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  1. #1
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    equations

    prove that the line with the equation x + 2y = 2 and the curve with equation y = 3x^2 - 2x + 2 have no points of intersection

    i started by putting the first equation into the second after making the first equation x = 2 - 2y. so it became 3(2 - 2y)^2 - 2(2 - 2y) + 2 then expanded the brackets. so first i squared the first bracketed term and made it 3(4 - 8y + 4y^2) then expanded the second and made it (4 - 4y) so i was left with (12 - 24y + 12y^2) - (4 - 4y) + 2 which then came to 12y^2 - 20y + 10 and used the discriminant theory to make sure it came to less than 0. i came up with 400 - 480 for the discriminant. so -80 was the discriminant. but my book says its -39. can someone explain where i went wrong with this? thanks
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  2. #2
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    Did you forget the "=y" from the original equation into which you'd substituted?

    Also, did the book do the same substitution you did, or did it plug in for y instead of for x?
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  3. #3
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    yeah i did forget to write = y, does it matter though? also, the book didn't show or ask for a specific way of substituting, i just chose that way
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  4. #4
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    Quote Originally Posted by mark View Post
    prove that the line with the equation x + 2y = 2 and the curve with equation y = 3x^2 - 2x + 2 have no points of intersection

    i started by putting the first equation into the second after making the first equation x = 2 - 2y. so it became 3(2 - 2y)^2 - 2(2 - 2y) + 2 then expanded the brackets. so first i squared the first bracketed term and made it 3(4 - 8y + 4y^2) then expanded the second and made it (4 - 4y) so i was left with (12 - 24y + 12y^2) - (4 - 4y) + 2 which then came to 12y^2 - 20y + 10 and used the discriminant theory to make sure it came to less than 0. i came up with 400 - 480 for the discriminant. so -80 was the discriminant. but my book says its -39. can someone explain where i went wrong with this? thanks
    Your working out is fine.

    I think the book has found y in terms of x and then substituted for y. You did it the other way around.

    Doesn't matter though, you proved that there are not any intersections.
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    surely they should come out with the same result though no matter which way round you do it? or is that not the case?
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    Quote Originally Posted by mark View Post
    yeah i did forget to write = y, does it matter though?
    Without the "=y", you didn't subtract that y to get "=0", so your discriminant would have been "off" a bit.

    Quote Originally Posted by mark View Post
    the book didn't show or ask for a specific way of substituting, i just chose that way
    If there is no solution, then there is no solution. But if they substituted the other way 'round, they should have ended up with a different quadratic. The discriminant still would have been negative, but that different equation might explain their different value. (I haven't checked.)
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  7. #7
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    ah i see, so in the end it would have come to y = 12y^2 - 20y + 10 so what would i have to do to figure out what y was?

    ah you minus the y making it 12y^2 -21y + 10 = 0 then quadratic formula or whatever
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  8. #8
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    these graphs may help, the ist 1 is y = (2 - x)/2 and y = 3x^2 - 2x - 2

    the second one is the parameterized one, you compare
    Attached Thumbnails Attached Thumbnails equations-a1.gif   equations-a2.gif  
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  9. #9
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    yo pacman thanks for the post, do you like manny pacqiao then? you reckon he can beat mayweather?
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