if a = 6-√35 find the value of a^2 + 1/a^2
asked by guest on Apr 10, 2025 at 3:03 am
MathBot Answer:
a2+1a2=(6−35)2+1(6−35)2{a}^{2} + \frac{1}{{a}^{2}} = \left(6 - \sqrt{35}\right)^{2} + \frac{1}{\left(6 - \sqrt{35}\right)^{2}}a2+a21=(6−35)2+(6−35)21
a2+1a2=(6−35)2+(6−35)−2=((6−35)2)−1+(6−35)2=(6−35)2+1(6−35)2\begin{aligned}a^{2} + \frac{1}{a^{2}}&=\left(6 - \sqrt{35}\right)^{2} + \left(6 - \sqrt{35}\right)^{-2}\\&=\left(\left(6 - \sqrt{35}\right)^{2}\right)^{-1} + \left(6 - \sqrt{35}\right)^{2}\\&=\left(6 - \sqrt{35}\right)^{2} + \frac{1}{\left(6 - \sqrt{35}\right)^{2}}\end{aligned}a2+a21=(6−35)2+(6−35)−2=((6−35)2)−1+(6−35)2=(6−35)2+(6−35)21
