# Thread: equations

1. ## 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

2. 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?

3. 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

4. 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
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.

5. surely they should come out with the same result though no matter which way round you do it? or is that not the case?

6. 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.

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.)

7. 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

8. 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

9. yo pacman thanks for the post, do you like manny pacqiao then? you reckon he can beat mayweather?