rationalise the denominator of the following: 5+2root3 / 7+4root3
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rationalise the denominator of the following: 5+2root3 / 7+4root3
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Step-by-step explanation:
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[tex]⟹11 - 6 \sqrt{3} \: \: \: \tt \red{ Ans}. \\ \\ [/tex]
Step-by-step explanation:
Given that:-
[tex] \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ [/tex]
What to do:-
To rationalise the denominator
Solution:-
We have,
[tex] \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ [/tex]
The denominator is 5+2√3. Multiplying the numerator and denominator by 7-4√3,
we get,
[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]
⬤ Applying Algebraic Identity
(a+b)(a-b) = a² - b² to the denominator
We get,
[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]
Hence the denominator is rationalised.
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