if m+n=8 and mn=8 then find out the value of (m-n)^2
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MathBot Answer:
The system of equations has \(2\) solutions.
\[m = 4 - 2 \sqrt{2}, n = 2 \sqrt{2} + 4\]\[m = 2 \sqrt{2} + 4, n = 4 - 2 \sqrt{2}\]
By substituting these values into the expression, we find that: $$\begin{aligned}\left(m - n\right)^{2} &= 32\end{aligned}$$
Solve \(m + n = 8\) for \(m\). \[m = 8 - n\]Substitute \(8 - n\) for \(m\) in \(m n = 8\) and simplify. $$\begin{aligned}m n &= 8 \\ \left(8 - n\right) n &= 8 \\ n^{2} - 8 n &= -8 \\n^{2} - 8 n + 8 &= 0 \\ n &= \frac{-(-8) \pm \sqrt{(-8)^{2} - 4(1)(8)}}{2(-8)} \\ n = 4 - 2 \sqrt{2}&, n = 2 \sqrt{2} + 4\end{aligned}$$Substitute \(4 - 2 \sqrt{2}\) into \(m + n = 8\) to solve for \(m\). \[\begin{aligned}m - 2 \sqrt{2} + 4 &= 8\\m + \left(4 - 2 \sqrt{2}\right) &= 8\\m &= 2 \sqrt{2} + 4\end{aligned}\]This yields the following solution. $$\begin{aligned}m = 2 \sqrt{2} + 4,\,n = 4 - 2 \sqrt{2}\end{aligned}$$Substitute \(2 \sqrt{2} + 4\) into \(m + n = 8\) to solve for \(m\). \[\begin{aligned}m + 2 \sqrt{2} + 4 &= 8\\m + \left(2 \sqrt{2} + 4\right) &= 8\\m &= 4 - 2 \sqrt{2}\end{aligned}\]This yields the following solution. $$\begin{aligned}m = 4 - 2 \sqrt{2},\,n = 2 \sqrt{2} + 4\end{aligned}$$
By substituting these values into the expression, we find that: $$\begin{aligned}\left(m - n\right)^{2} &= 32\end{aligned}$$