You want factors of 4 and -3 since they multiply to equal 12 and add to be the 1 in front of the x.
you should have x = 3 and x = -4
When I worked with the first and third, I got (x-3)(x+2)
x = 3 x = -2
so the x=3 showed up twice. Substitute it in each of the 3 equations to see if you get y=33 in each case.
How do you find the coordinates that the curves y= 2x^2 +5x, y=x^2+4x+12 and y=3x^2+4x-6 have in common?
The answer should be (3,33)
I know that you have to solve a simultaneous equations. What I've done so far:
y=2x^2+5x
y=x^2+4x+12 --> 2x^2+5x=x^2+4x+12
=x^2+x-12=0
=(x+6)(x-5)
= x=-6 or x=5
2 answers
In general, of course, three curves need not intersect in a single point. But, as you say,
2x^2+5x = x^2+4x+12: x = -4 or 3
2x^2+5x = 3x^2+4x-6: x = -2 or 3
x^2+4x+12 = 3x^2+4x-6: x = -3 or 3
So, it looks like (3,33) works fine
You messed up in your solution. 6*5 = 30, not 12!
http://www.wolframalpha.com/input/?i=plot+y%3D2x%5E2%2B5x,+y%3Dx%5E2%2B4x%2B12,+y+%3D+3x%5E2%2B4x-6,+-5+%3C%3D+x+%3C%3D+4
2x^2+5x = x^2+4x+12: x = -4 or 3
2x^2+5x = 3x^2+4x-6: x = -2 or 3
x^2+4x+12 = 3x^2+4x-6: x = -3 or 3
So, it looks like (3,33) works fine
You messed up in your solution. 6*5 = 30, not 12!
http://www.wolframalpha.com/input/?i=plot+y%3D2x%5E2%2B5x,+y%3Dx%5E2%2B4x%2B12,+y+%3D+3x%5E2%2B4x-6,+-5+%3C%3D+x+%3C%3D+4
3 answers