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: Symmetric Systems and Vieta's Formulas
Using roots of a cubic polynomial to solve symmetric systems.

Step 1: Identify symmetric sums

We are given the system of equations:

$$
LATEXBLOCK0
$$

We can relate these variables to the coefficients of a cubic polynomial whose roots are \(x\), \(y\), and \(z\). Let this polynomial be:

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

where the elementary symmetric polynomials are:

$$
LATEXBLOCK1
$$

From the system, we already have:

$$ e_1 = 0 $$

$$ e_3 = -6 $$

Step 2: Find the pairwise sum \(e_2\)

We use 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: Construct and solve the cubic equation

The cubic polynomial with roots \(x\), \(y\), and \(z\) is:

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

$$ t^3 - 7t + 6 = 0 $$

We find the roots of this polynomial. By inspection, \(t = 1\) is a root because:

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

We can factor out \((t - 1)\) using polynomial division or synthetic division:

$$ t^3 - 7t + 6 = (t - 1)(t^2 + t - 6) = 0 $$

Factor the quadratic part:

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

Thus, the roots of the cubic equation are:

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

Step 4: Determine the number of real solutions

The values of \(x\), \(y\), and \(z\) must be a permutation of the set of roots \(\{1, 2, -3\}\).

Since all three roots are distinct and real, any permutation of these three values yields a unique real solution triplet \((x, y, z)\).

The number of permutations of 3 distinct elements is:

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

The 6 real solutions are:

$$ (1, 2, -3), \quad (1, -3, 2), \quad (2, 1, -3), \quad (2, -3, 1), \quad (-3, 1, 2), \quad (-3, 2, 1) $$

Answer:

6