How to find b when 1/b-5 - (10/b^2-5b+25) = 1/b+5

asked by guest
on Oct 20, 2024 at 10:21 pm



You asked:

Solve the equation \(\frac{1}{b} - 5 - \left(\frac{10}{{b}^{2}} - 5 b + 25\right) = \frac{1}{b} + 5\) for the variable \(b\).

MathBot Answer:

The 3 solutions to the equation are: \[\begin{aligned}b &= \frac{7}{3} + \sqrt[3]{\frac{370}{27} + \frac{\sqrt{2139}}{9}} + \frac{49}{9 \sqrt[3]{\frac{370}{27} + \frac{\sqrt{2139}}{9}}} \approx 7.0403498\\b &= - \frac{\sqrt[3]{\frac{\sqrt{2139}}{9} + \frac{370}{27}}}{2} - \frac{49}{18 \sqrt[3]{\frac{\sqrt{2139}}{9} + \frac{370}{27}}} + \frac{7}{3} + i \left(- \frac{49 \sqrt{3}}{18 \sqrt[3]{\frac{\sqrt{2139}}{9} + \frac{370}{27}}} + \frac{\sqrt{3} \sqrt[3]{\frac{\sqrt{2139}}{9} + \frac{370}{27}}}{2}\right) \approx -0.020174906 + 0.53260658 i\\b &= - \frac{\sqrt[3]{\frac{\sqrt{2139}}{9} + \frac{370}{27}}}{2} - \frac{49}{18 \sqrt[3]{\frac{\sqrt{2139}}{9} + \frac{370}{27}}} + \frac{7}{3} + i \left(- \frac{\sqrt{3} \sqrt[3]{\frac{\sqrt{2139}}{9} + \frac{370}{27}}}{2} + \frac{49 \sqrt{3}}{18 \sqrt[3]{\frac{\sqrt{2139}}{9} + \frac{370}{27}}}\right) \approx -0.020174906 -0.53260658 i\end{aligned}\]

\(i\) is the imaginary unit, defined as \(i^2 = -1\).

+ Submit Question