Question

The square root of $\dfrac{{{{\left( {3\dfrac{1}{4}} \right)}^4} - {{\left( {4\dfrac{1}{3}} \right)}^4}}}{{{{\left( {3\dfrac{1}{4}} \right)}^2} - {{\left( {4\dfrac{1}{3}} \right)}^2}}}$
A) $7\dfrac{1}{2}$
B) $5\dfrac{5}{{12}}$
C) $1\dfrac{1}{{12}}$

Answer 1

Hint:
Simplify the numerator using the relation ${a^4} = {\left( {{a^2}} \right)^2}$. Then factorise the numerator using the identity $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$. Further, cancel the common terms from numerator and denominator. Now, simplify the last expression and take square root of it to get the required answer.

Complete step by step solution:
As, we know that, ${a^4}$ can be written as ${\left( {{a^2}} \right)^2}$
We will rewrite the numerator of the given expression, ${\left( {3\dfrac{1}{4}} \right)^4} - {\left( {4\dfrac{1}{3}} \right)^4} = {\left( {{{\left( {3\dfrac{1}{4}} \right)}^2}} \right)^2} - {\left( {{{\left( {4\dfrac{1}{3}} \right)}^2}} \right)^2}$
Simplify the numerator using the formula, $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$
Therefore, the numerator can be written as,
${\left( {{{\left( {3\dfrac{1}{4}} \right)}^2}} \right)^2} - {\left( {{{\left( {4\dfrac{1}{3}} \right)}^2}} \right)^2} = \left( {{{\left( {3\dfrac{1}{4}} \right)}^2} + {{\left( {4\dfrac{1}{3}} \right)}^2}} \right)\left( {{{\left( {3\dfrac{1}{4}} \right)}^2} - {{\left( {4\dfrac{1}{3}} \right)}^2}} \right)$
Thus, the given expression is,
$\dfrac{{{{\left( {3\dfrac{1}{4}} \right)}^4} - {{\left( {4\dfrac{1}{3}} \right)}^4}}}{{{{\left( {3\dfrac{1}{4}} \right)}^2} - {{\left( {4\dfrac{1}{3}} \right)}^2}}} = \dfrac{{\left( {{{\left( {3\dfrac{1}{4}} \right)}^2} + {{\left( {4\dfrac{1}{3}} \right)}^2}} \right)\left( {{{\left( {3\dfrac{1}{4}} \right)}^2} - {{\left( {4\dfrac{1}{3}} \right)}^2}} \right)}}{{{{\left( {3\dfrac{1}{4}} \right)}^2} - {{\left( {4\dfrac{1}{3}} \right)}^2}}}$
Cancel the term ${\left( {3\dfrac{1}{4}} \right)^2} - {\left( {4\dfrac{1}{3}} \right)^2}$ from numerator and denominator.
This implies,
$\dfrac{{{{\left( {3\dfrac{1}{4}} \right)}^4} - {{\left( {4\dfrac{1}{3}} \right)}^4}}}{{{{\left( {3\dfrac{1}{4}} \right)}^2} - {{\left( {4\dfrac{1}{3}} \right)}^2}}} = {\left( {3\dfrac{1}{4}} \right)^2} + {\left( {4\dfrac{1}{3}} \right)^2}$
Solve the RHS using the concept, $a\dfrac{b}{c} = \dfrac{{a\left( c \right) + b}}{c}$
Therefore, we have
$
  {\left( {3\dfrac{1}{4}} \right)^2} + {\left( {4\dfrac{1}{3}} \right)^2} = {\left( {\dfrac{{3\left( 4 \right) + 1}}{4}} \right)^2} + {\left( {\dfrac{{4\left( 3 \right) + 1}}{3}} \right)^2} \\
   \Rightarrow {\left( {3\dfrac{1}{4}} \right)^2} + {\left( {4\dfrac{1}{3}} \right)^2} = {\left( {\dfrac{{13}}{4}} \right)^2} + {\left( {\dfrac{{13}}{3}} \right)^2} \\
 $
Take ${13^2}$common from the numerator and simplify it further.
$
  {\left( {\dfrac{{13}}{4}} \right)^2} + {\left( {\dfrac{{13}}{3}} \right)^2} = {\left( {13} \right)^2}\left( {{{\left( {\dfrac{1}{4}} \right)}^2} + {{\left( {\dfrac{1}{3}} \right)}^2}} \right) \\
   \Rightarrow {\left( {13} \right)^2}\left( {\dfrac{1}{{16}} + \dfrac{1}{9}} \right) \\
   \Rightarrow {\left( {13} \right)^2}\left( {\dfrac{{9 + 16}}{{16\left( 9 \right)}}} \right) \\
   \Rightarrow {\left( {13} \right)^2}\left( {\dfrac{{25}}{{144}}} \right) \\
   \Rightarrow {\left( {13} \right)^2}{\left( {\dfrac{5}{{12}}} \right)^2} \\
$
We have to calculate the square root of the expression. Taking square root implies taking power $\dfrac{1}{2}$ half of the given expression. Thus, we get,
$
  \sqrt {{{\left( {13} \right)}^2}{{\left( {\dfrac{5}{{12}}} \right)}^2}} = 13 \times \dfrac{5}{{12}} \\
   \Rightarrow \dfrac{{65}}{{12}} \\
   \Rightarrow 5\dfrac{5}{{12}} \\
$
Therefore, the square root of $\dfrac{{{{\left( {3\dfrac{1}{4}} \right)}^4} - {{\left( {4\dfrac{1}{3}} \right)}^4}}}{{{{\left( {3\dfrac{1}{4}} \right)}^2} - {{\left( {4\dfrac{1}{3}} \right)}^2}}}$ is $5\dfrac{5}{{12}}$

Hence, option B is correct.

Note:
In this type of questions, simplification of expression makes the solution easy and short. If the student solves each bracket by calculating the square of each given number, it will lead to lengthy calculations. For these types of questions, always try to factorise the terms and then if possible, cancel the common terms and then solve it further.