Question

If a= (3-√5)/(3+√5) and b=(3+√5)/(3-√5) then find a²+b²​

Answer 1

Answer :-

Let's first find the value of ' a² ' :-

$\implies\sf a = \dfrac{3-\sqrt5}{3+\sqrt5}$

Rationalizing the denominator :-

$\implies\sf a = \dfrac{(3-\sqrt5)}{(3+\sqrt5)} \times \dfrac{(3-\sqrt5)}{(3-\sqrt5)}$

$\implies\sf a = \dfrac{(3- \sqrt5)^2}{3^2 - (\sqrt5)^2}$

$\implies\sf a= \dfrac{9 + 5 - 2 \times 3 \times \sqrt5}{9-5}$

$\implies\sf a = \dfrac{14 - 6\sqrt5}{4}$

$\implies\sf a = \dfrac{2(7-3\sqrt5)}{4}$

$\implies\sf a = \dfrac{7-3\sqrt5}{2}$

Squaring on both side :-

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

$\implies\sf a^2 = \dfrac{7^2 + (3\sqrt5)^2 - 2 \times 7 \times 3\sqrt5}{4}$

$\implies\sf a^2 = \dfrac{49 + 48 - 42\sqrt5}{4}$

$\implies\sf a^2 = \dfrac{97-42\sqrt5}{4}$

Calculating the value of ' b² ' :-

$\implies\sf b = \dfrac{3+\sqrt5}{3-\sqrt5}$

Rationalizing the denominator :-

$\implies\sf b = \dfrac{(3+\sqrt5)}{(3-\sqrt5)} \times \dfrac{(3+\sqrt5)}{(3+\sqrt5)}$

$\implies\sf b = \dfrac{(3+ \sqrt5)^2}{3^2 - (\sqrt5)^2}$

$\implies\sf b = \dfrac{9 + 5 + 2 \times 3 \times \sqrt5}{9-5}$

$\implies\sf b = \dfrac{14 + 6\sqrt5}{4}$

$\implies\sf b = \dfrac{2(7+3\sqrt5)}{4}$

$\implies\sf b = \dfrac{7+3\sqrt5}{2}$

Squaring on both side :-

$\implies\sf b^2 = \dfrac{(7+3\sqrt5)^2}{(2)^2}$

$\implies\sf b^2 = \dfrac{7^2 + (3\sqrt5)^2 + 2 \times 7 \times 3\sqrt5}{4}$

$\implies\sf b^2 = \dfrac{49 + 48 + 42\sqrt5}{4}$

$\implies\sf b^2 = \dfrac{97 + 42\sqrt5}{4}$

Now adding the value of a² and b² :-

$\implies\sf a^2 + b^2 = \dfrac{97-42\sqrt5}{4} + \dfrac{97 + 42\sqrt5}{4}$

$\implies\sf a^2 + b^2 = \dfrac{97-42\sqrt5 + 97 + 42\sqrt5}{4}$

$\implies\sf a^2 + b^2 = \dfrac{194}{4}$

$\implies\boxed{\sf a^2 + b^2 = \dfrac{97}{2}}$

Answer 2

To find :-

• a²+b²

First we find the value of a²

$\implies \rm \: a = \dfrac{3-\sqrt5}{3+\sqrt5}$

$\implies \: a = \dfrac{(3-\sqrt5)}{(3+\sqrt5)} \times \dfrac{(3-\sqrt5)}{(3-\sqrt5)}$

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

$\implies a= \dfrac{9 + 5 - 2 \times 3 \times \sqrt5}{9-5}$

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

$\implies a = \dfrac{2(7-3\sqrt5)}{4}$

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

Now, Squaring on both the sides :-

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

$\implies\rm a^2 = \dfrac{7^2 + (3\sqrt5)^2 - 2 \times 7 \times 3\sqrt5}{4}$

$\implies\rm a^2 = \dfrac{49 + 48 - 42\sqrt5}{4}$

$\implies\rm \: a^2 = \dfrac{97-42\sqrt5}{4} \\ \\ \sf \: Calculating \: the \: value \: of \: b² :-$

$\implies\sf b = \dfrac{3+\sqrt5}{3-\sqrt5}$

$\implies\sf b = \dfrac{(3+\sqrt5)}{(3-\sqrt5)} \times \dfrac{(3+\sqrt5)}{(3+\sqrt5)}$

$\sf \implies b = \dfrac{(3+ \sqrt5)^2}{3^2 - (\sqrt5)^2 }$

$\implies\sf b^2 = \dfrac{97 + 42\sqrt5}{4}$

Now adding the value of a² and b²

$\implies\sf a^2 + b^2 = \dfrac{97-42\sqrt5}{4} + \dfrac{97 + 42\sqrt5}{4} \\ \\ \implies\sf a^2 + b^2 = \dfrac{97-42\sqrt5 + 97 + 42\sqrt5}{4}$

$\implies\sf a^2 + b^2 = \dfrac{194}{4} \\ \\ \implies\boxed {\underline{\sf \red{a^2 + b^2 = \dfrac{97}{2}}}}$