Question

Solving Quadratic Systems

x^2+y^2=20
x^2-6y=28

PLEASE SHOW WORK I'M HAVING TROUBLE SOLVING IT

Answer 1

we have a circle with radius 2 sqrt5 and center at the origin
and
a parabola that looks like:
6 y = x^2-28
y = (x^2-28)/6
opens up (holds water)
zeros at x = +/- sqrt 28
so vertex at x = 0
then y of vertex at -28/6
NOW SKETCH A GRAPH

now do algebra
x^2 + y^2 = 20
x^2 - 6 y = 28
-------------- subtract
y^2 + 6 y + 8 = 0

y = [ -6 +/- sqrt (36-32) ]/2

= -3 +/-(1/2) sqrt 4
= -3 +/- 1
= -2 or - 4

DRAW THOSE HORIZONTAL LINES ON YOUR SKETCH
see how the parabola cuts the circle below the x axis?

when y = -2
x = 4 or -4
when y = -4
x = 2 or -2

so
(2,-4) (-2,-4) (-4,-2)(+4,-2)

Answer 2

x^2 + y^2 = 20
-
x^2 - 6y = 28
________________
y^2 + 6y = -8

y^2 + 6y + 8 = 0

(y+4)(y+2)

y= -4,-2

plug those solution into first equation:
x^2 + (-4)^2 =20
x^2 + 16 = 20
x^2 = 4

x=+- 2

Answer 3

How did you get 8?
I'm so confused...

Answer 4

x^2 + y^2 = 20
-
x^2 - 6y = 28
________________
y^2 + 6y = -8

y^2 + 6y + 8 = 0