If \frac{\sqrt3-1}{\sqrt3+1}=a+b\sqrt3; then a^2+b^2=?

RRB Group D 10-Dec-2018 Shift 1 Solved Paper

Question : 79 of 100

Marks: +1, -0

{√3-1}/{√3+1}=a+b√3

; then

a^2+b^2=?

Solution:

{√3-1}/{√3+1}=a+b√3

{√3-1}/{√3+1}={√3-1}/{√3-1}=a+b√3

{(√3-1)^2}/{√3-1}=a+b√3

2-√3=a+b√3

a=2,b=-1

⇒a^2+b^2=5

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