Question

If a+1/a = 1/(√3)-1, then the value of a²+1/a² is -
a. 3
b. -1/√3/2
c. 2√3
d.4√3​

Answer 1

c. 2√3

a² + 1/a² = 7 Ans.

$a2−1a2=(a+1a)(a−1a)=5(a+1a) a−1a=5 ⟹a2+1a2−2=25 ⟹a2+1a2−2+4=25+4 ⟹a2+1a2+2=29 ⟹a+1a=29−−√ ∴a2−1a2=529−−√ .$

Answer 2

Given:

The value of$a + \frac{1}{a} = \frac{1}{\sqrt{3}-1 }$.

To Find:

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

Solution:

1. It is given that the value of$a + \frac{1}{a} = \frac{1}{\sqrt{3}-1 }$.

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

Consider RHS of the above equation, It can also be written as follows,

=>$\frac{1}{\sqrt{3}-1 }$,

=>$\frac{\sqrt{3}+1 }{(\sqrt{3}-1)*\sqrt{3}+1 }$   (Multiplying the numerator and denominator with$\sqrt{3}+1$.

=>$\frac{\sqrt{3}+1 }{2}$. ( Rationalisation)

2. Now the value$a+\frac{1}{a}$ becomes$\frac{\sqrt{3} +1}{2}$,

=> Apply square on both sides, we get

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

=>$a^2 + \frac{1}{a^2} +2 = \frac{3+1+2\sqrt{3} }{4}$,

=>$a^2 + \frac{1}{a^2} = 1+\frac{\sqrt{3} }{2}-2$,

=>$a^2+\frac{1}{a^2}= -1+\frac{\sqrt{3} }{2}$.

Therefore, The value of$a^2+\frac{1}{a^2}$ is$-1 +\frac{\sqrt{3} }{2}$. Option B is the correct option.