Question

plz.....anyone can solve this​

Answer 1

Answer

7

Explanation

Given that

$\sf a = \frac{3 + \sqrt{5}}{2}$

So,

$\sf \frac{1}{a} = \frac{2}{3 + \sqrt{5}}$

Rationalize it by multiplying with (3 - √5) on both numerator and denominator

$\sf \frac{1}{a} = \frac{2(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}$

This becomes

$\sf\frac{1}{a} = \frac{2(3-\sqrt{5})}{9-5}$

$\sf\frac{1}{a} = \frac{2(3-\sqrt{5})}{4}$

$\sf\frac{1}{a} = \frac{3 - \sqrt{5}}{2}$

Now,

a² + 1/a² = (a + 1/a)² - 2a × 1/a

→ a² + 1/a² = (a + 1/a)² - 2

So, $\sf a^2 + \frac{1}{a^2} = (\frac{3+\sqrt{5}}{2} + \frac{3-\sqrt{5}}{2})^2 - 2$

$\sf a^2 + \frac{1}{a^2} = (\frac{6}{2})^2 - 2$

$\sf a^2 + \frac{1}{a^2} = (3)^2 - 2$

$\sf a^2 + \frac{1}{a^2} = 7$

Answer 2

$a = \frac{3+\sqrt{5} }{2}$

$2a = 3+\sqrt{5}$

$2a - 3 = \sqrt{5}$

Squaring on both sides

$(2a-3)^2 = 5\\4a^2+9-12a = 5\\4a^2 = 12a-4$

$a^2 = 3a-1$

We need to find

$a^2+\frac{1}{a^2}$

Substitute the value of a²

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

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

            $= \frac{9a^2-6a+1+1}{3a-1}$

            $= \frac{9(3a-1)-6a+2}{3a-1}$

            $= \frac{21a-7}{3a-1}$

            $= \frac{7(3a-1)}{3a-1}$

            $= 7$

So the answer is 7