Question

ukupan broj realnih rješenja sistema

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$$\begin{cases} x + y + z = 0 \\\\ x^2 + y^2 + z^2 = 14 \\\\ xyz = -6 \\end{cases}$$

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je:

Explanation:

🆕 New Concept Discovered: Solving Systems using Symmetric Polynomials
Using roots of a cubic equation to find system solutions.

Step 1: Relate the variables to symmetric polynomials

Let \( x \), \( y \), and \( z \) be the real roots of a cubic polynomial:

$$ P(t) = t^3 - e_1 t^2 + e_2 t - e_3 $$

where the elementary symmetric polynomials are defined as:

$$ e_1 = x + y + z $$

$$ e_2 = xy + yz + zx $$

$$ e_3 = xyz $$

From the given system of equations, we have:

$$ e_1 = 0 $$

$$ e_3 = -6 $$

Step 2: Find the value of the second symmetric polynomial

We can find \( e_2 \) using the algebraic identity:

$$ (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) $$

Substitute the known values into the identity:

$$ 0^2 = 14 + 2e_2 $$

$$ 2e_2 = -14 $$

$$ e_2 = -7 $$

Step 3: Set up and solve the cubic equation

Now, substitute \( e_1 \), \( e_2 \), and \( e_3 \) back into the cubic polynomial \( P(t) \):

$$ P(t) = t^3 - 0 \cdot t^2 + (-7)t - (-6) $$

$$ P(t) = t^3 - 7t + 6 $$

To find the roots of \( t^3 - 7t + 6 = 0 \), we can test small integer values.
For \( t = 1 \):

$$ 1^3 - 7(1) + 6 = 1 - 7 + 6 = 0 $$

So, \( t = 1 \) is a root.

We can factor out \( (t - 1) \) from the polynomial:

$$ t^3 - 7t + 6 = (t - 1)(t^2 + t - 6) $$

Now, factor the quadratic part \( t^2 + t - 6 \):

$$ t^2 + t - 6 = (t + 3)(t - 2) $$

Thus, the three roots of the cubic equation are:

$$ t_1 = 1, \quad t_2 = 2, \quad t_3 = -3 $$

Since all three roots are real numbers, any permutation of these three values \( \{1, 2, -3\} \) is a valid real solution \( (x, y, z) \) for the system.

Step 4: Count the total number of real solutions

The solutions \( (x, y, z) \) are the permutations of the set of roots \( \{1, 2, -3\} \).

The number of permutations of 3 distinct elements is:

$$ 3! = 3 \times 2 \times 1 = 6 $$

The 6 distinct real solutions are:

1. \( (1, 2, -3) \)
2. \( (1, -3, 2) \)
3. \( (2, 1, -3) \)
4. \( (2, -3, 1) \)
5. \( (-3, 1, 2) \)
6. \( (-3, 2, 1) \)

Answer:

6