Question

$7 + 3 \sqrt{5 } \div 3 + \sqrt{5 }$
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Answer 1

$\bf{\large{\underline{\underline{Answer:-}}}}$

$\boxed{\sf{ \dfrac{7 + 3 \sqrt{5} }{3 + \sqrt{5} }= \dfrac{3 + \sqrt{5} }{2} }}$

$\bf{\large{\underline{\underline{Explanation:-}}}}$

$\dfrac{7 + 3 \sqrt{5} }{3 + \sqrt{5} }$

To find the value rationize the denominator.

The Rationalising factor of 3 + √5 is 3 - √5. So, multiply numerator and denominator with rationalizing factor.

$\dfrac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } \times \dfrac{3 - \sqrt{5} }{3 - \sqrt{5} }$

$= \dfrac{7 + 3 \sqrt{5}(3 - \sqrt{5})}{ {3}^{2} - {( \sqrt{5})}^{2} }$

[Since (x + y)(x - y) = x² - y² and here x = 3, y = √5]

$= \dfrac{7(3 - \sqrt{5}) + 3 \sqrt{5}(3 - \sqrt{5}) }{9 - 5}$

[Since 3² = 3 * 3 = 9, (√5)² = 5]

$= \dfrac{21 - 7 \sqrt{5} + 9 \sqrt{5} - 3 \sqrt{5 \times 5} }{4}$

$= \dfrac{21 + 2 \sqrt{5} - 3 \sqrt{ {5}^{2} } }{4}$

$= \dfrac{21 + 2 \sqrt{5} - 3(5)}{4}$

$= \dfrac{21 + 2 \sqrt{5} - 15}{4}$

$= \dfrac{6 + 2 \sqrt{5} }{4}$

It can be written as :-

$= \dfrac{2(3 + \sqrt{5})}{4}$

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

$\boxed{\bf{ \dfrac{7 + 3 \sqrt{5} }{3 + \sqrt{5} }= \dfrac{3 + \sqrt{5} }{2} }}$

Answer 2

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