Question

if x= sqrt 3 + sqrt 2 /sqrt 3 - sqrt 2 and y= sqrt 3 - sqrt 2 /sqrt 3 + sqrt 2 . then find the value of x ^ 2 + y ^ 2​

Answer 1

Step-by-step explanation:

hope you understand this answer...

Answer 2

We need to recall the following definition of conjugate.

Conjugate is a change in the sign in the middle of two terms.

($+$ to $-$ ) or ($-$ to $+$)

This problem is about the conjugation of a term.

Given:

$x=\frac{\sqrt{3}+\sqrt{2} }{\sqrt{3}-\sqrt{2} }$  and  $y=\frac{\sqrt{3}-\sqrt{2} }{\sqrt{3}+\sqrt{2} }$

Let's consider,

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

Multiply the numerator and denominator of each term by the conjugate of the denominator.

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

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

$x^{2} +y^2=(5+2\sqrt{6})^2+(5-2\sqrt{6})^2$

$x^{2} +y^2=25+24+20\sqrt{6}+25+24-20\sqrt{6}$

$x^{2} +y^2=98$