Question

If $ \sqrt 2 = 1.414 $ , then the square root of $ \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} $ is nearest to ?
(A) $ 0.172 $
(B) $ 0.414 $
(C) $ 0.586 $
(D) $ 1.414 $

Answer 1

Hint: In the given question, we are given the value of $ \sqrt 2 $ in decimal form. We are required to find the square root of the expression $ \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} $ taking the value of $ \sqrt 2 $ as given in the problem itself. In the given rational function, we have an irrational number in the denominator. To simplify the fraction, first we rationalise the denominator in the fraction so as to eliminate radical from the denominator.

Complete step-by-step answer:
In the given problem, we have to find the value of $ \sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} $ . So, in order to find the value of this square root, we have to first rationalise the denominator of the rational function inside the square root so as to simplify the calculations. The denominator of the rational function $ \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} $ is $ \sqrt 2 + 1 $ .
So, we multiply the rational function by the conjugate of the denominator $ \sqrt 2 + 1 $ , that is $ \sqrt 2 - 1 $ .
So, multiplying both the numerator and denominator of the rational function by $ \sqrt 2 - 1 $ , we get,
\[ \Rightarrow \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} \times \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 - 1}}\]
Now, we can easily evaluate the numerator and denominator of the resultant fraction with the use of algebraic identities.
So, we know that $ {a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right) $ . So, applying this algebraic identity in the denominator of the fraction, we get,
\[ \Rightarrow \dfrac{{{{\left( {\sqrt 2 - 1} \right)}^2}}}{{{{\left( {\sqrt 2 } \right)}^2} - {{\left( 1 \right)}^2}}}\]
Simplifying the expression further, we get,
\[ \Rightarrow \dfrac{{{{\left( {\sqrt 2 - 1} \right)}^2}}}{{2 - 1}}\]
\[ \Rightarrow {\left( {\sqrt 2 - 1} \right)^2}\]
So, we get that the rational function given to us as $ \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} $ is same as \[{\left( {\sqrt 2 - 1} \right)^2}\].
Now, we have to evaluate the square root of the function $ \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} $ as asked in the problem.
So, we get,
 $ \Rightarrow \sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} = \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} $
Now, evaluating the square root, we get,
 $ \Rightarrow \sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} = \left| {\sqrt 2 - 1} \right| $
Now, we are given the value of $ \sqrt 2 $ as $ 1.414 $ in the problem. So, we get,
 $ \Rightarrow \sqrt {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} = 1.414 - 1 = 0.414 $
Hence, the value of the square root of $ \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} $ is nearest to $ 0.414 $ . Therefore, option (B) is correct.
So, the correct answer is “Option B”.

Note: If the denominator of a fraction has square root, we rationalise the rational number. We multiply the numerator and denominator by the same number so as not to change the value of the fraction and get rid of the irrational number in the denominator