Question

Rationalise the denominator of 7+3√3 by 2+√5 ​

Answer 1

ANSWER:

To Rationalize:

• (7+3√3)/(2+√5)

Solution:

We are given that,

$\implies\dfrac{7+3\sqrt3}{2+\sqrt5}$

To rationalise, we'll multiply and divide by 2 - √5.

So,

$\implies\dfrac{7+3\sqrt3}{2+\sqrt5}\times\dfrac{2-\sqrt5}{2-\sqrt5}$

$\implies\dfrac{(7+3\sqrt3)(2-\sqrt5)}{(2+\sqrt5)(2-\sqrt5)}$

We know that,

$\hookrightarrow (a-b)(a+b)=a^2-b^2$

$\implies\dfrac{(7+3\sqrt3)(2-\sqrt5)}{(2+\sqrt5)(2-\sqrt5)}$

$\implies\dfrac{14-7\sqrt5+6\sqrt3-3\sqrt{15}}{(2)^2-(\sqrt5)^2}$

$\implies\dfrac{14-7\sqrt5+6\sqrt3-3\sqrt{15}}{4-5}$

$\implies\dfrac{14-7\sqrt5+6\sqrt3-3\sqrt{15}}{-1}$

$\implies -14+7\sqrt5-6\sqrt3+3\sqrt{15}$

So,

$\implies\bf 7\sqrt5-6\sqrt3+3\sqrt{15}-14$

Formula Used:

• $\hookrightarrow (a-b)(a+b)=a^2-b^2$

Answer 2

Rationalise the denominator of 7+3√3 by 2+√5 is

7 √5 −6 √3 +3 √15 −14