Question

The value of \[3\sqrt 2 + \sqrt[4]{{64}} + \sqrt[4]{{2500}} + \sqrt[6]{8}\] is equal to
A.\[11\sqrt 2 \]
B.\[11\sqrt[3]{2}\]
C.\[\sqrt[3]{2}\]
D.\[11\sqrt 4 \]

Answer 1

Hint: In this problem that looks difficult is actually very easy to solve. First we will try to simplify it as much as we can. Simplification is nothing but we will take the perfect square roots ,cube roots etc. then we will get the terms with the same coefficient \[\sqrt 2 \]. This is the step where we will add the terms to get our final answer.

Complete step-by-step answer:
Given that,
\[3\sqrt 2 + \sqrt[4]{{64}} + \sqrt[4]{{2500}} + \sqrt[6]{8}\]
This can be written as
\[ \Rightarrow 3\sqrt 2 + {\left( {64} \right)^{\dfrac{1}{4}}} + {\left( {2500} \right)^{\dfrac{1}{4}}} + {\left( 8 \right)^{\dfrac{1}{6}}}\]
Now we know that \[64 = {8^2},2500 = {50^2}\] and \[8 = {2^3}\]. So let’s replace the above sum with these simplified terms.
\[ \Rightarrow 3\sqrt 2 + {\left( {{8^2}} \right)^{\dfrac{1}{4}}} + {\left( {{{50}^2}} \right)^{\dfrac{1}{4}}} + {\left( {{2^3}} \right)^{\dfrac{1}{6}}}\]
Using the law of indices \[ \to {\left( {{a^m}} \right)^n} = {a^{mn}}\]
\[ \Rightarrow 3\sqrt 2 + \left( {{8^{2 \times \dfrac{1}{4}}}} \right) + \left( {{{50}^{2 \times \dfrac{1}{4}}}} \right) + \left( {{2^{3 \times \dfrac{1}{6}}}} \right)\]
Let’s simplify the indices
\[ \Rightarrow 3\sqrt 2 + \left( {{8^{\dfrac{1}{2}}}} \right) + \left( {{{50}^{\dfrac{1}{2}}}} \right) + \left( {{2^{\dfrac{1}{2}}}} \right)\]
This can be written as
\[ \Rightarrow 3\sqrt 2 + \left( {{{(4 \times 2)}^{\dfrac{1}{2}}}} \right) + \left( {{{\left( {25 \times 2} \right)}^{\dfrac{1}{2}}}} \right) + \left( {{2^{\dfrac{1}{2}}}} \right)\]
But we know that \[{a^{\dfrac{1}{2}}} = \sqrt a \]. So above terms can be written as
\[ \Rightarrow 3\sqrt 2 + \sqrt {4 \times 2} + \sqrt {25 \times 2} + \sqrt 2 \]
Here \[\sqrt 4 = 2\& \sqrt {25} = 5\]. So
\[ \Rightarrow 3\sqrt 2 + 2\sqrt 2 + 5\sqrt 2 + \sqrt 2 \]
Now all terms have \[\sqrt 2 \] in common .Take it out and add the remaining terms.
\[ \Rightarrow \sqrt 2 (3 + 2 + 5 + 1)\]
Add the terms in the bracket we get our answer as
\[ \Rightarrow 11\sqrt 2 \].
This is our answer \[ \Rightarrow 11\sqrt 2 \]
So the correct option is A.

Note: Here only confusion can be in the options regarding the root. Because option A and D have only difference in the number under root. But remember option D has answer 22 not \[11\sqrt 2 \] because \[11 \times 2 = 22\].