Question

rationalise 7/ 2√3+√5​

Answer 1

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

• Rationalise = $\sf\:\frac{7}{2\sqrt{3} + \sqrt{5}}$

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

• Rationalise the given expression

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

In order to rationalise an expression we need to multiply and divide the expression by the conjugate of its denominator.

Here :

• Numerator = 7

• Denominator = $\sf\:2\sqrt{3} + \sqrt{5}$

So conjugate of $\sf\:2\sqrt{3} + \sqrt{5}$ = $\sf\:2\sqrt{3} - \sqrt{5}$

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Now let's multiply and divide the given expression by $\sf\:2\sqrt{3} - \sqrt{5}$

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$\sf\to\:\frac{7}{2\sqrt{3} + \sqrt{5}} \times \frac{2 \sqrt{3} - \sqrt{5}}{2\sqrt{3} - \sqrt{5}}$

$\sf\to\:\frac{7(2 \sqrt{3} - \sqrt{5})}{(2\sqrt{3} + \sqrt{5}) (2 \sqrt{3} - \sqrt{5})}$

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• Multiplying the terms in numerator and applying an identity in the denominator

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Identity used in denominator :

$\small{\underline{\boxed{\mathrm\orange{(a+b)(a-b)=a^{2}-b^{2}}}}}$

Here :

• a = 2√3

• b = √5

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$\sf\to\:\frac{14 \sqrt{3} -7 \sqrt{5}}{(2\sqrt{3}) ^{2}- (\sqrt{5})^{2}}$

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• In the denominator root symbol and square gets cancelled out and 2 comes out in its square form

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$\sf\to\:\frac{14 \sqrt{3} -7 \sqrt{5}}{(4 \times 3)- 5}$

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• Doing simple calculations in denominator

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$\sf\to\:\frac{14 \sqrt{3} -7 \sqrt{5}}{12- 5}$

$\sf\to\:\frac{14 \sqrt{3} -7 \sqrt{5}}{7}$

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• Taking 7 as common in numerator

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$\sf\to\:\frac{7(2 \sqrt{3} - \sqrt{5})}{7}$

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• Cancelling 7 from numerator as well as denominator

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$\sf\to\:\frac{\cancel{7}(2 \sqrt{3} - \sqrt{5})}{\cancel{7}}$

$\sf\to\:\frac{(2 \sqrt{3} - \sqrt{5})}{1}$

$\large{\mathfrak\red{\to\:2 \sqrt{3} - \sqrt{5}}}$ ★

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!! Hope it helps !!

Answer 2

$\sqrt[2]{3} - 5$