Math, asked by jiyu8732, 9 months ago

5 + 2 root 3 / 7 + 4 root 3​

Answers

Answered by Salmonpanna2022
1

Answer:

⟹11 - 6 \sqrt{3} \: \:  \:   \tt \red{ Ans}. \\  \\

Step-by-step explanation:

Given that:-

 \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} }  \\  \\

What to do:-

To rationalise the denominator

Solution:-

We have,

 \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} }  \\  \\

The denominator is 5+2√3. Multiplying the numerator and denominator by 7-4√3,

we get,

 ⟹\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} }  \times  \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} }  \\  \\

⟹ \frac{(5 + 2 \sqrt{3} )(7 - 4 \sqrt{3}) }{(7 + 4 \sqrt{3} )(7 - 4 \sqrt{3}) }  \\  \\

⬤ Applying Algebraic Identity

(a+b)(a-b) = a² - b² to the denominator

We get,

⟹ \frac{(5 + 2 \sqrt{3} )(7 - 4 \sqrt{3}) }{(7 {)}^{2} - (4 \sqrt{3} {)}^{2}   }  \\  \\

⟹ \frac{35 + 14 \sqrt{3} - 20 \sqrt{3} - 8 \sqrt{3 \times 3}   }{49 - 48}  \\  \\

⟹ \frac{35 + 14 \sqrt{3} - 20 \sqrt{3}  - 24 }{1}  \\  \\

⟹(35 - 24) - 6 \sqrt{3}  \\  \\

⟹11 - 6 \sqrt{3} \: \:  \:   \tt \red{ Ans}. \\  \\

Hence the denominator is rationalised.

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