y^2+1/4y^2=10,y^5-1/32y^5=?

asked by guest
on Nov 28, 2024 at 5:46 am



You asked:

Given \({y}^{2} + \frac{1}{4 {y}^{2}} = 10\) evaluate the expression \({y}^{5} - \frac{1}{32 {y}^{5}}\).

MathBot Answer:

\(y^{5} - \frac{1}{32 y^{5}} \in \left\{- \frac{1}{32 \left(- \frac{3}{2} + \frac{\sqrt{11}}{2}\right)^{5}} + \left(- \frac{3}{2} + \frac{\sqrt{11}}{2}\right)^{5}, \left(\frac{3}{2} - \frac{\sqrt{11}}{2}\right)^{5} - \frac{1}{32 \left(\frac{3}{2} - \frac{\sqrt{11}}{2}\right)^{5}}, - \frac{1}{32 \left(\frac{3}{2} + \frac{\sqrt{11}}{2}\right)^{5}} + \left(\frac{3}{2} + \frac{\sqrt{11}}{2}\right)^{5}, \left(- \frac{\sqrt{11}}{2} - \frac{3}{2}\right)^{5} - \frac{1}{32 \left(- \frac{\sqrt{11}}{2} - \frac{3}{2}\right)^{5}}\right\}\)

The 4 solutions to the equation are: \[\begin{aligned}y &= - \frac{3}{2} + \frac{\sqrt{11}}{2} \approx 0.1583124\\y &= \frac{3}{2} - \frac{\sqrt{11}}{2} \approx -0.1583124\\y &= \frac{3}{2} + \frac{\sqrt{11}}{2} \approx 3.1583124\\y &= - \frac{3}{2} - \frac{\sqrt{11}}{2} \approx -3.1583124\end{aligned}\]

By substituting this value into the expression, we find that: $$\begin{aligned}y^{5} - \frac{1}{32 y^{5}} &= - \frac{1}{32 \left(- \frac{3}{2} + \frac{\sqrt{11}}{2}\right)^{5}} + \left(- \frac{3}{2} + \frac{\sqrt{11}}{2}\right)^{5}\\&= - \frac{1}{32 \left(- \frac{3}{2} + \frac{\sqrt{11}}{2}\right)^{5}} + \left(- \frac{3}{2} + \frac{\sqrt{11}}{2}\right)^{5}\end{aligned}$$

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