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Equation ... finding x
- Thread starter t*s*
- Start date
Hello, t*s*!
I don't think it can be done . . .
\(\displaystyle \text{We have: }\:\phi \;=\;\dfrac{b^2x}{u}\cdot\left(\dfrac{b^2x}{b(1+2x)}\right)^{\frac{2}{3}}\cdot \sqrt{y}\)
. . . . . . . . \(\displaystyle \displaystyle \phi \;=\;\frac{b^2x}{u}\cdot\frac{b^{\frac{4}{3}}x^{\frac{2}{3}}}{b^{\frac{2}{3}}(1+2x)^{\frac{2}{3}}}\cdot \sqrt{y} \)
. . . . . . . . \(\displaystyle \displaystyle \phi \;=\;\frac{b^{\frac{8}{3}}x^{\frac{5}{3}}y^{\frac{1}{2}}}{u(1+2x)^{\frac{2}{3}}} \)
. . . . . . . . \(\displaystyle \phi u (1+2x)^{\frac{2}{3}} \;=\;b^{\frac{8}{3}}y^{\frac{1}{2}}x^{\frac{5}{3}} \)
I see no way to solve for \(\displaystyle x\).
I don't think it can be done . . .
\(\displaystyle \phi \;=\; \dfrac{b^2x}{u} \cdot \left(\dfrac{b^2x}{b+2bx}\right)^{\frac{2}{3}}\cdot\sqrt{y} \qquad \text{ Solve for }x. \)
\(\displaystyle \text{We have: }\:\phi \;=\;\dfrac{b^2x}{u}\cdot\left(\dfrac{b^2x}{b(1+2x)}\right)^{\frac{2}{3}}\cdot \sqrt{y}\)
. . . . . . . . \(\displaystyle \displaystyle \phi \;=\;\frac{b^2x}{u}\cdot\frac{b^{\frac{4}{3}}x^{\frac{2}{3}}}{b^{\frac{2}{3}}(1+2x)^{\frac{2}{3}}}\cdot \sqrt{y} \)
. . . . . . . . \(\displaystyle \displaystyle \phi \;=\;\frac{b^{\frac{8}{3}}x^{\frac{5}{3}}y^{\frac{1}{2}}}{u(1+2x)^{\frac{2}{3}}} \)
. . . . . . . . \(\displaystyle \phi u (1+2x)^{\frac{2}{3}} \;=\;b^{\frac{8}{3}}y^{\frac{1}{2}}x^{\frac{5}{3}} \)
I see no way to solve for \(\displaystyle x\).
D
Deleted member 4993
Guest
If you "cube" both sides - then you would get a "quintic" and there is no general analytic (closed form) solution for those types of equations.