quadratic formula should do the trick, or you could factor into \(-(x-6)(x+1)\) and solve for x that way
Ah, double check your math for when you solved x=-1
no they are both wrong i used a graphing calculator and the zeroes are x=0 and x=-1.2
it wants me to solve this system algebraically
Lol your answers x=6 and x=1 are both correct.
If they want you to solve it algebraically, then why did you use a graphing calculator? :D
I would first multiply both sides and all terms by -1 (because I don't like -x^2) what do you get ?
because the second part is for me to graph it to prove my answers
now try factoring it
How I wish we were allowed to use a graphing calculator when we graph something. Do you know how to plot points like this? |dw:1439243109664:dw|
but it looks like you did it correctly up top x= 6 and x=-1
(x^2-2x)-(3x-6)=0 x(x-2)-3(x-2)=0 (x-3)(x-2)=0 x=3 and x=2
wait i messed up on the sign there
(x^2 -6x) + (x-6) = 0
the solution is where the two curves cross (intersect) it is true when we are given a single curve (or line), we are often interested in where it crosses the x-axis (the x value is called the zero) but not in this case.
and thats all?
solving two equations is the same as asking "what point lies on both curves"
ok thank you!
you had the right idea except for the final step... once you found x= -1 or x=6 you find the y value (using either equation) and the (x,y) pairs are where the two curves intersect. i.e. are the "solutions"
oh ok thank you i have to go :)