What are the intersection points of the equations p(x,y)=(t,t^2 + 3) and x+y=9? Best explanation gets medal!
Lets first make an equation in x and y from the parameterized equation in t. x,y = t, t^2 +3 x= t, y = t^2+3 so, y= x^2 + 3 now solve this simultaneously with x+y = 9 plug in y = x^2+3 here x + x^2 + 3 = 9 this is a simple quadratic in x which will give you 2 roots :) can you get those?
need help getting 2 values of x ??
It would turn into t^2 + t + 12 = 0 right?
12? we subtract 9 from both sides, x^2 +x + 3 - 9 = 0 x^2 + x - 6= 0
Oh yeah you subtract not add
I'm just confused because it needs to multiply to equal 1 but adds up to -6
2 numbers which multiply to -6 and adds upto +1
sorry that's what i meant can't be 3 & 3 or 6 and 1
they aren't gonna be whole numbers?
one of the number is negative
Ugh sorry I've been out sick from school for two weeks lol
This is so simple why isn't it clicking?
oh my god i'm an idiot
okay so it's -2 and 3
yes!
okay wow that was quite a journey, so now that i have that, i need to find numerical values right?
x^2 + x - 6= 0 factorization method z^2 + 3x - 2x - 6= 0
x^2 + 3x - 2x - 6= 0 ***
x(x+3) - 2 (x+3) = 0 want to continue?
sorry how did you get there again?
from x^2 +3x , i factored out x so i got x(x+3)
and where did the x+3 come from? one of the possible t values i factored out?
oh you just factored it out and then that is part of it
the +3x?? i splitted +x into 3x -2x
ohhhh!!
what happens now with x(x+3) ??
wait, can't I just plug in the x values into the x+y=9 equation as an x= ? like -2 = -y + 9 and 3 = -y + 9 ??
.... because we still don't have the values of x! so we are here x(x+3) - 2 (x+3) = 0 factor out x+3 (x+3)(x-2) = 0
okay i've got that now
so, x = -3 or x = 2 now you can plug these values in x+y = 9 :)
oh okay okay and that will be 12 and 7?
yes so your solutions are -3,12 and 2,7
(-3,12) AND (7,7) ?
i meant 2 yeah okay yay that's it?
yes, thats it :)
thank you so much sorry for being so slow lol
welcome ^_^ and no problem
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