Question

solve: 30/root 20+ root 5​

Answer 1

$\bold{\dfrac{30}{\sqrt{20} + \sqrt{5}}}$

Rationalizing the denominator

$= \dfrac{30}{\sqrt{20} + \sqrt{5}}$

$= \dfrac{30}{\sqrt{20} + \sqrt{5}} \times \dfrac{\sqrt{20} - \sqrt{5}}{\sqrt{20} - \sqrt{5}}$

$= \dfrac{30(\sqrt{20} - \sqrt{5})}{{(\sqrt{20})}^{2} - {(\sqrt{5})}^{2}}$

$= \dfrac{30(\sqrt{20} - \sqrt{5})}{15}$

$= 2(\sqrt{20} - \sqrt{5})$

$= 4\sqrt{5} - 2\sqrt{5}$

$\boxed{= 2\sqrt{5}}$

___________________________________________

$\bold{\dfrac{30}{\sqrt{20}} + \sqrt{5}}$

$= \dfrac{30}{2\sqrt{5}}+\sqrt{5}$

By taking LCM

$= \dfrac{30}{2\sqrt{5}}+\dfrac{\sqrt{5}(2\sqrt{5})}{2\sqrt{5}}$

$= \dfrac{30 + 10}{2\sqrt{5}}$

$= \dfrac{40}{2\sqrt{5}}$

$\boxed{= \dfrac{20}{\sqrt{5}}}$

Answer 2

$\mathfrak{\large{\underline{\underline{Correct \: Question:-}}}}$

$\sf{Solve \: \dfrac{30}{ \sqrt{20} + \sqrt{5} } }$

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

$\boxed{\bf{ \dfrac{30}{ \sqrt{20} + \sqrt{5} } = 2 \sqrt{5} }}$

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

$\dfrac{30}{ \sqrt{20} + \sqrt{5} }$

To solve it first we need to rationalise the denominator.

To rationalise denominator first we have to know what is its rationalising factor.

The rationalising factor of √20 + √5 is √20 - √5. So, multiply both denominator and denom8nator with rationalising factor.

$= \dfrac{30}{ \sqrt{20} + \sqrt{5} } \times \dfrac{ \sqrt{20} - \sqrt{5} }{ \sqrt{20} - \sqrt{5} }$

$= \dfrac{30( \sqrt{20} - \sqrt{5})}{ {( \sqrt{20})}^{2} - {( \sqrt{5})}^{2} }$

[In denominator (x + y)(x - y) = x² - y² and here x = √20 and y = √5]

$= \dfrac{30( \sqrt{4 \times 5} - \sqrt{5} }{20 - 5}$

[Here √20 can be written as √4 * √5 ]

$= \dfrac{30(2 \sqrt{5} - \sqrt{5})}{15}$

[Here in numerator √4 is written as 2 because √4 = 2]

$=2(2 \sqrt{5} - \sqrt{5})$

$= 4 \sqrt{5} - 2 \sqrt{5}$

$= 2 \sqrt{5}$

$\boxed{\bf{ \dfrac{30}{ \sqrt{20} + \sqrt{5} } = 2 \sqrt{5} }}$

$\mathfrak{\large{\underline{\underline{Identity \: Used:-}}}}$

(x + y)(x - y) = x² - y²

$\mathfrak{\large{\underline{\underline{Extra \: Information:-}}}}$

What is rationalising factor ?

If the product of two irrational number is rational number then each of the other is rationalising factor.

• To rationalise the denomiantor of $\sf{ \dfrac{1}{ \sqrt{a} + b} }$ , we multiply this by $\sf{ \dfrac{ \sqrt{a} - b}{ \sqrt{a} - b} }$ , where a, b are integers.