X^(4/3)-10X^(2/3)+9=0
asked by guest on Nov 16, 2024 at 4:56 am
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MathBot Answer:
The real solutions are: X={1,27}≈{1.0,27.0}X={X | X∈{−27,−1}∧9+X43−10X23=0}≈∅\begin{aligned}X &= \left\{1, 27\right\} \approx \left\{1.0, 27.0\right\}\\X &= \left\{X\; \middle|\; X \in \left\{-27, -1\right\} \wedge 9 + X^{\frac{4}{3}} - 10 X^{\frac{2}{3}} = 0 \right\} \approx \emptyset\end{aligned}XX={1,27}≈{1.0,27.0}={XX∈{−27,−1}∧9+X34−10X32=0}≈∅