Find the coordinates of the points of intersection of the parabola y=x^2 and y=x+3.

So.. im stuck at this question. Was wondering if anyone of you could help me work it out.

If you want to find where they intersect, you just set them equal to each other:

x^2=x+3
x^2-x-3=0

x=(1±√(1-4(1)(-3)))/2(1)
x=(1±√(1+12))/2
x=(1±√13)/2

So they intersect when x=(1+√13)/2 and (1-√13)/2, which when calculated out come out to be approximately 2.303 and -1.303, respectively.

Once you have these x-values, you can solve for y using either equation (although, I'd suggest the y=x+3):

y=(2.303)+3
y=5.303

y=(-1.303)+3
y= 1.697

So your points of intersection are (-1.303, 1.697) and (2.303, 5.303). Hope this helps.

x² = x + 3
x² - x - 3 = 0
x = [ - b ± √ ( b ² - 4 a c ) ] / 2 a
x = [ 1 ± √ ( 1 + 12 ) ] / 2
x = [ 1 ± √13 ] / 2
x = 2.3 , x = - 1.3
y = 5.3 , y = 1.7
Points (2.3,5.3) , (- 1.3,1.7)

Okay,
so we know that y=x^2
therefore, x^2= x +3
rearrange, so that x^2 -x-3=0
and solve
X=-1.303, and 2.3027,
then to find y co-ord corresp. to these, put into either original eq, i.e Y=x^2

that would be points that satisfy both equations.
just do y=y which in your case is

x^2=x+3
x^2-x-3=0

then solve quadratic equation to get

x1=-3/2

x2=5/2

so

y1=3/2
y2=11/2

points are
(-3/2,3/2), (5/2,11/2)

y = y
x^2 - x = 3
x^2 - x + 1/4 = 3 + 1/4
(x - 1/2)^2 = 13/4
x - 1/2 = ±√13 / 2
x = 1/2(1 ± √13)
y =1/2(7 ± √13)