Solving Quadratic Systems
x^2+y^2=20
x^2-6y=28
PLEASE SHOW WORK I'M HAVING TROUBLE SOLVING IT
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)
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
How did you get 8?
I'm so confused...
x^2 + y^2 = 20
-
x^2 - 6y = 28
________________
y^2 + 6y = -8
y^2 + 6y + 8 = 0
To solve the given quadratic system of equations:
1. Start by rearranging Equation 2 to express x in terms of y:
x^2 - 6y = 28
x^2 = 6y + 28
x = ± √(6y + 28)
2. Substitute the value of x from Equation 2 into Equation 1:
(±√(6y + 28))^2 + y^2 = 20
6y + 28 + y^2 = 20
3. Rearrange the equation to make it a quadratic equation in standard form:
y^2 + 6y - 8 = 0
4. Solve the quadratic equation to find the values of y:
You can use factoring, completing the square, or the quadratic formula to solve for y. Let's use factoring for this example:
y^2 + 6y - 8 = 0
(y - 1)(y + 8) = 0
Set each factor equal to zero and solve for y:
y - 1 = 0 or y + 8 = 0
y = 1 or y = -8
5. Substitute the values of y back into Equation 2 to find the corresponding values of x:
For y = 1:
x = ± √(6(1) + 28)
x = ± √(34)
For y = -8:
x = ± √(6(-8) + 28)
x = ± √(20)
6. Simplify the square roots to get the final values of x:
For y = 1:
x = ± √34
For y = -8:
x = ± √20
Thus, the solutions to the quadratic system of equations are:
(x, y) = (± √34, 1)
(x, y) = (± √20, -8)