( 0.112/y)^3/2 = 4y/4+2y

asked by guest
on Mar 21, 2025 at 11:14 am



You asked:

Solve the equation (0.112y)32=4y4+2y\frac{{\left( \frac{0.112}{y} \right)}^{3}}{2} = \frac{4 y}{4} + 2 y for the variable yy.

MathBot Answer:

The 4 solutions to the equation are: y=−2⋅10534375≈−0.12370171y=2⋅10534375≈0.12370171y=−105342375i≈−0.12370171iy=105342375i≈0.12370171i\begin{aligned}y &= - \frac{\sqrt{2} \cdot 105^{\frac{3}{4}}}{375} \approx -0.12370171\\y &= \frac{\sqrt{2} \cdot 105^{\frac{3}{4}}}{375} \approx 0.12370171\\y &= - \frac{105^{\frac{3}{4}} \sqrt{2}}{375} i \approx -0.12370171 i\\y &= \frac{105^{\frac{3}{4}} \sqrt{2}}{375} i \approx 0.12370171 i\end{aligned}

ii is the imaginary unit, defined as i2=−1i^2 = -1.

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