Question

Find the square root in the form of binomial surd: $ 56 - 24\sqrt 5 $

Answer 1

Hint: Here first of all we will take the given binomial in the form of the binomial surd considering the sign in between the two terms and will take square and square root concepts to get the values in the required form.

Complete step-by-step answer:
Given expression: $ 56 - 24\sqrt 5 $
Square root of $ 56 - 24\sqrt 5 $
 $ \sqrt {56 - 24\sqrt 5 } = \sqrt a - \sqrt b $ …. (A)
Take square on both the sides of the above equation –
 $ {\left( {\sqrt {56 - 24\sqrt 5 } } \right)^2} = {\left( {\sqrt a - \sqrt b } \right)^2} $
Square and square root cancels each other on the left hand side of the equation. Apply the identity of difference of two terms and its whole square which is expressed as $ {(a - b)^2} = {a^2} - 2ab + {b^2} $
 $ 56 - 24\sqrt 5 = \left( {{{\left( {\sqrt a } \right)}^2} - 2\sqrt a \sqrt b + {{\left( {\sqrt b } \right)}^2}} \right) $
Again, square and square root cancel each other.
 $ 56 - 24\sqrt 5 = \left( {a - 2\sqrt a \sqrt b + b} \right) $
The above expression can be re-written as –
 $ 56 - 24\sqrt 5 = \left( {a + b - 2\sqrt {ab} } \right) $
 $ 56 - 2\sqrt {144 \times 5} = \left( {a + b - 2\sqrt {ab} } \right) $ (when any number goes inside the square it is written twice to keep its original value)
Now compare the terms –
 $ \Rightarrow a + b = 56 $ and $ - 2\sqrt {144 \times 5} = - 2\sqrt {ab} $
 $ - 2\sqrt {12 \times 12 \times 5} = - 2\sqrt {ab} $
Factorize the terms in the above expression –
\[ - 2\sqrt {\underline {6 \times 6} \times \underline {2 \times 2 \times 5} } = - 2\sqrt {ab} \]
Hence, $ a = 36 $ and $ b = 20 $
Placing the above expression in the equation (A)
 $ \sqrt {56 - 24\sqrt 5 } = \sqrt {36} - \sqrt {20} $
This is the required solution.
So, the correct answer is “ $ \sqrt {36} - \sqrt {20} $ ”.

Note: Be good in squares, square roots and the factorization of the terms. Square is the number multiplied with the same number twice. Always remember that the square and square root cancel each other.