# how to solve an equation by using trial and error method?

์กฐํ ์: 20 (์ต๊ทผ 30์ผ)
Ali Deniz 2022๋ 5์ 16์ผ
๋๊ธ: Ali Deniz 2022๋ 5์ 16์ผ
๐โ๐๐๐๐ก (๐ดโ)=647.5 ๐๐2
A=1027.5 mm^2
How can I calculate ๐๐ฅ value by using trial and error method or how can I find the value of Mx by another method. Thank you.
##### ๋๊ธ ์: 0์ด์  ๋๊ธ -2๊ฐ ํ์์ด์  ๋๊ธ -2๊ฐ ์จ๊ธฐ๊ธฐ

๋๊ธ์ ๋ฌ๋ ค๋ฉด ๋ก๊ทธ์ธํ์ญ์์ค.

### ์ฑํ๋ ๋ต๋ณ

Sam Chak 2022๋ 5์ 16์ผ
I'm not sure what the trial-and-error method is for this, but I think that basic Algebraic methods can be used to rearrange the equation. I followed the example in fzero and make an assumption:
% Assume
gamma = 1.55526564459735;
% function to be solved
f = @(M) (1/M)*((2*(1 + (gamma - 1)/2*M^2))/(gamma + 1))^((gamma + 1)/(2*(gamma - 1))) - Ax/Astar;
% initial guess
M0 = 3;
% root-finding solver
Msol = fzero(f, M0)
There should be two real roots.
##### ๋๊ธ ์: 1์ด์  ๋๊ธ -1๊ฐ ํ์์ด์  ๋๊ธ -1๊ฐ ์จ๊ธฐ๊ธฐ
Ali Deniz 2022๋ 5์ 16์ผ
Thank you.
-Ali

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### ์ถ๊ฐ ๋ต๋ณ (1๊ฐ)

Walter Roberson 2022๋ 5์ 16์ผ
Multiply both sides by Mx. Subtract the left side from the right. Expand out. You now have a polynomial of degree 2 in Mx.
##### ๋๊ธ ์: 2์์ ํ์์์ ์จ๊ธฐ๊ธฐ
Ali Deniz 2022๋ 5์ 16์ผ
I think you neglect the power of the RHS.
Walter Roberson 2022๋ 5์ 16์ผ
Ah, I am on mobile. I had tried scrolling the image but it didn't scroll so I didn't see the power.

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