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

Question

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

Accepted Answer

∻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 $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.⚹◊What are the possible values of $m$ and $n$ if $m^2 = m + 1$ and $n^2 = n + 1$⍭ Generate me a similar question◊

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