Question

Simplify the given expression $\dfrac{{3 + \sqrt 6 }}{{\sqrt {75} - \sqrt {48} - \sqrt {32} + \sqrt {50} }}$
(A)$\sqrt 2 $
(B)$\sqrt 3 $
(C) $\sqrt 3 + \sqrt 2 $
(D)$\sqrt 3 - \sqrt 2 $

Answer 1

Hint: To solve the root of number find out the prime factorization of numbers and then rationalize the denominator.

Complete step-by-step answer:
Step 1: Solve the roots given in denominator-
$\sqrt {75} = \sqrt {3 \times 5 \times 5} = \sqrt {3 \times (5 \times 5)} = 5\sqrt 3 $
$\sqrt {48} = \sqrt {2 \times 2 \times 2 \times 2 \times 3} = \sqrt {(2 \times 2) \times (2 \times 2) \times 3} = (2) \times (2)\sqrt 3 = 4\sqrt 3 $
$\sqrt {32} = \sqrt {2 \times 2 \times 2 \times 2 \times 2} = \sqrt {(2 \times 2) \times (2 \times 2) \times 2} = (2) \times (2)\sqrt 3 = 4\sqrt 2 $
$\sqrt {50} = \sqrt {2 \times 5 \times 5} = \sqrt {2 \times (5 \times 5)} = 5\sqrt 2 $
Step 2: Put the values of above roots in given expression-
$\dfrac{{3 + \sqrt 6 }}{{\sqrt {75} - \sqrt {48} - \sqrt {32} + \sqrt {50} }} = \dfrac{{3 + \sqrt 6 }}{{5\sqrt 3 - 4\sqrt 3 - 4\sqrt 2 + 5\sqrt 2 }}$
Step 3: Solve the like terms-
$\dfrac{{3 + \sqrt 6 }}{{(5\sqrt 3 - 4\sqrt 3 ) + ( - 4\sqrt 2 + 5\sqrt 2 )}} = \dfrac{{3 + \sqrt 6 }}{{\sqrt 3 + \sqrt 2 }}$
Step 4: Rationalize the denominator-
$\dfrac{{3 + \sqrt 6 }}{{\sqrt 3 + \sqrt 2 }} = \dfrac{{3 + \sqrt 6 }}{{\sqrt 3 + \sqrt 2 }} \times \dfrac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$$ = \dfrac{{(3 + \sqrt 6 )(\sqrt 3 - \sqrt 2 )}}{{(\sqrt 3 + \sqrt 2 )(\sqrt 3 - \sqrt 2 )}}$
Using Formula in denominator $(a - b)(a + b) = {a^2} - {b^2}$
Therefore, $\dfrac{{3 + \sqrt 6 }}{{\sqrt 3 + \sqrt 2 }} = \dfrac{{3 \cdot \sqrt 3 - 3 \cdot \sqrt 2 + \sqrt 6 \sqrt 3 - \sqrt 6 \sqrt 2 }}{{{{(\sqrt 3 )}^2} - {{(\sqrt 2 )}^2}}}$
$ = \dfrac{{3\sqrt 3 - 3\sqrt 2 + \sqrt {2 \times 3} \sqrt 3 - \sqrt {2 \times 3} \sqrt 2 }}{{3 - 2}}$
$ = \sqrt 3 (3 - 2) = \sqrt 3 $
Hence, Option (B) is correct.

Additional Information:
1. BASIC FORMULAS -
Some important basic formulas need to remember-
${(a + b)^2} = {a^2} + {b^2} + 2a \cdot b$
${(a - b)^2} = {a^2} + {b^2} - 2a \cdot b$
$(a - b)(a + b) = {a^2} - {b^2}$
2. SQUARE ROOT USING PRIME FACTORIZATION LAW -
To find the square root of the given number through prime factorization method we follow the following steps:
(i) First we divide the given number into its prime factor.
(ii) Make the pair of similar factors such that the both factors in each pair are equal.
(iii) Take one factor from the pair.
(iv) Find the product of factors obtained by taking one factor from each pair.
That product is the square root of the given number.
For Example:
To find the square root of 256.