Question

Math

Posted about 1 month ago

```
$\begin{array}{c}\text { 若 } m^{2}=m+1 \quad n^{2}=n+1 \\ \text { 求 } m+n=\end{array}$
```

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Answer from Sia

Posted about 1 month ago

Solution by Steps

step 1

Solve the equation $m^2 = m + 1$ for $m$

step 2

The solutions to the quadratic equation $m^2 - m - 1 = 0$ are $m = \frac{1}{2}(1 - \sqrt{5})$ and $m = \frac{1}{2}(1 + \sqrt{5})$

step 3

Solve the equation $n^2 = n + 1$ for $n$

step 4

The solutions to the quadratic equation $n^2 - n - 1 = 0$ are $n = \frac{1}{2}(1 - \sqrt{5})$ and $n = \frac{1}{2}(1 + \sqrt{5})$

step 5

Since $m$ and $n$ have the same solutions, we can write $m + n = \left(\frac{1}{2}(1 - \sqrt{5})\right) + \left(\frac{1}{2}(1 + \sqrt{5})\right)$

step 6

Simplifying, we get $m + n = \frac{1}{2}(1 - \sqrt{5} + 1 + \sqrt{5}) = \frac{1}{2}(2) = 1$

Answer

$m + n = 1$

Key Concept

Solving quadratic equations

Explanation

The key concept here is solving quadratic equations of the form $ax^2 + bx + c = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

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