Question

If a = 2 + root 3, find the value of a+1/a and a 2 +1/a 2

Answer 1

a=2+√3
∴, 1/a=1/(2+√3)=(2-√3)/(2+√3)(2-√3)=(2-√3)/(4-3)=2-√3
∴, a+1/a=2+√3+2-√3=4
now, a²=(2+√3)²=4+4√3+3=7+4√3
1/a²=(1/a)²=(2-√3)²=4-4√3+3=7-4√3
∴, a²+1/a²=7+4√3+7-4√3=14

Answer 2

Answer:

The value of $a + \frac { 1 } { a } \text { and } a ^ { 2 } + \frac { 1 } { a ^ { 2 } }$ is 4 and 14 respectively.

Solution:

Given the value of  

$a=2+\sqrt3$

First to find the value of $a+\frac {1}{a}$

We know that,  

$\begin{aligned} a & = 2 + \sqrt { 3 } \\\\ \frac { 1 } { a } & = \frac { 1 } { 2 + \sqrt { 3 } } \end{aligned}$

Multiplying and dividing it by $(2-\sqrt3)$

$\begin{array} { c } { \frac { 1 } { a } = \frac { 1 } { 2 + \sqrt { 3 } } \times \frac { ( 2 - \sqrt { 3 } ) } { 2 - \sqrt { 3 } } } \\\\ { = \frac { ( 2 - \sqrt { 3 } ) } { 2 ^ { 2 } - ( \sqrt { 3 } ) ^ { 2 } } } \\\\ { = \frac { ( 2 - \sqrt { 3 } ) } { 4 - 3 } } \\\\ { = ( 2 - \sqrt { 3 } ) } \end{array}$

Hence, the value of $a+\frac {1}{a}$

$a + \frac { 1 } { a } = 2 + \sqrt { 3 } + 2 - \sqrt { 3 } = 4$

Now, the value of $a^2+\frac {1}{a^2}$

$\begin{aligned} a ^ { 2 } = ( 2 + \sqrt { 3 } ) ^ { 2 } & = 4 + 3 + 4 \sqrt { 3 } = 7 + 4 \sqrt { 3 } \\\\ \frac { 1 } { a ^ { 2 } } & = \frac { 1 } { 7 + 4 \sqrt { 3 } } \end{aligned}$

Multiplying and Dividing it by $7-4\sqrt3$

$\begin{array} { c } { \frac { 1 } { a ^ { 2 } } = \frac { 1 } { 7 + 4 \sqrt { 3 } } \times \frac { ( 7 - 4 \sqrt { 3 } ) } { 7 - 4 \sqrt { 3 } } } \\\\ { = \frac { 7 - 4 \sqrt { 3 } } { 7 ^ { 2 } - ( 4 \sqrt { 3 } ) ^ { 2 } } } \\\\ { = \frac { ( 7 - 4 \sqrt { 3 } ) } { 49 - 48 } } \\\\ { = \frac { 7 - 4 \sqrt { 3 } } { 1 } } \\\\ { = 7 - 4 \sqrt { 3 } } \end{array}$

Thus, the value of

$a ^ { 2 } + \frac { 1 } { a ^ { 2 } } = 7 + 4 \sqrt { 3 } + 7 - 4 \sqrt { 3 } = 14$

Therefore, the values of $a + \frac { 1 } { a } \text { and } a ^ { 2 } + \frac { 1 } { a ^ { 2 } }$ is 4 and 14 respectively.