Question

Solve $ \sqrt {12 - \sqrt {68 + 48\sqrt 2 } } = $

Answer 1

Hint: When a number is multiplied with itself, the bigger number obtained is called the square of the number that was multiplied with itself. For example, $ 8 \times 8 = {(8)^2} = 64 $ , here 64 is the square of 8. The number which is multiplied with itself to get the bigger number is called the square root of the bigger number, so we can also say that in the example mentioned above 8 is the square root of 64. The square root of an integer can be found out simply but in the given question, we have to find the square root of an irrational number, so we have to first simplify the question to find the square root of \[12 - \sqrt {68 + 48\sqrt 2 } \]. We try to convert all the terms inside the root in the form of $ {a^2} + {b^2} + 2ab = {(a + b)^2}\,and\,{a^2} + {b^2} - 2ab = {(a - b)^2} $ by adding and subtracting appropriate terms.

Complete step-by-step answer:
The most important step here is ,in the simplification of terms under the root we manipulate and try to convert all the terms in the form of
$ {a^2} + {b^2} + 2ab = {(a + b)^2}\,and\,{a^2} + {b^2} - 2ab = {(a - b)^2} $
so that root and square gets cancelled and the simplification becomes easier.
To find the square root of \[12 - \sqrt {68 + 48\sqrt 2 } \] , we have to first find the square root of $ 68 + 48\sqrt 2 $ .
So to find it we rewrite it –
 $
  \sqrt {12 - \sqrt {68 + 48\sqrt 2 } } \\
   = \sqrt {12 - \sqrt {36 + 32 + 2 \times 6 \times 4\sqrt 2 } } \\
   = \sqrt {12 - \sqrt {{{(6)}^2} + {{(4\sqrt 2 )}^2} + 2 \times 6 \times 4\sqrt 2 } } \\
   = \sqrt {12 - \sqrt {{{(6 + 4\sqrt 2 )}^2}} } \\
   = \sqrt {12 - 6 - 4\sqrt 2 } \\
   = \sqrt {6 - 4\sqrt 2 } \;
  $
Again rewriting $ 6 - 4\sqrt 2 $ like $ 68 + 48\sqrt 2 $ , we get –
 $ \sqrt {6 - 4\sqrt 2 } = \sqrt {4 + 2 - 2 \times 2 \times \sqrt 2 } = \sqrt {{{(2)}^2} + {{(\sqrt 2 )}^2} - 2 \times 2 \times \sqrt 2 } = \sqrt {{{(2 - \sqrt 2 )}^2}} = 2 - \sqrt 2 $
Thus, $ \sqrt {12 - \sqrt {68 + 48\sqrt 2 } } = 2 - \sqrt 2 $
So, the correct answer is “$2 - \sqrt 2 $”.

Note: There are many identities for finding the square and cube of two and three terms. In the above question, we have to find the square root so we have to express the term under the square root in the form of a square of a number that is $ \sqrt {{{(a)}^2}} $ so that the square root can be easily obtained that is $ \sqrt {{a^2}} = a $ , but in the question, we are given two terms under the square root so we have to express them in the form of $ \sqrt {{{(a + b)}^2}} $ , that’s why we rewrite the terms as $ {a^2} + {b^2} + 2ab\,or\,{a^2} + {b^2} - 2ab $ because
$ {a^2} + {b^2} + 2ab = {(a + b)^2}\,and\,{a^2} + {b^2} - 2ab = {(a - b)^2} $ .
Thus, the correct answer can be obtained easily.