Question

Rationalise the denominator : 3 + √5 ÷ 3 - √5.​

Answer 1

Heya !!!

3+✓5 / 3 - ✓5

=> 3 + ✓5 / 3 - ✓5 × 3 + ✓5/ 3 + ✓5

=> ( 3 + ✓5) ( 3 + ✓5) / ( 3 - ✓5) ( 3 + ✓5)

=> ( 3 + ✓5)² / (3)² - ( ✓5)²

=> (3)² + ( ✓5)² + 2 × 3 × ✓5 / 9 -5

=> 9 + 5 + 6✓5 / 4

=> 14 + 6✓5/4

=> 2 ( 7 + 3✓5 ) /4

=> 7 + 3✓5/2

HOPE IT WILL HELP YOU...... :-)

Answer 2

Question :-

Rationalize the Denominator : $\frac{3 + \sqrt{5}}{3 - \sqrt{5}}$

Solution :-

The Rationalizing Factor is $\bold {3 + \sqrt{5}}$

∴ $\frac{3 + \sqrt{5}}{3 - \sqrt{5}} \times \frac{3 + \sqrt{5}}{3 + \sqrt{5}} = \frac{(3 + \sqrt{5})^2}{(3 - \sqrt{5})(3 + \sqrt{5})}$

Using the Identities, $\boxed {\bold {(a + b)^2 = a^2 + 2ab + b^2}}$ and $\boxed {\bold {(a + b)(a - b) = a^2 - b^2}}$ , we get -

$\Rightarrow \frac{(3 + \sqrt{5})^2}{(3 - \sqrt{5})(3 + \sqrt{5})} = \frac{3^2 + (2 \times 3 \times \sqrt{5}) + (\sqrt{5})^2 }{3^2 - (\sqrt{5})^2}\\\\\\\Rightarrow \frac{9 + 6\sqrt{5} + 5}{9 - 5} = \bold {\frac{14 + 6\sqrt{5}}{4}} = \frac{2(7 + 3\sqrt{5})}{4} = \bold {\frac{7 + 3\sqrt{5}}{2}}$

Conclusion :-

The Rationalized Form of $\frac{3 + \sqrt{5}}{3 - \sqrt{5}}$ is $\bold {\frac{7 + 3\sqrt{5}}{2}}$