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    Hey folks, I'm a little stumped on this question - (it doesn't show any workings in the textbook, it only gives you the answers).

    A curve has equation y = ax^2 + bx, where a and b are integers.

    a) The point (3,21) lies on the curve. Show that 7 = 3a + b

    b) The point (2,8) also lies on the curve. Find another linear equation satisfied by a and b.

    I'll be fine once I find the two equations, I'm just not too sure how to solve these.

    Thanks.
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    (Original post by Srxjer)
    Hey folks, I'm a little stumped on this question - (it doesn't show any workings in the textbook, it only gives you the answers).

    A curve has equation y = ax^2 + bx, where a and b are integers.

    a) The point (3,21) lies on the curve. Show that 7 = 3a + b

    b) The point (2,8) also lies on the curve. Find another linear equation satisfied by a and b.

    I'll be fine once I find the two equations, I'm just not too sure how to solve these.

    Thanks.
    7=3a+b--1

    4=2a+b --2

    subtract 1 from 2 and can you find the value of a (the b's cancel out), and sub back in to get b..
    The question however doesnt seem to ask to find these values though
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    (Original post by Srxjer)
    Hey folks, I'm a little stumped on this question - (it doesn't show any workings in the textbook, it only gives you the answers).

    A curve has equation y = ax^2 + bx, where a and b are integers.

    a) The point (3,21) lies on the curve. Show that 7 = 3a + b

    b) The point (2,8) also lies on the curve. Find another linear equation satisfied by a and b.

    I'll be fine once I find the two equations, I'm just not too sure how to solve these.

    Thanks.
    what does (3,21) mean?
    what does (2,8) mean?

    do you have anything lying around in your question with those particular variables, with which you could make a sneaky substitution?
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    srxjer's done it wrong as the equation is y=ax^\emph{2}+bx
    so actually it's

    21=9a+b (1)

    and

    8=4a+b (2)

    as your simultaneous equations

    rearrange one of them to get b=something (i suggest equation 2)then substitute that into the other equation and solve for a.
    Then you can sub that value of a into your eq for b=something and solve easily

    Once you have that, sub in x=1 to your equation to find another value for y and i THINK that answers your question, it's a strange one
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    (Original post by munn)
    srxjer's done it wrong as the equation is y=ax^\emph{2}+bx
    so actually it's

    21=9a+b (1)

    and

    8=4a+b (2)

    as your simultaneous equations

    rearrange one of them to get b=something (i suggest equation 2)then substitute that into the other equation and solve for a.
    Then you can sub that value of a into your eq for b=something and solve easily

    Once you have that, sub in x=1 to your equation to find another value for y and i THINK that answers your question, it's a strange one
    no.. if y = ax^2 + bx then setting y = 21, x = 3 leads to:

    21 = a(3*3) + b(3)

    dividing both sides by 3

    7 = 3a + b
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    (Original post by zdo0o)
    no.. if y = ax^2 + bx then setting y = 21, x = 3 leads to:

    21 = a(3*3) + b(3)

    dividing both sides by 3

    7 = 3a + b
    oh yeah my mistake i thought it said ax^2 + b, not bx (and not only after chiding someone for messing up the square, i actually wrote down the correct equation too! instant karma)
    yeah listen to this
 
 
 
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