1/X^2 - 1/(X+1)^2 =3/4

asked by guest
on Oct 08, 2024 at 7:52 am



You asked:

Solve the equation \(\frac{1}{{X}^{2}} - \frac{1}{{\left( X + 1 \right)}^{2}} = \frac{3}{4}\) for the variable \(X\).

MathBot Answer:

The 4 solutions to the equation are: \[\begin{aligned}X &= 1\\X &= -1 + \sqrt[3]{\frac{1}{3} + \frac{2 \sqrt{3}}{9}} - \frac{1}{3 \sqrt[3]{\frac{1}{3} + \frac{2 \sqrt{3}}{9}}} \approx -0.47666449\\X &= -1 - \frac{\sqrt[3]{\frac{1}{3} + \frac{2 \sqrt{3}}{9}}}{2} + \frac{1}{6 \sqrt[3]{\frac{1}{3} + \frac{2 \sqrt{3}}{9}}} + i \left(\frac{\sqrt{3}}{6 \sqrt[3]{\frac{1}{3} + \frac{2 \sqrt{3}}{9}}} + \frac{\sqrt{3} \sqrt[3]{\frac{1}{3} + \frac{2 \sqrt{3}}{9}}}{2}\right) \approx -1.2616678 + 1.0979117 i\\X &= -1 - \frac{\sqrt[3]{\frac{1}{3} + \frac{2 \sqrt{3}}{9}}}{2} + \frac{1}{6 \sqrt[3]{\frac{1}{3} + \frac{2 \sqrt{3}}{9}}} + i \left(- \frac{\sqrt{3} \sqrt[3]{\frac{1}{3} + \frac{2 \sqrt{3}}{9}}}{2} - \frac{\sqrt{3}}{6 \sqrt[3]{\frac{1}{3} + \frac{2 \sqrt{3}}{9}}}\right) \approx -1.2616678 -1.0979117 i\end{aligned}\]

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

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