Question

$if a=3+root5/2, then find the value of a^2+1/a^2$

Answer 1

a = ( 3 +√5)/2

then ,
1/a = (3-√5)/2

now,
a² + 1/a² = ( a +1/a)² -2
={( 3 +√5)/2+(3-√5)/2}² -2
={3}² -2
=7

Answer 2

Answer:

$a^2+\frac{1}{a^2}=\frac{55+21\sqrt5}{7+3\sqrt5}$

Step-by-step explanation:

Given : If $a=\frac{3+\sqrt5}{2}$

To find : The value of $a^2+\frac{1}{a^2}$

Solution :

Expression  $a^2+\frac{1}{a^2}$

Substitute,  $a=\frac{3+\sqrt5}{2}$

$=(\frac{3+\sqrt5}{2})^2+\frac{1}{(\frac{3+\sqrt5}{2})^2}$

$=\frac{9+5+6\sqrt5}{4}+\frac{1}{\frac{9+5+6\sqrt5}{4}}$

$=\frac{7+3\sqrt5}{2}+\frac{1}{\frac{7+3\sqrt5}{2}}$

$=\frac{7+3\sqrt5}{2}+\frac{2}{7+3\sqrt5}$

$=\frac{(7+3\sqrt5)(7+3\sqrt5)+4\times 4}{2(7+3\sqrt5)}$

$=\frac{49+45+42\sqrt5+16}{2(7+3\sqrt5)}$

$=\frac{110+42\sqrt5}{2(7+3\sqrt5)}$

$=\frac{2(55+21\sqrt5)}{2(7+3\sqrt5)}$

$=\frac{55+21\sqrt5}{7+3\sqrt5}$

Therefore, $a^2+\frac{1}{a^2}=\frac{55+21\sqrt5}{7+3\sqrt5}$