Hi,
I can't seem to figure out the answer to this problem.
Thanks to anyone who can help!
*Solve the system of equations. Write your solution(s) in the form (x,y).
y = x^2
x^2 + y^2 = 20
x^2+x^4-20=0
(x^2-4)(x^2+5)=0
(x+2)(x-2)(x^2+5)=0
x= +-2, x= +-sqrt(-5
(2,4)(-2,4)(sqrt-5,-5)(-sqrt-5, -5)
y + y^2 = 20
y^2 + y - 20 = 0
(y-4)(y+5) = 0
y = 4 or y = -5
y = 4 is easy, then x = -2 or +2
y = -5 means x = i sqrt 5
now go back and see which of those solutions work in the original
if y = 4 and x = 2 then
yes, y = x^2
and
yes, 4 + 16 = 20
if y = 4 and x = -2 then
yes, y = x^2
yes, 4 + 16 = 20
if y = -5 and x = i sqrt 5
yes, y = x^2
yes, -5 + 25 = 20
I left out y = -5, x = -i sqrt 5
y = (-isqrt5)^2 = -5 yes
-5 + 25 = 20 yes
To solve this system of equations, you can use the method of substitution. Let's walk through the steps to find the solution.
1. Start with the first equation: y = x^2.
2. Substitute this expression for y in the second equation: x^2 + (x^2)^2 = 20.
3. Simplify the equation: x^2 + x^4 = 20.
4. Rearrange the equation in standard form: x^4 + x^2 - 20 = 0.
5. This is a quadratic equation in terms of x^2. Let's solve it by factoring. Start by trying to factor it as (x^2 + a)(x^2 + b) = 0, where a and b are constants. Try different combinations of a and b until you find one that works.
By trial and error, we find (x^2 + 5)(x^2 - 4) = 0 satisfies the equation.
6. Set each factor equal to zero and solve for x:
a) x^2 + 5 = 0
x^2 = -5
x = ±√(-5)
b) x^2 - 4 = 0
x^2 = 4
x = ±√4
x = ±2
So, we have four possible solutions for x: x = √(-5), -√(-5), 2, -2.
7. Substitute these values of x back into the first equation y = x^2 to find the corresponding y-values.
For x = √(-5):
y = (√(-5))^2
= -5
For x = -√(-5):
y = (-√(-5))^2
= -5
For x = 2:
y = 2^2
= 4
For x = -2:
y = (-2)^2
= 4
8. Write the final solutions in the form (x, y):
The solutions are ( √(-5), -5), (-√(-5), -5), (2, 4), and (-2, 4).
And that's how you solve the system of equations!