Question

rationalize the denominator of 6 upon root 5 + root 2​

Answer 1

6/√5+√2

6/√5+√2×√5-√2/√5-√2

6(√5-√2)/5-2...(a+b)(a-b)=a^2-b^2

6(√5-√2)/3

2(√5-√2).

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Answer 2

Answer:

After rationalization the given expression is $2( \sqrt{5} - \sqrt{2} )$

Step-by-step explanation:

Given expression is 6 upon root 5 + root 2 $= \frac{6}{ \sqrt{5} + \sqrt{2} }$

This is a problem of Power of indices.

Here we want to rationalize the denominator of given expression.

For rationalization we are multiplying denominator and numerator by $( \sqrt{5} - \sqrt{2} )$and get,

$\frac{6 \times ( \sqrt{5} - \sqrt{2} )}{( \sqrt{5} + \sqrt{2} )( \sqrt{5} - \sqrt{2}) } = \frac{6 \times ( \sqrt{5} - \sqrt{2} )}{ { \sqrt{5} }^{2} - { \sqrt{2} }^{2} } = \frac{6 \times ( \sqrt{5} - \sqrt{2} )}{5 - 2} = \frac{6 \times ( \sqrt{5} - \sqrt{2} )}{3} = 2\times ( \sqrt{5} - \sqrt{2} )$

After rationalization the given expression is

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

Here applied formula,

${x}^{2} - {y}^{2} = (x + y)(x - y)$

Some important formulas of Power of indices,

${x}^{0} = 1 \\ {x}^{1} = x \\ {x}^{a} \times {x}^{b} = {x}^{a + b} \\ \frac{ {x}^{a} }{ {x}^{b} } = {x}^{a - b} \\ {x}^{ {y}^{a} } = {x}^{ya} \\ {x}^{ - 1} = \frac{1}{x} \\ {x}^{a} \times {y}^{a} = {(xy)}^{a}$