Question

rationalise the denominator 7+√3. /7-√3​

Answer 1

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$\sf\implies\:\frac{7 + \sqrt{3} }{7 - \sqrt{3} } \\ \\ rationalizing \: the \: denominator \: \\ \\\sf\implies\: \frac{7 + \sqrt{3} }{7 - \sqrt{3} } \times \frac{7 + \sqrt{3} }{7 + \sqrt{3} } \\ \\ \sf\implies\: \frac{ {(7 + \sqrt{3}) }^{2} }{ {(7)}^{2} - {( \sqrt{3} )}^{2} } \\ \\\sf\implies\: \frac{49 + 14 \sqrt{3} + 3}{49 - 3} \\ \\ \sf\implies\:\frac{52 + 14 \sqrt{3} }{46} \\ \\ \sf\implies\:\frac{2(26 + 7 \sqrt{3}) }{46} \\ \\ \sf\implies\: \frac{26 + 7\sqrt{3} }{23}$

Answer 2

Answer:

$\frac{26+7\sqrt{3}}{23}$

Step-by-step explanation:

Given Expression :

                          $\frac{7+\sqrt3}{7-\sqrt3}$

To Find :

Rationalize the denominator

Solution :

Given expression

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

denominator of the fraction is ${7-\sqrt{3}}$

and the conjugate of the denominator = $7+\sqrt3$

multiply by $7+\sqrt3$ on numerator and denominator on the fraction

                        $\frac{7+\sqrt{3}}{7-\sqrt{3}}=\frac{7+\sqrt{3}}{7-\sqrt{3}}\ \text{x}\ \frac{7+\sqrt{3}}{7+\sqrt{3}}$

                                 $=\frac{(7+\sqrt3)^2}{(7+\sqrt3)(7-\sqrt3}$

since we now that

$(a+b)^2=a^2+b^2+2ab$     and $a^2-b^2=(a-b)(a+b)$

using above inequalities we get

                              $=\frac{49+3+14\sqrt3}{49-3}$

                              $=\frac{52+14\sqrt3}{46}$

                              $=\frac{26+7\sqrt3}{23}$

Hence by rationalizing the expression we get

                                $=\frac{26+7\sqrt3}{23}$

#SPJ2