Question
If $$\sqrt 2 = 1.414{\text{,}}$$ the square root of $$\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}$$ is nearest to = ?
Answer: Option B
$$\eqalign{
& = \frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} \cr
& = \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)}} \times \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 - 1} \right)}} \cr
& = {\left( {\sqrt 2 - 1} \right)^2} \cr
& \therefore \sqrt {\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \cr
& = \left( {\sqrt 2 - 1} \right) \cr
& = \left( {1.414 - 1} \right) \cr
& = 0.414 \cr} $$
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$$\eqalign{
& = \frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} \cr
& = \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)}} \times \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 - 1} \right)}} \cr
& = {\left( {\sqrt 2 - 1} \right)^2} \cr
& \therefore \sqrt {\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \cr
& = \left( {\sqrt 2 - 1} \right) \cr
& = \left( {1.414 - 1} \right) \cr
& = 0.414 \cr} $$
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