Question

find the value of x and y: square root 5 + square root 3 by square root 5 minus square root 3= x+y sqrt 15

Answer 1

Step-by-step explanation:

The value of x+y = \frac{14-\sqrt{5}}{11}1114−5 .

Given equation is \frac{5+\sqrt{3}}{5-\sqrt{3}}5−35+3 = x-y√15

We have to find the value of x+y

We solve the equality first

\frac{5+\sqrt{3}}{5-\sqrt{3}}5−35+3 = x-y\sqrt{15}x−y15

We solve the Left Hand side of the equation,

\frac{5+\sqrt{3}}{5-\sqrt{3}}5−35+3

Multiplying numerator and denominator with the conjugate of the denominator,

= \frac{5+\sqrt{3}}{5-\sqrt{3}}5−35+3 × \frac{5+\sqrt{3}}{5+\sqrt{3}}5+35+3

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

= \frac{(5+\sqrt{3})^2}{5^2 - {(\sqrt{3})}^2}52−(3)2(5+3)2

= \frac{25+10\sqrt{3}+3}{25 - 3}25−325+103+3

= \frac{28+10\sqrt{3}}{22}2228+103

= \frac{28}{22} + \frac{10\sqrt{3}}{22}2228+22103

Now equating with the Right hand side of the equation,

\frac{28}{22} + \frac{10\sqrt{3}}{22}2228+22103 = x-y\sqrt{15}x−y15

Therefore, x = \frac{28}{22}2228 and -y√15 = \frac{10\sqrt{3}}{22}22103

x = \frac{14}{11}1114 and y√15 =  - \frac{\sqrt{5}\sqrt{15}}{11}11515

x = \frac{14}{11}1114 and y =  - \frac{\sqrt{5}}{11}115

Now, x+y = \frac{14-\sqrt{5}}{11}1114−5