Question

a=3+underoot5/2, then find the value of a2+1/a2

Answer 1

Question:

$Find \: the \: value \: of \: {a}^{2} + \frac{1}{ {a}^{2} }$

Answer:

$The \: value \: of \: \boxed{{a}^{2} + \frac{1}{ {a}^{2} } \: is \: 7}$

Step-by-step explanation:

$Given, \: a = \frac{3 + \sqrt{5} }{2}$

$\frac{1}{a} = \frac{2}{3 + \sqrt{5} }$

Now, rationalising the denominator

$= \frac{2(3 - \sqrt{5}) }{(3 + \sqrt{5})(3 - \sqrt{5}) }$

$= \frac{2(3 - \sqrt{5}) }{ {3}^{2} - {( \sqrt{5}) }^{2} }$

$= \frac{2(3 - \sqrt{5}) }{ 9 - 5 }$

$= \frac{2(3 - \sqrt{5}) }{ 4}$

$\boxed{\frac{1}{a} = \frac{(3 - \sqrt{5}) }{ 2}}$

Now,

$a + \frac{1}{a} = \frac{3 + \sqrt{5} }{2} + \frac{3 - \sqrt{5} }{2}$

$= \frac{3 + \sqrt{5} + 3 - \sqrt{5} }{2}$

$= \frac{6}{2}$

$a + \frac{1}{a} = 3$

Squaring on both sides

${(a + \frac{1}{a}) }^{2} = {3}^{2}$

${a}^{2} + 2 \times a \times \frac{1}{a} + \frac{1}{ {a}^{2} } = 9$

${a}^{2} + 2 + \frac{1}{ {a}^{2} } = 9$

${a}^{2} + \frac{1}{ {a}^{2} } = 9 - 2$

$\boxed{{a}^{2} + \frac{1}{ {a}^{2} } = 7}$