Question

rationalise the denominator of 5+√3÷√7-√11​

Answer 1

Answer :

$= \sf \dfrac{-(5\sqrt{7}+5\sqrt{11}+\sqrt{21}+\sqrt{33})}{4}$

Step-by-step explanation :

$\underline{\underline{\bf Rationalizing \ factor:}}$

• The factor of multiplication by which rationalization is done, is called as rationalizing factor.

• If the product of two surds is a rational number, then each surd is a rationalizing factor to other.

• To find the rationalizing factor,

        =>  If the denominator contains 2 terms, just change the sign between the two terms.

            For example, rationalizing factor of (3 + √2) is (3 - √2)

        => If the denominator contains 1 term, the radical found in the denominator is the factor.

            For example, rationalizing factor of √2 is √2

____________________________

Given,

$\dfrac{5+\sqrt{3}}{\sqrt{7}-\sqrt{11}}$

Rationalizing factor = √7 + √11

Multiply and divide the given fraction by (√7 + √11)

  $=\dfrac{5+\sqrt{3}}{\sqrt{7}-\sqrt{11}} \times \dfrac{\sqrt{7} +\sqrt{11}}{\sqrt{7}+\sqrt{11}} \\\\\\ = \dfrac{(5+\sqrt{3})(\sqrt{7}+\sqrt{11})}{(\sqrt{7}-\sqrt{11})(\sqrt{7}+\sqrt{11})} \\\\\\ = \dfrac{5(\sqrt{7}+\sqrt{11})+\sqrt{3}(\sqrt{7}+\sqrt{11})}{\sqrt{7}(\sqrt{7}+\sqrt{11})-\sqrt{11}(\sqrt{7}+\sqrt{11})} \\\\\\ =\dfrac{5\sqrt{7}+5\sqrt{11}+\sqrt{21}+\sqrt{33}}{\sqrt{7}^2+\sqrt{77}-\sqrt{77}-\sqrt{11}^2} \\\\\\ =\dfrac{5\sqrt{7}+5\sqrt{11}+\sqrt{21}+\sqrt{33}}{7-11}$

 $=\dfrac{5\sqrt{7}+5\sqrt{11}+\sqrt{21}+\sqrt{33}}{-4} \\\\\\ = \dfrac{-(5\sqrt{7}+5\sqrt{11}+\sqrt{21}+\sqrt{33})}{4}$

the denominator is rationalized!

Answer 2

Answer :

$\sf\dfrac{-(5\sqrt{7}+5\sqrt{11}+\sqrt{21}+\sqrt{33})}{4}$

Step-by-step explanation :

$\underline{\underline{\bf Rationalizing \ factor:}}$

The factor of multiplication by which rationalization is done, is called as rationalizing factor.

If the product of two surds is a rational number, then each surd is a rationalizing factor to other.

To find the rationalizing factor,

=> If the denominator contains 2 terms, just change the sign between the two terms.

For example, rationalizing factor of (3 + √2) is (3 - √2)

=> If the denominator contains 1 term, the radical found in the denominator is the factor.

For example, rationalizing factor of √2 is √2

____________________________

Given,

$\dfrac{5+\sqrt{3}}{\sqrt{7}-\sqrt{11}}$

Rationalizing factor = √7 + √11

Multiply and divide the given fraction by (√7 + √11)

$\begin{gathered}=\dfrac{5+\sqrt{3}}{\sqrt{7}-\sqrt{11}} \times \dfrac{\sqrt{7} +\sqrt{11}}{\sqrt{7}+\sqrt{11}} \\\\\\ = \dfrac{(5+\sqrt{3})(\sqrt{7}+\sqrt{11})}{(\sqrt{7}-\sqrt{11})(\sqrt{7}+\sqrt{11})} \\\\\\ = \dfrac{5(\sqrt{7}+\sqrt{11})+\sqrt{3}(\sqrt{7}+\sqrt{11})}{\sqrt{7}(\sqrt{7}+\sqrt{11})-\sqrt{11}(\sqrt{7}+\sqrt{11})} \\\\\\ =\dfrac{5\sqrt{7}+5\sqrt{11}+\sqrt{21}+\sqrt{33}}{\sqrt{7}^2+\sqrt{77}-\sqrt{77}-\sqrt{11}^2} \\\\\\ =\dfrac{5\sqrt{7}+5\sqrt{11}+\sqrt{21}+\sqrt{33}}{7-11}\end{gathered}$

$\begin{gathered}=\dfrac{5\sqrt{7}+5\sqrt{11}+\sqrt{21}+\sqrt{33}}{-4} \\\\\\ = \dfrac{-(5\sqrt{7}+5\sqrt{11}+\sqrt{21}+\sqrt{33})}{4}\end{gathered}$

the denominator is rationalized!