Question

if a=3+√5/2 then find the value of a2 + 1/a2 I need full explanation

Answer 1

A = 3+✓5/2 

Therefore,

a²+1/a² = (3+✓5)²+1/(3+✓5)² 

= 9+5/4+1/9+5/4

= 14+4/4/9+5/4

= 18/14

= 9/7

Answer 2

Answer:

Value of $a^2+\frac{1}{a^2}\:\:is\:\:\frac{133+45\sqrt{5}}{34}$

Step-by-step explanation:

Given value of $a=\frac{3+\sqrt{5}}{2}$

We need to find value of $a^2+\frac{1}{a^2}$

First we find ,

$a^2=(\frac{3+\sqrt{5}}{2})^2=\frac{(3+\sqrt{5})^2}{4}=\frac{3^2+(\sqrt{5})^2+2\times3\times\sqrt{5}}{4}=\frac{9+5+6\sqrt{5}}{4}=\frac{14+6\sqrt{5}}{4}=\frac{7+3\sqrt{5}}{2}$

Now, $\frac{1}{a^2}=\frac{2}{7+3\sqrt{5}}=\frac{2}{7+3\sqrt{5}}\times\frac{7-3\sqrt{5}}{7-3\sqrt{5}}=\frac{2(7-3\sqrt{5})}{(7+3\sqrt{5})(7-3\sqrt{5})}$

$=\frac{2(7-3\sqrt{5})}{49-3\times5}=\frac{2(7-3\sqrt{5})}{34}=\frac{7-3\sqrt{5}}{17}$

Fionally,

$a^2+\frac{1}{a^2}=\frac{7+3\sqrt{5}}{2}+\frac{7-3\sqrt{5}}{17}=\frac{17(7+3\sqrt{5})+2(7-3\sqrt{5})}{34}=\frac{119+51\sqrt{5}+14-6\sqrt{5}}{34}=\frac{133+45\sqrt{5}}{34}$

Therefore, Value of $a^2+\frac{1}{a^2}\:\:is\:\:\frac{133+45\sqrt{5}}{34}$