If p-1 upon p = 3 then find the value of p^2+1 upon p^2

asked by guest
on Sep 21, 2024 at 7:38 am



You asked:

Given \(p - \frac{1}{p} = 3\) evaluate the expression \({p}^{2} + \frac{1}{{p}^{2}}\).

MathBot Answer:

\(p^{2} + \frac{1}{p^{2}} \in \left\{\left(\frac{3}{2} - \frac{\sqrt{13}}{2}\right)^{2} + \frac{1}{\left(\frac{3}{2} - \frac{\sqrt{13}}{2}\right)^{2}}, \frac{1}{\left(\frac{3}{2} + \frac{\sqrt{13}}{2}\right)^{2}} + \left(\frac{3}{2} + \frac{\sqrt{13}}{2}\right)^{2}\right\}\)


The 2 solutions to the equation are: \[\begin{aligned}p &= \frac{3}{2} - \frac{\sqrt{13}}{2} \approx -0.30277564\\p &= \frac{3}{2} + \frac{\sqrt{13}}{2} \approx 3.3027756\end{aligned}\]

By substituting this value into the expression, we find that: $$\begin{aligned}p^{2} + \frac{1}{p^{2}} &= \left(\frac{3}{2} - \frac{\sqrt{13}}{2}\right)^{2} + \frac{1}{\left(\frac{3}{2} - \frac{\sqrt{13}}{2}\right)^{2}}\\&= \left(\frac{3}{2} - \frac{\sqrt{13}}{2}\right)^{2} + \frac{1}{\left(\frac{3}{2} - \frac{\sqrt{13}}{2}\right)^{2}}\end{aligned}$$