Question

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$$\begin{aligned} x + y &= 4 \\\\ xy &= 16 \\end{aligned}$$

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Explanation:

构造一元二次方程

$$ t^2 - (x+y)t + xy = 0 $$

$$ t^2 - 4t + 16 = 0 $$

利用求根公式求解

$$ t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times 16}}{2 \times 1} $$

$$ t = \frac{4 \pm \sqrt{16 - 64}}{2} $$

$$ t = \frac{4 \pm \sqrt{-48}}{2} $$

$$ t = \frac{4 \pm 4\sqrt{3}i}{2} $$

$$ t = 2 \pm 2\sqrt{3}i $$

确定方程组的解

$$
LATEXBLOCK0
\quad \text{或} \quad
LATEXBLOCK1
$$

Answer:

$$
LATEXBLOCK0
\quad \text{或} \quad
LATEXBLOCK1
$$