Question

rationalise the denominator of 1 by 7+4root 3

Answer 1

$\displaystyle \sf{ \frac{1}{7 + 4 \sqrt{3} } = 7 - 4 \sqrt{3} }$

Given :

The expression

$\displaystyle \sf{ \frac{1}{7 + 4 \sqrt{3} } }$

To find :

To rationalize the expression

Solution :

Step 1 of 2 :

Write down the given expression

The given expression is

$\displaystyle \sf{ \frac{1}{7 + 4 \sqrt{3} } }$

Step 2 of 2 :

Rationalize the denominator

$\displaystyle \sf{ \frac{1}{7 + 4 \sqrt{3} } }$

$\displaystyle \sf{ = \frac{(7 - 4 \sqrt{3}) }{(7 + 4 \sqrt{3} )(7 - 4 \sqrt{3}) } }$

$\displaystyle \sf{ = \frac{(7 - 4 \sqrt{3}) }{ {7}^{2} - {(4 \sqrt{3} )}^{2} } }$

$\displaystyle \sf{ = \frac{(7 - 4 \sqrt{3}) }{49 - 48} }$

$\displaystyle \sf{ = \frac{(7 - 4 \sqrt{3}) }{1} }$

$\displaystyle \sf{ = 7 - 4 \sqrt{3} }$

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