multiply to give -14 and adds to give +9 .. okay there isnt any

    step.3 since there are no common factors use the quadratic formula


    b=+9 c=-7 a = 2

    step4. sub values in and get a solution and through a house party in celebration
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    Thanks all. The other equation was with the 'factorisation' questions on the exam, I guess thats the hint. If there is a lone quadratric equation I wont waste time and just use the equation
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    (Original post by Tinkletea)
    Thanks all. The other equation was with the 'factorisation' questions on the exam, I guess thats the hint. If there is a lone quadratric equation I wont waste time and just use the equation
    have a quick check to see if you can factorise if you don't notice it within say 20 seconds then just use the formula
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    (Original post by madfish)
    have a quick check to see if you can factorise if you don't notice it within say 20 seconds then just use the formula
    Will do, thanks
 
 
 
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