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    i dont want you to just do the question but could you actually explain how you're doing it. that'd be great.
    find the value of x for which:
    2x^2 - 7x + 3 > 0

    Thanks if you could help..... i just dont know what to do when there is a squared in there! :|
    Step by step help please
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    Factorise?
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    (Original post by boromir9111)
    Factorise?
    ok i know what factorising it but i dont knw how you'd factorise this?! im really bad at maths !
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    two numbers that multiply to gove 3 (the constant term at the end) and add to give 14 (it is 14 because 2 x 7 [7 is from the middle]. and the 2 comes from the 2 in front of the x^2)

    I don't know how you're going to remember it, so just make up some weird mnemonic for it.

    the answer is (2x - 1) (x-3)

    so x= 1/2, 3.

    do you want me to explain anything else you're unclear about? quote me so I see reply
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    (Original post by ilyking)
    two numbers that multiply to gove 3 (the constant term at the end) and add to give 14 (it is 14 because 2 x 7 [7 is from the middle]. and the 2 comes from the 2 in front of the x^2)

    I don't know how you're going to remember it, so just make up some weird mnemonic for it.

    the answer is (2x - 1) (x-3)

    so x= 1/2, 3.

    do you want me to explain anything else you're unclear about? quote me so I see reply
    Thanks for that! i understand it now
    i am now stuck on another question! GRAPHS....... i hate them
    it says:
    the points A and B have co-ordinates (1, 2) and (5, 8)
    find, in the form y=mx+c an equation for the line through both points A and B
    ...i dont have a clue!
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    (Original post by Jay1012)
    Thanks for that! i understand it now
    i am now stuck on another question! GRAPHS....... i hate them
    it says:
    the points A and B have co-ordinates (1, 2) and (5, 8)
    find, in the form y=mx+c an equation for the line through both points A and B
    ...i dont have a clue!
    What is m? how do you find it? what is c? how do you find it?
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    (Original post by boromir9111)
    What is m? how do you find it? what is c? how do you find it?
    m and c are just numbers i think....
    you have to find an equation of the line that goes through those 2 points and just give it in the form y=mx+c m and c are just integers
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    (Original post by Jay1012)
    Thanks for that! i understand it now
    i am now stuck on another question! GRAPHS....... i hate them
    it says:
    the points A and B have co-ordinates (1, 2) and (5, 8)
    find, in the form y=mx+c an equation for the line through both points A and B
    ...i dont have a clue!
    hmmm you seem to have a good understanding so I'll try and explain it through tsr

    you are given two points, (1,2) and (5,8). basically, take the x and y values and put it into gradient formula:

    m (gradient) = change in y / change in x so.....

    m = 2 - 8 / 1 - 5 = -3/2

    then you put that into another formula for the actual equation, because now you have two points and a gradient, so....

    formula is: y - y1 = m(x-x1) <<<(equation of line formula, in order to use this we need to have ONE coordiante and a gradient [which we just figured out to be -3/2]

    which is, by considering only ONE coordiante (gonna take (1,2) it's easier to work with) and the -3/2 gradient, we simply sub in:

    y - 2 = -3/2 (x-1)

    so... y = -3/2x + 7/2

    and done.

    seem long, but not really...
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    (Original post by ilyking)
    hmmm you seem to have a good understanding so I'll try and explain it through tsr

    you are given two points, (1,2) and (5,8). basically, take the x and y values and put it into gradient formula:

    m (gradient) = change in y / change in x so.....

    m = 2 - 8 / 1 - 5 = -3/2

    then you put that into another formula for the actual equation, because now you have two points and a gradient, so....

    formula is: y - y1 = m(x-x1)

    which is, by considering only ONE coordiante (gonna take (1,2) it's easier to work with) and the -3/2 gradient, we simply sub in:

    y - 2 = -3/2 (x-1)

    so... y = -3/2x + 7/2

    and done.

    seem long, but not really...
    You know the forum rules. Don't post full solutions!
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    (Original post by boromir9111)
    You know the forum rules. Don't post full solutions!
    oh yeah and my name is aragorn
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    (Original post by ilyking)
    oh yeah and my name is aragorn
    I had to +rep you for that even though you broke the forum rules
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    (Original post by ilyking)
    hmmm you seem to have a good understanding so I'll try and explain it through tsr

    you are given two points, (1,2) and (5,8). basically, take the x and y values and put it into gradient formula:

    m (gradient) = change in y / change in x so.....

    m = 2 - 8 / 1 - 5 = -3/2

    then you put that into another formula for the actual equation, because now you have two points and a gradient, so....

    formula is: y - y1 = m(x-x1) <<<(equation of line formula, in order to use this we need to have ONE coordiante and a gradient [which we just figured out to be -3/2]

    which is, by considering only ONE coordiante (gonna take (1,2) it's easier to work with) and the -3/2 gradient, we simply sub in:

    y - 2 = -3/2 (x-1)

    so... y = -3/2x + 7/2

    and done.

    seem long, but not really...
    thanks! youre great help! i understand what you're telling me but it's just knowing what to do to solve to question!
    i hate Graph stuff, anything to do with graphs im awful at.
    Could you explain this one last question.... maybe just the steps to take and start me off

    The line y=41 meets C at point R
    find the x coordinate of R, giving your answer in the form p+q(square root of 2) where p and q are integers..... so surd form i think?
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    (Original post by ilyking)
    hmmm you seem to have a good understanding so I'll try and explain it through tsr

    you are given two points, (1,2) and (5,8). basically, take the x and y values and put it into gradient formula:

    m (gradient) = change in y / change in x so.....

    m = 2 - 8 / 1 - 5 = -3/2

    then you put that into another formula for the actual equation, because now you have two points and a gradient, so....

    formula is: y - y1 = m(x-x1) <<<(equation of line formula, in order to use this we need to have ONE coordiante and a gradient [which we just figured out to be -3/2]

    which is, by considering only ONE coordiante (gonna take (1,2) it's easier to work with) and the -3/2 gradient, we simply sub in:

    y - 2 = -3/2 (x-1)

    so... y = -3/2x + 7/2

    and done.

    seem long, but not really...
    another question im stuck on........
    the curve C with equation y=x^2 - 6x + 18 meets the y-axis at P and has a minimum point at Q.
    sketch the graph of C, showing the coordinates of P and Q...
    how would you work these 2 points out and what would the graph look like?
    thanks if you could help....... i wont bother you after this one
 
 
 
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