equations

• Sep 28th 2009, 06:43 AM
mark
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
• Sep 28th 2009, 06:46 AM
stapel
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?
• Sep 28th 2009, 06:50 AM
mark
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
• Sep 28th 2009, 06:51 AM
Prove It
Quote:

Originally Posted by mark
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

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.
• Sep 28th 2009, 06:52 AM
mark
surely they should come out with the same result though no matter which way round you do it? or is that not the case?
• Sep 28th 2009, 07:11 AM
stapel
Quote:

Originally Posted by mark
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
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.)
• Sep 28th 2009, 07:26 AM
mark
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
• Sep 28th 2009, 07:36 AM
pacman
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 (Hi)
• Sep 28th 2009, 07:39 AM
mark
yo pacman thanks for the post, do you like manny pacqiao then? you reckon he can beat mayweather?