4r^3 = (10^2)*30

asked by guest
on Oct 24, 2024 at 8:31 pm



You asked:

Solve the equation \(4 \cdot {r}^{3} = {10}^{2} \cdot 30\) for the variable \(r\).

MathBot Answer:

The solutions are given by the formula: \[r = 5 \cdot \sqrt[3]{6} \cos{\left(\frac{2 \pi k}{3} \right)} + 5 \cdot \sqrt[3]{6} \sin{\left(\frac{2 \pi k}{3} \right)} i\] where \(k\) is an integer in \(0 \le k < 3\).

There is \(1\) real solution.

There are \(2\) solutions with nonzero imaginary part.

Real solutions: \[\begin{aligned}r &= 5 \cdot \sqrt[3]{6} \\&\approx 9.085603\end{aligned}\]

Solutions with nonzero imaginary part: \[\begin{aligned}r &= - \frac{5 \cdot \sqrt[3]{6}}{2} + \frac{5 \cdot \sqrt[3]{2} \cdot 3^{\frac{5}{6}} i}{2} \\&\approx -4.5428015 + 7.868363 i\\r &= - \frac{5 \cdot \sqrt[3]{6}}{2} - \frac{5 \cdot \sqrt[3]{2} \cdot 3^{\frac{5}{6}} i}{2} \\&\approx -4.5428015 -7.868363 i\end{aligned}\]

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

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