Question

Simplify: $3\sqrt 2 + \sqrt[4]{{64}} + \sqrt[4]{{2500}} + \sqrt[6]{8}$
A. $11\sqrt 2 $
B. $2\sqrt {11} $
C. $11\sqrt[2]{2}$
D. $2\sqrt[2]{{11}}$

Answer 1

Hint: This is a simplification question. To solve the above question, we will simplify each term of the question using exponential formulae. Then we will substitute simpler forms of terms in the question to get the answer.

Complete step-by-step answer:
We are given with the question that,
$3\sqrt 2 + \sqrt[4]{{64}} + \sqrt[4]{{2500}} + \sqrt[6]{8}$
First term $3\sqrt 2 $is already in simpler form.
Simplifying the second term, $A = \sqrt[4]{{64}}$
We can also write it as,
$A = {(64)^{\dfrac{1}{4}}}$
We can write 64 as ${2^6}$, substituting it in the above equation we get,
$A = {({2^6})^{\dfrac{1}{4}}}$
Using formula ${({x^a})^{\dfrac{1}{b}}} = {x^{a \times \dfrac{1}{b}}} = {x^{\dfrac{a}{b}}}$in the above equation we get,
$A = {2^{6 \times \dfrac{1}{4}}} = {2^{\dfrac{6}{4}}}$
Simplifying the power of the above equation we get,
$A = {2^{\dfrac{3}{2}}}$
We can also write the above equation putting the formula ${x^{a + b}} = {x^a} \times {x^b}$ as,
$A = 2 \times {2^{\dfrac{1}{2}}}$
$A = 2\sqrt 2 $
Simplifying the third term, $B = \sqrt[4]{{2500}}$
We can also write it as,
$B = {(2500)^{\dfrac{1}{4}}}$
We can write 2500 as ${50^2}$, substituting it in the above equation we get,
$B = {({50^2})^{\dfrac{1}{4}}}$
Using formula ${({x^a})^{\dfrac{1}{b}}} = {x^{a \times \dfrac{1}{b}}} = {x^{\dfrac{a}{b}}}$in the above equation we get,
$B = {50^{2 \times \dfrac{1}{4}}} = {50^{\dfrac{2}{4}}}$
Simplifying the power of the above equation we get,
$B = {50^{\dfrac{1}{2}}}$
We can also write the above equation as,
$B = \sqrt {50} $
Expanding the above question we get,
$B = \sqrt {25 \times 2} $
Taking 25 out of the root we get,
$B = 5\sqrt 2 $
Simplifying the fourth term, $C = \sqrt[6]{8}$
We can also write it as,
$C = {(8)^{\dfrac{1}{6}}}$
We can write 8 as ${2^3}$, substituting it in the above equation we get,
$C = {({2^3})^{\dfrac{1}{6}}}$
Using formula ${({x^a})^{\dfrac{1}{b}}} = {x^{a \times \dfrac{1}{b}}} = {x^{\dfrac{a}{b}}}$in the above equation we get,
$C = {2^{3 \times \dfrac{1}{6}}} = {2^{\dfrac{3}{6}}}$
Simplifying the power of the above equation we get,
$C = {2^{\dfrac{1}{2}}}$
$C = \sqrt 2 $
We are given with the question that,
$3\sqrt 2 + \sqrt[4]{{64}} + \sqrt[4]{{2500}} + \sqrt[6]{8}$
Substituting simpler forms of each term we get,
$ = 3\sqrt 2 + 2\sqrt 2 + 5\sqrt 2 + \sqrt 2 $
Taking $\sqrt 2 $common we get,
$ = \sqrt 2 (3 + 2 + 5 + 1)$
$ = 11\sqrt 2 $
The answer is $11\sqrt 2 $.

Option A is correct.

Note: Students many times write the powers in fraction in complex form, keep in mind the power should be in lowest form i.e. there should be no common factor between the numerator and the denominator.
Whenever adding the values like multiple of square roots or cube roots always take the respective root common and then add its coefficients, otherwise you are suspected to make a calculation mistake. Also, also substitute the value of under root in the end as per requirement of the question.