Question

rationalise the denominator 5+2root3 / 7+4root3 .
I will mark as brainliest if anyone answers correctly to my question ...

Answer 1

Hope it helps.................

Answer 2

Answer:

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

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.

I hope it's help with...☺