If x =\frac{\sqrt3+1}{\sqrt3-1} and y=\frac{\sqrt3-1}{\sqrt3+1}; then find the value of (x^2+y^2).

Powers and Roots Practice Test 1

Question : 39 of 40

Marks: +1, -0

x ={√3+1}/{√3-1}

and

y={√3-1}/{√3+1}

; then find the value of

(x^2+y^2).

Solution:

Given,

x ={√3+1}/{√3-1}

and

y={√3-1}/{√3+1}

Then,

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

= [{√3+1}/{√3-1}+{√3-1}/{√3+1}]^2-2×{√3+1}/{√3-1}×{√3-1}/{√3+1}

= [({{√3+1})^2+( {√3-1})^2}/{({√3-1}) ({√3+1})}]^2- 2

=({3+1+2√3+3+1-2√3}/{3-1})^2-2

=(8/2)^2-2=16-2=14

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