Question

Find 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) $7\dfrac{5}{{12}}$
(B) $7\dfrac{7}{{12}}$
(C) $5\dfrac{5}{{12}}$
(D) $5\dfrac{7}{{12}}$

Answer 1

Hint:Transform the given expression into its simpler form using the identity ${m^{ab}} = {\left( {{m^a}} \right)^b}$ and ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$. Now cancel out the common terms from numerator and denominator. This will give you a simpler term. Take the square root of this expression. Take out all the perfect squares from the radical sign. Take LCM and find a square root. Change the answer into a mixed fraction to find the correct option.

Complete step-by-step answer:
We need to find square root of $\dfrac{{{{\left( {3\dfrac{1}{4}} \right)}^4} - {{\left( {4\dfrac{1}{3}} \right)}^4}}}{{{{\left( {3\dfrac{1}{4}} \right)}^2} - {{\left( {4\dfrac{1}{3}} \right)}^2}}}$.
The given expression can also be written using ${m^{ab}} = {\left( {{m^a}} \right)^b}$ as:
$ \Rightarrow \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}} \right]}^2} - {{\left[ {{{\left( {4\dfrac{1}{3}} \right)}^2}} \right]}^2}}}{{{{\left( {3\dfrac{1}{4}} \right)}^2} - {{\left( {4\dfrac{1}{3}} \right)}^2}}}$
As we know the identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$, using that:
$ \Rightarrow \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}} \right]}^2} - {{\left[ {{{\left( {4\dfrac{1}{3}} \right)}^2}} \right]}^2}}}{{{{\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}}}$
Now the term $\left[ {{{\left( {3\dfrac{1}{4}} \right)}^2} - {{\left( {4\dfrac{1}{3}} \right)}^2}} \right]$ can be cancelled from numerator and denominator:
$ \Rightarrow \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}}} = {\left( {3\dfrac{1}{4}} \right)^2} + {\left( {4\dfrac{1}{3}} \right)^2}$
Therefore, from this transformation we get:
$ \Rightarrow \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}$
If we evaluate this mixed fraction using $p\dfrac{m}{n} = \dfrac{{n \times p + m}}{n}$ , we will get:
$ \Rightarrow \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} = {\left( {\dfrac{{13}}{4}} \right)^2} + {\left( {\dfrac{{13}}{3}} \right)^2}$
According to the question, we need to find the square root of this expression:
$ \Rightarrow \sqrt {{{\left( {\dfrac{{13}}{4}} \right)}^2} + {{\left( {\dfrac{{13}}{3}} \right)}^2}} = \sqrt {{{13}^2} \times \left( {\dfrac{1}{{16}} + \dfrac{1}{9}} \right)} $
Let’s bring out ${13^2}$from the radical sign:
$ \Rightarrow \sqrt {{{\left( {\dfrac{{13}}{4}} \right)}^2} + {{\left( {\dfrac{{13}}{3}} \right)}^2}} = \sqrt {{{13}^2} \times \left( {\dfrac{1}{{16}} + \dfrac{1}{9}} \right)} = 13\sqrt {\dfrac{1}{{16}} + \dfrac{1}{9}} $
Now we can take LCM inside the radical sign, this will give us:
$ \Rightarrow \sqrt {{{\left( {\dfrac{{13}}{4}} \right)}^2} + {{\left( {\dfrac{{13}}{3}} \right)}^2}} = 13\sqrt {\dfrac{1}{{16}} + \dfrac{1}{9}} = 13\sqrt {\dfrac{{9 + 16}}{{16 \times 9}}} = 13\sqrt {\dfrac{{25}}{{16 \times 9}}} $
So, numbers $25,16{\text{ and 9}}$ are a perfect square, and can be brought out of radical sign:
$ \Rightarrow \sqrt {{{\left( {\dfrac{{13}}{4}} \right)}^2} + {{\left( {\dfrac{{13}}{3}} \right)}^2}} = 13\sqrt {\dfrac{{25}}{{16 \times 9}}} = 13 \times \dfrac{5}{{4 \times 3}} = \dfrac{{65}}{{12}}$
We got the answer but the options given are in mixed fraction. This can be converted into mixed fraction using $p\dfrac{m}{n} = \dfrac{{n \times p + m}}{n}$
$ \Rightarrow \sqrt {{{\left( {\dfrac{{13}}{4}} \right)}^2} + {{\left( {\dfrac{{13}}{3}} \right)}^2}} = \dfrac{{65}}{{12}} = \dfrac{{12 \times 5 + 5}}{{12}} = 5\dfrac{5}{{12}}$
Therefore, we get the square root of $\dfrac{{{{\left( {3\dfrac{1}{4}} \right)}^4} - {{\left( {4\dfrac{1}{3}} \right)}^4}}}{{{{\left( {3\dfrac{1}{4}} \right)}^2} - {{\left( {4\dfrac{1}{3}} \right)}^2}}}$ as $5\dfrac{5}{{12}}$.

So, the correct answer is “Option C”.

Note:Notice that the use of identities of algebra was a crucial part of the solution. An alternative approach to this problem can be the use of identity ${a^4} - {b^4} = \left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right)$. This will change the given expression into a simpler form. After this, you can take the square root and solve mixed fractions