Question

If x =5-√3/5+√3. and y= 5+√3/5-√3,show that x²-y²=-10√3/11

Answer 1

rationalising x we get
x=
$(5 - \sqrt{3) } ^{2} \div 2$

y =
$(5 + \sqrt{3)} {}^{2} \div 2$
${x}^{2} - {y}^{2} = (x + y)(x - y)$
put value get result

Answer 2

Answer:

$x^2-y^2= -\frac{280\sqrt{3}}{121}$

Step-by-step explanation:

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

Rationalize the right side and we get

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

$x=\frac{(5-\sqrt{3})^2}{22}$

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

Rationalize the right side and we get

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

$y=\frac{(5+\sqrt{3})^2}{22}$

$x+y=\frac{(5-\sqrt{3})^2}{22}+\frac{(5+\sqrt{3})^2}{22}\Rightarrow \frac{56}{22}$

$x-y=\frac{(5-\sqrt{3})^2}{22}-\frac{(5+\sqrt{3})^2}{22}\Rightarrow \frac{-20\sqrt{3}}{22}$

As we know,

$x^2-y^2=(x+y)(x-y)$

Substitute x+y and x-y

$x^2-y^2=\frac{56}{22}\cdot \frac{-20\sqrt{3}}{22}$

$x^2-y^2= -\frac{280\sqrt{3}}{121}$