Question

if x=√5+√3/√5-√3 and y=√5-√3/√5+√3,find the value of(x^2+y^2​

Answer 1

Answer:

l think I m not sure but I solve it

Answer 2

Answer:

62

Step-by-step explanation:

$x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

Let us first solve for x and then y:

$\frac{a+b}{a-b}=\frac{(a+b)^{2}}{a^{2}-b^{2}}$

We know that 'a' is √5 and 'b'  is √3

$Numerator:\\(a+b)^{2}=a^{2}+2ab+b^{2}\\(\sqrt{5}+\sqrt{3})^{2}=(\sqrt5)^{2}+2\sqrt{5}\sqrt{3}+(\sqrt{3})^{2}\\5+2\sqrt{15}+3\\8+2\sqrt{15}\\\\Denominator:\\a^{2}-b^{2}\\(\sqrt{5})^{2}-(\sqrt{3})^2\\5-3\\2\\\\x = \frac{8+2\sqrt{15}}{2}$

$y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\\\frac{a-b}{a+b} = \frac{(a-b)^{2}}{a^{2}-b^{2}}\\\\a^{2}-b^{2}=2\\(a-b)^{2}=a^{2}-2ab+b^{2}\\\\numerator:\\(\sqrt{5}-\sqrt{3})^{2}=(\sqrt{5})^{2}-2\sqrt{5}\sqrt{3}+(\sqrt{3})^{2}\\5-2\sqrt{15}+3 \\8-2\sqrt{15}\\\\denominator:\\2$

Now, we know that x = (8+2√15)/2 and y = (8-2√15)/2

x² = (124 + 32√15 )/4 , y² = (124 - 32√15)/4

x²+y² = (124 + 32√15) + (124 - 32√15)/4

x²+y² = 248/4

x²+y² =  62