a=6-root 35 find a^2+1/a^2

Here, a = 6 - √35

$\begin{aligned} & \Rightarrow 1 / a=1 /(6-\sqrt{35}) \\ & \Rightarrow 1 / a=(6+\sqrt{35}) /[(6+\sqrt{35})(6-\sqrt{35})] \end{aligned}$

[By multiplying (6 + √35) in both numerator & denominator]

$\begin{aligned} & \Rightarrow 1 / a=(6+\sqrt{35}) /\left(6^2-(\sqrt{35})^2\right) \\ & \Rightarrow 1 / a=(6+\sqrt{35})(36-35) \\ & \Rightarrow 1 / a=(6+\sqrt{35}) / 1 \\ & \Rightarrow 1 / a=6+\sqrt{35} \end{aligned}$

Now we have to find value of a² + 1/a²

$\begin{aligned} & \Rightarrow a^2+1 / a^2=(a)^2+(1 / a)^2 \\ & \Rightarrow(6-\sqrt{ } 35)^2+(6+\sqrt{ } 35)^2 \\ & \Rightarrow\left(6^2-2 \times 6 \sqrt{35}+35\right)+\left(6^2+2 \times 6 \sqrt{35}+35\right) \\ & \Rightarrow 36-12 \sqrt{ } 35+35+36+12 \sqrt{ } 35+35 \\ & \Rightarrow 72+70 \\ & \Rightarrow 142 \end{aligned}$