solve the system of equations x^2 = y and y^2 = x 

Let's solve the system of equations:

1. \( x^2 = y \)
2. \( y^2 = x \)

We will solve this system by substitution and by considering different cases for the values of \(x\) and \(y\).

### Step 1: Substitution

From equation (1), we have \( y = x^2 \). Substituting this into equation (2), we get:
\[
(x^2)^2 = x
\]
Simplifying, this becomes:
\[
x^4 = x
\]
This equation can be rewritten as:
\[
x^4 - x = 0
\]
Factoring out \(x\), we get:
\[
x(x^3 - 1) = 0
\]
This further factors into:
\[
x(x-1)(x^2 + x + 1) = 0
\]

### Step 2: Finding Solutions

This equation gives us three potential solutions for \(x\):

1. \( x = 0 \)
2. \( x = 1 \)
3. \( x^2 + x + 1 = 0 \)

#### Case 1: \( x = 0 \)
If \( x = 0 \), substituting into equation (1) gives:
\[
y = 0^2 = 0
\]
So, one solution is \( (0, 0) \).

#### Case 2: \( x = 1 \)
If \( x = 1 \), substituting into equation (1) gives:
\[
y = 1^2 = 1
\]
So, another solution is \( (1, 1) \).

#### Case 3: \( x^2 + x + 1 = 0 \)
To solve \( x^2 + x + 1 = 0 \), we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = 1 \), and \( c = 1 \):
\[
x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}
\]
\[
x = \frac{-1 \pm \sqrt{1 - 4}}{2}
\]
\[
x = \frac{-1 \pm \sqrt{-3}}{2}
\]
\[
x = \frac{-1 \pm i\sqrt{3}}{2}
\]
This gives us two complex solutions for \( x \):
\[
x = \frac{-1 + i\sqrt{3}}{2} \quad \text{and} \quad x = \frac{-1 - i\sqrt{3}}{2}
\]
Let's denote these solutions as \( \omega \) and \( \omega^2 \), which are the complex cube roots of unity (where \( \omega = e^{2\pi i / 3} \) and \( \omega^2 = e^{4\pi i / 3} \)).

Since \( y = x^2 \), we find the corresponding \( y \) values:
1. For \( x = \omega \), \( y = \omega^2 \).
2. For \( x = \omega^2 \), \( y = \omega \).

### Final Solutions

The solutions to the system of equations are:

1. \( (0, 0) \)
2. \( (1, 1) \)
3. \( \left( \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2} \right) \)
4. \( \left( \frac{-1 - i\sqrt{3}}{2}, \frac{-1 + i\sqrt{3}}{2} \right) \)

---

Do you want more details on any of these solutions or have any questions?

Here are 5 related questions to consider:

1. How do we solve similar systems involving square roots or higher powers?
2. What is the geometric interpretation of the real solutions \((0, 0)\) and \((1, 1)\)?
3. How can we graphically represent the complex solutions in the complex plane?
4. What are the properties of the complex cube roots of unity?
5. How does this problem generalize to higher powers, such as \(x^n = y\) and \(y^n = x\)?

**Tip:** When solving polynomial equations with complex solutions, consider the symmetries and properties of roots of unity.

Solve the system of equations x^2 = y and y^2 = x.

Explore how to solve the system of equations x^2 = y and y^2 = x through substitution and factoring. This detailed solution covers real solutions like (0,0) and (1,1), as well as complex solutions involving the cube roots of unity. Ideal for Grades 11-12 students learning algebra and complex numbers.

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