Question

Rationalise the denominator
$7 + \sqrt{2} \div \sqrt{3} + \sqrt{2}$
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Answer 1

Solution

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$\small\sf\red{ \frac{7 + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } }$

rationalize the denominator by multiplying the numerator and denominator by √3-√2

We get as,

$\small\sf\red{ \frac{(7 + \sqrt{2}) \: \times ( \sqrt{3} - \sqrt{2} ) }{ (\sqrt{3} + \sqrt{2}) \: \times ( \sqrt{3} - \sqrt{2} )} }$

Consider (√3+√2)(√3-√2),

multiplication can be transformed into difference of squares using the rule:(a-b) (a+b) =a²-b²

So, we get as,

$\small\sf\red{ \frac{(7 + \sqrt{2}) \times ( \sqrt{3 } - \sqrt{2} ) }{( \sqrt{3} ) {}^{2} - ( \sqrt{2} ) {}^{2} } }$

Square √3.Square √2.

$\small\sf\red{ \frac{(7 + \sqrt{2}) \times ( \sqrt{3} - \sqrt{2} ) }{3 - 2} }$

Subtract 2 from 3 to get 1

$\small\sf\red{ \frac{(7 + \sqrt{2} ) \times ( \sqrt{3} - \sqrt{2} )}{1} }$

Anything divided by 1 gives itself.

$\small\sf\red{(7 + \sqrt{2)} \times ( \ \sqrt{3} - \sqrt{2} )}$

Apply the distributive property by multiplying each term of 7+√2 by each term of √3-√2

$\small\sf\red{7 \sqrt{3} - 7 \sqrt{2} + \sqrt{2 \times 3} - ( \sqrt{2}) {}^{2} }$

multiply √2 with √3 to get √6 (2×3=6) .

And square root of 2 is 2 .

We get as,

$\small\sf\red{7 \sqrt{3} - 7 \sqrt{2} + \sqrt{6} - 2}$

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Hence, your answer is

$\small\sf\red{7 \sqrt{3} - 7 \sqrt{2} + \sqrt{6} - 2}$

Answer 2

Answer:

$\frac{7 + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } =7 \sqrt{3} - 7 \sqrt{2} + \sqrt{6} - 2$

Step-by-step explanation:

$\frac{7 + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

$= \frac{7 + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$= \frac{7( \sqrt{3} - \sqrt{2} ) + \sqrt{2}( \sqrt{3} - \sqrt{2}) }{3 - 2}$

$= 7 \sqrt{3} - 7 \sqrt{2} + \sqrt{6} - 2$