Question

$if \: a = 6 - \sqrt{35} \: find \: the \: value \: of \: a {}^{2} + \frac{1}{a { }^{2} }$

Answer 1

Hi friend, Harish here.

Here is your answer:

Given that,

a = 6 - √35.

To find,

$a^{2} + \frac{1}{a^{2}}$

Solution:

We know that:

$a = 6 - \sqrt{35}$

So,

$\frac{1}{a} = \frac{1}{6- \sqrt{35} }$  

(Now, rationalize the above equation).

To rationalize we must multiply and divide the number by it's conjugate.

Conjugate of the number is 6 + √35.

Then,

$\frac{1}{a}= \frac{1}{6- \sqrt{35} } \times \frac{6+ \sqrt{35} }{6+ \sqrt{35} }$

We know that (a+b) ( a-b) = a² - b².

So, (6 - √35) (6+√35) = 6² - (√35)² = 36 - 35 = 1.

Then,

$\frac{1}{a}= \frac{6+ \sqrt{35} }{(6- \sqrt{35})(6+ \sqrt{35})} = \frac{6+ \sqrt{35}}{1} = 6+ \sqrt{35}$

(Now,Adding both a & 1/a )

Then,

$(a+ \frac{1}{a}) = [(6- \sqrt{35}) + (6+ \sqrt{35})]$

→ $(a+ \frac{1}{a}) = 6 + 6$

→ $(a+ \frac{1}{a}) = 12$

Now, square on both sides.

Then,

$(a+ \frac{1}{a})^{2} = 12^{2}$

(Use ( a + b)² = a² + b² + 2ab)

→ $[a^{2} + \frac{1}{a^{2}}+2(a)( \frac{1}{a})] = 144$

→ $(a^{2} + \frac{1}{a^{2}}+2) = 144$

→ $(a^{2} + \frac{1}{a^{2}})=144- 2$

→ $(a^{2} + \frac{1}{a^{2}})= 142$

Therefore the value of it is 142.
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Hope my answer is helpful to you.