Question

Rationalise the following :-

$\begin{gathered} \\ \\ \large \red{ \frac{7 \sqrt{3} }{ \sqrt{10} + \sqrt{3} } - \frac{ 2\sqrt{5} }{ \sqrt{6} + \sqrt{5} } - \frac{3 \sqrt{2} }{ \sqrt{15} + 3 \sqrt{2} } }\end{gathered}$


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Answer 1

$\large\tt \: {\red{\frac{ 7 \sqrt{ 3 } }{ \sqrt{ 10 } + \sqrt{ 3 } } - \frac{ 2 \sqrt{ 5 } }{ \sqrt{ 6 } + \sqrt{ 5 } } - \frac{ 3 \sqrt{ 2 } }{ \sqrt{ 15 } +3 \sqrt{ 2 } }} }$

$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{\left(\sqrt{10}+\sqrt{3}\right)\left(\sqrt{10}-\sqrt{3}\right)}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}} \\ \\ \tt \: Consider \left(\sqrt{10}+\sqrt{3}\right)\left(\sqrt{10}-\sqrt{3}\right)$

• Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{\left(\sqrt{10}\right)^{2}-\left(\sqrt{3}\right)^{2}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}} \\ \\ \tt \:</p><p>\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{10-3}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}} \\ \\ \tt \:</p><p>\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}} \\ \\ \tt \:</p><p>\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-\frac{2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}\\ \\ \tt \: Consider \: \left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right).$

• Multiplication can be transformed into difference of squares using the rule:

$\tt \: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2} \\ \\ \tt \: \frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-\frac{2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)}{\left(\sqrt{6}\right)^{2}-\left(\sqrt{5}\right)^{2}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}} \\ \\ \tt \: </p><p>\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-\frac{2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)}{6-5}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}} \\ \\ \tt \: \frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-\frac{2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)}{6-5}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}} \\ \\ \tt \: \frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}} \\ \\ \tt \: </p><p>\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{\left(\sqrt{15}+3\sqrt{2}\right)\left(\sqrt{15}-3\sqrt{2}\right)}\\ \\ \tt \: Consider \: \left(\sqrt{15}+3\sqrt{2}\right)\left(\sqrt{15}-3\sqrt{2}\right).$

• Multiplication can be transformed into difference of squares using the rule:

$\tt \: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2} \\ \\ \tt \:\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{\left(\sqrt{15}\right)^{2}-\left(3\sqrt{2}\right)^{2}} \\ \\ \tt \: \frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{15-\left(3\sqrt{2}\right)^{2}} \\ \\ \tt \: \frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{15-\left(3\sqrt{2}\right)^{2}} \\ \\ \tt \: \frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{15-9\left(\sqrt{2}\right)^{2}} \\ \\ \tt \:\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{15-9\times 2} \\ \\ \tt \: \frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{15-18} \\ \\ \tt \: \frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\frac{3\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)}{-3} \\ \\ \tt \: </p><p>\frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)-\left(-\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right)\right) \\ \\ \tt \: \frac{7\sqrt{3}\left(\sqrt{10}-\sqrt{3}\right)}{7}-2\sqrt{5}\left(\sqrt{6}-\sqrt{5}\right)+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right) \\ \\ \tt \: </p><p>\sqrt{30}-3-2\sqrt{30}-\left(-10\right)+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right) \\ \\ \tt \: -\sqrt{30}-3+10+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right) \\ \\ \tt \: -\sqrt{30}+7+\sqrt{2}\left(\sqrt{15}-3\sqrt{2}\right) \\ \\ \tt \: -\sqrt{30}+7+\sqrt{2}\sqrt{15}-3\left(\sqrt{2}\right)^{2} \\ \\ \tt \: -\sqrt{30}+7+\sqrt{30}-3\left(\sqrt{2}\right)^{2} \\ \\ \tt \: -\sqrt{30}+7+\sqrt{30}-3\times 2 \\ \\ \tt \: -\sqrt{30}+7+\sqrt{30}-6\\ \\ \tt \: 7-6\\ \\ \tt \: 1$

$\sf \: @WildCat7083$

Answer 2

$\large\sf\underline{Given\::}$

• $\sf\:\frac{7 \sqrt{3}}{\sqrt{10} + \sqrt{3}} - \frac{2 \sqrt{5}}{\sqrt{6} + \sqrt{5}} - \frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}}$

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$\large\sf\underline{To\::}$

• Rationalise the given expression

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$\large\sf\underline{Things\:to\:remember\::}$

Rationalization means the process of getting rid of any surds in the denominator. Simplifying a fraction with surd simply refers to rationalization.

So now new question arises :

$\small\bf\:How\:do\:we\:get\:rid\:of\:the\:surds\:from\:the\:denominator\:?$

This is pretty simple we just need to multiply the fraction by the conjugate of its denominator.

Now :

$\small\bf\:What\:is\:meant\:by\:conjugate\:?$

Let's understand conjugate with the help of an example. If we are given x + y and said to find it's conjugate then, it's conjugate would be x - y.

So conjugate is an expression formed by changing the sign of the given expression.

Hope I sound clear now :D

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$\large\sf\underline{Concept\::}$

Now in this question we could rationalise whole expression at once but it might create confusion. Therefore we will rationalise the expression one by one in three steps . Doing so we will simplify the expression and could easily calculate the value of given expression. Let's begin!

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$\large\sf\underline{Solution\::}$

Taking $\sf\:\frac{7 \sqrt{3}}{\sqrt{10} + \sqrt{3}}$

• Conjugate of $\sf\:\sqrt{10} + \sqrt{3}$ is $\sf\:\sqrt{10} - \sqrt{3}$ . Multiplying the fraction by this conjugate.

$\sf\rightarrow\:\frac{7 \sqrt{3}}{\sqrt{10} + \sqrt{3}} \times \frac{\sqrt{10} - \sqrt{3}}{\sqrt{10} - \sqrt{3}}$

• Using identity (a + b) (a - b) = $\sf\:a^{2}-b^{2}$ in the denominator

$\sf\rightarrow\:\frac{7 \sqrt{3} \times (\sqrt{10} - \sqrt{3})}{(\sqrt{10})^{2} - (\sqrt{3})^{2}}$

$\sf\rightarrow\:\frac{7\sqrt{30} - 7\sqrt{9}}{10 - 3}$

$\sf\rightarrow\:\frac{7\sqrt{30} - 7 \times 3}{7}$

$\sf\rightarrow\:\frac{7\sqrt{30} - 21}{7}$

$\sf\rightarrow\:\frac{7(\sqrt{30} - 3)}{7}$

$\sf\rightarrow\:\frac{\cancel{7}(\sqrt{30} - 3)}{\cancel{7}}$

$\small{\underline{\boxed{\mathrm\red{\rightarrow\:\sqrt{30} - 3}}}}$ ★

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Taking $\sf\:\frac{2 \sqrt{5}}{\sqrt{6} + \sqrt{5}}$

• Conjugate of $\sf\:\sqrt{6} + \sqrt{5}$ is $\sf\:\sqrt{6} - \sqrt{5}$ . Multiplying the fraction by this conjugate.

$\sf\rightarrow\:\frac{2 \sqrt{5}}{\sqrt{6} + \sqrt{5}} \times \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}}$

• Using identity (a + b) (a - b) = $\sf\:a^{2}-b^{2}$ in the denominator

$\sf\rightarrow\:\frac{2 \sqrt{5} \times (\sqrt{6} - \sqrt{5})}{(\sqrt{6})^{2} - (\sqrt{5})^{2}}$

$\sf\rightarrow\:\frac{2\sqrt{30} - 2\sqrt{25}}{6 - 5}$

$\sf\rightarrow\:\frac{2\sqrt{30} - 2 \times 5}{1}$

$\small{\underline{\boxed{\mathrm\red{\rightarrow\:2\sqrt{30} - 10}}}}$ ★

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Taking $\sf\:\frac{3 \sqrt{2}}{\sqrt{15} + 3\sqrt{2}}$

• Conjugate of $\sf\:\sqrt{15} + 3\sqrt{2}$ is $\sf\:\sqrt{15} - 3\sqrt{2}$ . Multiplying the fraction by this conjugate.

$\sf\rightarrow\:\frac{3 \sqrt{2}}{\sqrt{15} + 3\sqrt{2}} \times \frac{\sqrt{15} - 3\sqrt{2}}{\sqrt{15} - 3\sqrt{2}}$

• Using identity (a + b) (a - b) = $\sf\:a^{2}-b^{2}$ in the denominator

$\sf\rightarrow\:\frac{3 \sqrt{2} \times (\sqrt{15} - 3\sqrt{2})}{(\sqrt{15})^{2} - (3\sqrt{2})^{2}}$

$\sf\rightarrow\:\frac{3\sqrt{30} - 9\sqrt{4}}{15 - 9 \times 2}$

$\sf\rightarrow\:\frac{3\sqrt{30} - 9 \times 2}{15-18}$

$\sf\rightarrow\:\frac{3\sqrt{30} - 18}{-3}$

$\sf\rightarrow\:\frac{-3(-\sqrt{30} +6)}{-3}$

$\sf\rightarrow\:\frac{\cancel{-3}(-\sqrt{30} +6)}{\cancel{-3}}$

$\small{\underline{\boxed{\mathrm\red{\rightarrow\:-\sqrt{30}+6}}}}$ ★

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Now :

$\sf\:\frac{7 \sqrt{3}}{\sqrt{10} + \sqrt{3}} - \frac{2 \sqrt{5}}{\sqrt{6} + \sqrt{5}} - \frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}}$

• Lets substitute the simplified values

$\sf\implies\:\sqrt{30} - 3 -(2\sqrt{30} - 10) - (-\sqrt{30}+6)$

$\sf\implies\:\sqrt{30} - 3 - 2\sqrt{30}+ 10 + \sqrt{30} - 6$

$\sf\implies\:\sqrt{30} + \sqrt{30}- 2\sqrt{30}- 3 + 10 - 6$

$\sf\implies\:2\sqrt{30} - 2\sqrt{30} + 7 - 6$

$\sf\implies\:\cancel{2\sqrt{30}} - \cancel{2\sqrt{30}}+ 7 - 6$

$\sf\implies\:7 - 6$

$\small{\underline{\boxed{\mathrm\red{\implies\:1}}}}$ ★

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!! Hope it helps !!

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