Question

if a= 2-√3 then find a^2 + 1/a^2​

Answer 1

Answer:

$\sf{The \ value \ of \ a^{2}+\dfrac{1}{a^{2}} \ is \ 14.}$

Given:

$\sf{a=2-\sqrt3}$

To find:

$\sf{The \ value \ of \ a^{2}+\dfrac{1}{a^{2}}.}$

Solution:

$\sf{a+\dfrac{1}{a}=2-\sqrt3+\dfrac{1}{2-\sqrt3}}$

$\sf{\therefore{a+\dfrac{1}{a}=2-\sqrt3+\dfrac{1(2+\sqrt3}{(2-\sqrt3)(2+\sqrt3)}}}$

$\sf{\therefore{a+\dfrac{1}{a}=2-\sqrt3+\dfrac{2+\sqrt3}{4-3}}}$

$\sf{\therefore{a+\dfrac{1}{a}=2-\sqrt3+2+\sqrt3}}$

$\sf{\therefore{a+\dfrac{1}{a}=4...(1)}}$

__________________

$\sf{a^{2}+b^{2}=(a+b)^{2}-2ab}$

$\sf{Here, \ a=a \ and \ b=\dfrac{1}{a}}$

$\sf{\therefore{a^{2}+\dfrac{1}{a^{2}}=(a+\dfrac{1}{a})^{2}-2(a\times\dfrac{1}{a})}}$

$\sf{...from \ equation(1)}$

$\sf{a^{2}+\dfrac{1}{a^{2}}=4^{2}-2}$

$\sf{\therefore{a^{2}+\dfrac{1}{a^{2}}=14}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ a^{2}+\dfrac{1}{a^{2}} \ is \ 14.}}}$