Question

if x=√3+√2/√3-√2 and y=√3-√2/√3+√2 then find x^2-y^2

Answer 1

Answer:

$40\sqrt{6}$

Step-by-step explanation:

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

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

Identity

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

$x+y=(\sqrt{3} +\sqrt{2} )^2+(\sqrt{3} -\sqrt{2} )^2=10\\x-y=(\sqrt{3} +\sqrt{2} )^2-(\sqrt{3} -\sqrt{2} )^2=4\sqrt{6}$

$x^2-y^2=10*4\sqrt{6} =40\sqrt{6}$

∴$40\sqrt{6}$