(√(m^2 −4m+4)) + ((m−2n+8))^(1/4) = 0 n = ?


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Question Number 182999 by Shrinava last updated on 18/Dec/22

$\sqrt{m^{2} - 4 m + 4} + \sqrt[4]{m - 2 n + 8} = 0$ $n = ?$
	Answered by cortano1 last updated on 18/Dec/22
	$∣m−2∣ +((m−2n+8))^(1/4) = 0 { ((m=2)),((((10−2n))^(1/4) = 0 ⇒n=5 )) :}$
	$∣ m - 2 ∣ + \sqrt[4]{m - 2 n + 8} = 0$ ${\begin{cases} m = 2 \\ \sqrt[4]{10 - 2 n} = 0 \Rightarrow n = 5 \end{cases}$
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