Question

simplify

$7 + 3 \sqrt{5}$
upon
$7 - 3 \sqrt{5}$
rationalise
​

Answer 1

$\sf\huge {\underline{\underline{{Solution}}}}$

$\sf\implies \dfrac{7 + 3 \sqrt{5} }{7 - 3 \sqrt{5} }$

Here,we will Multiply with opposite sign of denominator to both numerator and denominator.

$\sf\implies \dfrac{7 + 3 \sqrt{5} }{7 - 3 \sqrt{5} } \times \dfrac{7 + 3 \sqrt{5} }{7 + 3 \sqrt{5} }$

Here ,an identity is used :

$\sf\: \bold{{(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}$

$\sf \: \bold{ {x}^{2} - {y}^{2} = (x + y)(x - y)}$

$\sf\implies \dfrac{ {(7 + 3 \sqrt{5} )}^{2} }{ {7}^{2} - {(3 \sqrt{5}) }^{2} } = \dfrac{ {7}^{2} + 2(7)(3 \sqrt{5}) + {(3 \sqrt{5}) }^{2} }{ {7}^{2} - {(3 \sqrt{5}) }^{2} }$

$\sf\implies \dfrac{49 + 2(7)(3 \sqrt{5}) + 45}{49 - 45}$

$\sf\implies \dfrac{94 + 42 \sqrt{5} }{4}$

Taking 2 common:

$\sf\implies \dfrac{ 2(47 + 21 \sqrt{5} )}{4}$

$\sf \: = \dfrac{47 + 21 \sqrt{5} }{2}$

Extra information=>

Rationalisation is a method of removing root values from denominator and shifts towards numerator.

Rationalisation is used to simplify the fractional root values.