Question

3+√5 ÷ 3-√5 rationalize the denominator

Answer 1

given :-

(3 + √5)/(3 - √5)

$\tt = \frac{3 + \sqrt{5} }{3 - \sqrt{5} } \times \frac{3 + \sqrt{5} }{3 + \sqrt{5} } \\ \\ \tt = \frac{(3 + \sqrt{5} )(3 + \sqrt{5} )}{(3 - \sqrt{5} )(3 + \sqrt{5}) } \\ \\ \tt = \frac{ {(3 + \sqrt{5}) }^{2} }{( {3})^{2} - ( { \sqrt{5} })^{2} } \\ \\ \tt = \frac{( {3)}^{2} + 2(3)( \sqrt{5} ) + ( { \sqrt{5} )}^{2} }{9 - 5} \\ \\ \tt = \frac{9 + 6 \sqrt{5} + 5}{4} \\ \\ \tt = \frac{14 + 6 \sqrt{5} }{4} \\ \\ \tt = \frac{ \cancel2 \times 2 + \cancel2 \times 3 \sqrt{5} }{ \cancel2 \times 2} \\ \\ \tt = \boxed{ \frac{2 + 3 \sqrt{5} }{2} }$

identities used :-

• (a + b)² = a² + 2ab + b²

• (a - b)(a + b) = a² - b²

Answer 2

Solution:-

$\bold{ \frac{3 + \sqrt{5} }{3 - \sqrt{5} } }$

$\textsf{By rationalizating the denominater}$

$\bold{ = \frac{3 + \sqrt{5} }{3 - \sqrt{5} } \times \frac{3 + \sqrt{5} }{3 + \sqrt{5} } }$

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

By,using the identity (a+b)²= a²+2ab+b²

By, using the identity (a+b)(a-b)= a²-b²

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

$\bold{ = \frac{5 + 6 \sqrt{5} + 9}{9 - 5} }$

$\bold{ = \frac{14 + 6 \sqrt{5} }{4}}$

$\boxed{\red{ \bold{ = \frac{7 + 3 \sqrt{5} }{2}} }}$