Final answer:
The system of equations x = y^2/4 and x^2 + y^2 = 32 is solved using substitution, which leads to a quadratic equation in terms of y^2. The solutions found are (x = 4, y = 4) and (x = 4, y = -4).
Step-by-step explanation:
To solve the system of equations x = y^2/4 and x^2 + y^2 = 32, we can use substitution. Replacing x with y^2/4 in the second equation, we have:
(y^2/4)^2 + y^2 = 32
y^4/16 + y^2 = 32
Multiplying through by 16:
y^4 + 16y^2 = 512
Let's denote z = y^2 to make the equation simpler:
z^2 + 16z - 512 = 0
We now have a quadratic equation in terms of z. Factoring this equation:
(z + 32)(z - 16) = 0
So, z = -32 or z = 16. Since z represents y^2 and y^2 cannot be negative, we reject z = -32. Our solution for z is:
z = y^2 = 16
Therefore, y = ±4 (since we're dealing with real numbers and the square root of a positive number has two solutions).
Substituting the value of y back into the first equation to find x:
x = y^2/4 = 16/4 = 4
Thus, the solutions for the original system are:
x = 4, y = 4 and x = 4, y = -4