Question

If a=3+√5/2,then find the value of a^2+1/a^2

Answer 1

ANSWER:

Given:

• a = (3 + √5)/2

To Find:

• Value of a^2 + 1/a^2

Solution:

We are given that,

$\implies a=\dfrac{3+\sqrt5}{2}$

We need to find value of,

$\implies a^2+\dfrac{1}{a^2}$

So,

$\implies a^2$

$\implies \left(\dfrac{3+\sqrt5}{2}\right)^2$

Using, (a+b)^2 = a^2 + b^2 + 2ab,

$\implies \dfrac{(3)^2+(\sqrt5)^2+2(3)(\sqrt5)}{2^2}$

$\implies \dfrac{9+5+6\sqrt5}{4}$

$\implies \dfrac{14+6\sqrt5}{4}$

$\implies \dfrac{7+3\sqrt5}{2}$

So,

$\implies\bf a^2= \dfrac{7+3\sqrt5}{2}- - - -(1)$

And,

$\implies \dfrac{1}{a^2}$

$\implies \dfrac{1}{\left(\frac{7+3\sqrt5}{2}\right)}$

$\implies \dfrac{2}{7+3\sqrt5}$

Rationalising,

$\implies \dfrac{2}{7+3\sqrt5}\times\dfrac{7-3\sqrt5}{7-3\sqrt5}$

$\implies \dfrac{2(7-3\sqrt5)}{(7+3\sqrt5)(7-3\sqrt5)}$

Using, (a+b)(a-b)=a^2-b^2,

$\implies \dfrac{14-6\sqrt5}{(7)^2-(3\sqrt5)^2}$

$\implies \dfrac{14-6\sqrt5}{49-45}$

$\implies \dfrac{14-6\sqrt5}{4}$

$\implies \dfrac{7-3\sqrt5}{2}$

So,

$\implies\bf \dfrac{1}{a^2}= \dfrac{7-3\sqrt5}{2}- - - -(2)$

Now,

$\implies a^2+\dfrac{1}{a^2}$

Using (1) & (2),

$\implies \dfrac{7+3\sqrt5}{2}+ \dfrac{7-3\sqrt5}{2}$

$\implies \dfrac{7+3\sqrt5+7-3\sqrt5}{2}$

Cancelling 3√5,

$\implies \dfrac{14}{2}$

$\implies 7$

Therefore,

$\implies a^2+\dfrac{1}{a^2} = 7$