If a = 6 - [tex]\sqrt{35} [/tex], find the value of a² + [tex] \frac{1}{a²} [/tex]
a = 6 - [tex]\sqrt{35} [/tex] (Given)
Now, first of all we have to find the value of [tex] \frac{1}{a} [/tex].
So, [tex] \frac{1}{a} [/tex] = [tex] \frac{1}{6 \: - \: \sqrt{35}} [/tex]
[tex] \frac{1}{a} [/tex] = [tex] \frac{6 \: + \sqrt{35}}{(6 \: - \: \sqrt{35})(6 \: + \: \sqrt{35})} [/tex]
[tex] \frac{1}{a} [/tex] = [tex] \frac{6 \: + \sqrt{35}}{(6)² \: - \: (\sqrt{35})²} [/tex]
[tex] \frac{1}{a} [/tex] = [tex] \frac{6 \: + \sqrt{35}}{36 \: - \: 35} [/tex] [tex] (\because \sqrt{35} \: × \: \sqrt{35} \: = \: 35) [/tex]
[tex] \frac{1}{a} [/tex] = [tex] \frac{6 \: + \: \sqrt{35}}{1} [/tex]
[tex] \frac{1}{a} [/tex] = 6 + [tex]\sqrt{35} [/tex]
Now, we have to find the value of a² + [tex] \frac{1}{a²} [/tex]
So, a² + [tex] \frac{1}{a²} [/tex] = [tex] (a \: + \frac{1}{a})² [/tex] - 2.a.[tex] \frac{1}{a} [/tex]
a² + [tex] \frac{1}{a²} [/tex] = [tex] (6 \: - \sqrt{35} \: + \: 6 \: + \sqrt{35})² [/tex] - 2
a² + [tex] \frac{1}{a²} [/tex] = (12)² - 2
a² + [tex] \frac{1}{a²} [/tex] = 144 - 2
a² + [tex] \frac{1}{a²} [/tex] = 142
The value of a² + [tex] \frac{1}{a²} [/tex] is 142. [Answer]
