Question

rationalise the denomiator of 5+√3/5-√3​

Answer 1

Solution:-

We have

$\rm \implies \: \dfrac{5 + \sqrt{3} }{5 - \sqrt{3} }$

To find rationalization the denominator

Now take

$\implies \dfrac{5 + \sqrt{3} }{5 - \sqrt{3} } \times \dfrac{5 + \sqrt{3} }{5 + \sqrt{3} }$

Using the identities

$\rm \to \: (a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab$

$\rm \to \: (a - b)(a + b) = {a}^{2} - {b}^{2}$

Now we get

$\rm \implies \: \dfrac{(5 + \sqrt{3} ) {}^{2} }{(5 - \sqrt{3})(5 + \sqrt{3} )}$

$\rm \implies \: \dfrac{ {5}^{2} + ( \sqrt{3}) {}^{2} + 2 \times 5 \times \sqrt{3} }{5 {}^{2} - ( \sqrt{3} ) {}^{2} }$

$\rm \implies \dfrac{25 + 3 + 10 \sqrt{3} }{25 - 3}$

$\rm \implies \: \dfrac{28 + 10 \sqrt{3} }{22}$

Answer is

$\rm \implies \: \dfrac{28 + 10 \sqrt{3} }{22}$

Answer 2

(Question)

• Rationalise the denominator of $\sf \frac{5 + \sqrt{3} }{5 - \sqrt{3} }$

(Solution)

$\sf \frac{5 + \sqrt{3} }{5 - \sqrt{3} }$

$\sf = \frac{5 + \sqrt{3} }{5 - \sqrt{3} } \times \frac{5 + \sqrt{3} }{5 + \sqrt{3} }$

$\sf = \frac{ {(5 + \sqrt{3}) }^{2} }{ {(5)}^{2} - {( \sqrt{3} )}^{2} }$

$\sf = \frac{25 + 3 + 2 \times 5 \times \sqrt{3} }{25 - 3}$

$\sf = \frac{28 + 10 \sqrt{2} }{22}$

$\sf = \frac{14 + 5 \sqrt{2} }{11}$

Hence, the rationalises form is $\sf \frac{14 + 5 \sqrt{2} }{11}$

(Identity Used)

➡ (a + b)² = a² + 2ab + b²

➡ (a + b)(a - b) = a² - b²