Step-by-step explanation:
[tex] \bf \underline{Given-} \\ [/tex]
[tex] \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \\ [/tex]
[tex] \bf \underline{What\: to\: do-} \\ [/tex]
To rationalise the denominator
[tex] \bf \underline{Solution-} \\ [/tex]
[tex]\textsf{We have,}[/tex]
[tex] \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ [/tex]
[tex]\textsf{The denominator is 5+2√3. Multiplying the numerator and denominator by 7-4√3,}\\[/tex]
[tex]\textsf{we get,}\\[/tex]
[tex] ⟹\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ [/tex]
[tex]⟹ \frac{(5 + 2 \sqrt{3} )(7 - 4 \sqrt{3}) }{(7 + 4 \sqrt{3} )(7 - 4 \sqrt{3}) } \\ \\ [/tex]
[tex]\textsf{⬤ Applying Algebraic Identity
(a+b)(a-b) = a² - b² to the denominator}\\[/tex]
[tex]\textsf{We get,}\\[/tex]
[tex]⟹ \frac{(5 + 2 \sqrt{3} )(7 - 4 \sqrt{3}) }{(7 {)}^{2} - (4 \sqrt{3} {)}^{2} } \\ \\ [/tex]
[tex]⟹ \frac{35 + 14 \sqrt{3} - 20 \sqrt{3} - 8 \sqrt{3 \times 3} }{49 - 48} \\ \\ [/tex]
[tex]⟹ \frac{35 + 14 \sqrt{3} - 20 \sqrt{3} - 24 }{1} \\ \\ [/tex]
[tex]⟹(35 - 24) - 6 \sqrt{3} \\ \\ [/tex]
[tex]⟹11 - 6 \sqrt{3} \: \: \: \tt \red{ Ans}. \\ \\ [/tex]
[tex] \bf \underline{Hence \:the \:denominator\: is\: rationalised.}\\[/tex]
