Question

Solve it ASAP. Question number 3​

Answer 1

QUESTION:

$\tt If \ x= \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \ and \ y = \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}\ then$

$\tt x + y + xy=$

ANSWER:

• The value of x + y + xy = 9

GIVEN:

$\tt x= \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \ and \ y = \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$

TO FIND:

• The value of x + y + xy

EXPLANATION:

$\tt x= \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

$\tt Multiply\ and\ divide\ by \ \sqrt{5}+\sqrt{3}$

$\tt x= \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

$\tt x= \dfrac{(\sqrt{5}+\sqrt{3} {)}^{2} }{(\sqrt{5} {)}^{2} -(\sqrt{3} {)}^{2} }$

$\boxed{ \bold{ \large{ \gray{(A + B)^2 = A^2 + 2AB + B^2}}}}$

$\tt x= \dfrac{5 + 3 + 2 \sqrt{3} \sqrt{5} }{5 - 3 }$

$\tt x= \dfrac{8 + 2 \sqrt{15}}{2}$

$\tt x= 4 + \sqrt{15}$

$\tt y = \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$

$\tt Multiply\ and\ divide\ by \ \sqrt{5} - \sqrt{3}$

$\tt y = \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} \times \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}}$

$\tt y = \dfrac{(\sqrt{5} - \sqrt{3} {)}^{2} }{(\sqrt{5} {)}^{2} - ( \sqrt{3} {)}^{2} }$

$\boxed{ \bold{ \large{ \gray{(A - B)^2 = A^2 - 2AB + B^2}}}}$

$\tt y= \dfrac{5 + 3 - 2 \sqrt{3} \sqrt{5} }{5 - 3 }$

$\tt y= \dfrac{8 - 2 \sqrt{15} }{2 }$

$\tt y= 4 - \sqrt{15}$

$\tt xy= \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$

$\tt xy= \dfrac{(\sqrt{5} {)}^{2} - (\sqrt{3} {)}^{2} }{(\sqrt{5} {)}^{2} -(\sqrt{3} {)}^{2} }$

$\tt xy= \dfrac{5 - 3}{5 - 3}$

$\tt xy= \dfrac{2}{2}$

$\tt xy= 1$

$\tt x + y + xy= 4 + \sqrt{15} + 4 - \sqrt{15} + 1$

$\tt x + y + xy= 4 + 4+ 1$

$\tt x + y + xy=9$

Hence the value of x + y + xy = 9.