Question

Find the square root of $47 + 4\sqrt {33} $.
(A) $\sqrt {44} + \sqrt 3 $
(B) $\sqrt {42} + \sqrt 3 $
(C) $\sqrt {24} + \sqrt 3 $
(D) $\sqrt {44} + \sqrt 2 $

Answer 1

Hint: Square root of an irrational number is also an irrational number. The given number is $47 + 4\sqrt {33} $. For its square root, assume a variable irrational number $\sqrt a + \sqrt b $ for calculation. Square it and equate it to the given number. Then determine the values of $a$ and $b$ by framing equations.

Complete step-by-step answer:
According to the question, the given number is $47 + 4\sqrt {33} $ and it is an irrational number. We have to determine its square root.
We know that the square root of an irrational number is also an irrational number. So let’s assume an irrational number $\sqrt a + \sqrt b $ is the square root of $47 + 4\sqrt {33} $. So we have:
$ \Rightarrow \sqrt a + \sqrt b = \sqrt {47 + 4\sqrt {33} } $
On squaring both sides, we’ll get:
$ \Rightarrow {\left( {\sqrt a + \sqrt b } \right)^2} = 47 + 4\sqrt {33} $
According to an algebraic formula, ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. Using this formula in above equation, we’ll get:
$ \Rightarrow a + b + 2\sqrt {ab} = 47 + 4\sqrt {33} $
Now comparing rational and irrational parts, we have:
\[
   \Rightarrow a + b = 47{\text{ }}.....{\text{(1)}} \\
   \Rightarrow 2\sqrt {ab} = 4\sqrt {33} {\text{ }}.....{\text{(2)}}
 \]
Simplifying equation (2) further, we’ll get:
$ \Rightarrow \sqrt {ab} = 2\sqrt {33} $
Squaring both sides, we’ll get:
$
   \Rightarrow ab = 132 \\
   \Rightarrow b = \dfrac{{132}}{a}
 $
Putting this in equation (1), we have:
$
   \Rightarrow a + \dfrac{{132}}{a} = 47 \\
   \Rightarrow {a^2} - 47a + 132 = 0
 $
For solving this quadratic equation, we will factorize it by breaking the middle term into two terms:
\[
   \Rightarrow {a^2} - 44a - 3a + 132 = 0 \\
   \Rightarrow a\left( {a - 44} \right) - 3\left( {a - 44} \right) = 0 \\
   \Rightarrow \left( {a - 3} \right)\left( {a - 44} \right) = 0
 \]
We will get two different values of $a$:
$ \Rightarrow a = 3{\text{ or }}a = 44$
Putting $a = 3$ in equation (1), we’ll get:
$
   \Rightarrow 3 + b = 47 \\
   \Rightarrow b = 44
 $
Putting $a = 44$ in equation (1), we’ll get:
$
   \Rightarrow 44 + b = 47 \\
   \Rightarrow b = 3
 $
Thus we have:
$ \Rightarrow a = 3{\text{ and }}b = 44{\text{ or }}a = 44{\text{ and }}b = 3$
Therefore our assumed number $\sqrt a + \sqrt b $ will be:
$ \Rightarrow \sqrt a + \sqrt b = \sqrt 3 + \sqrt {44} {\text{ or }}\sqrt {44} + \sqrt 3 $
Hence the square root of $47 + 4\sqrt {33} $ is $\sqrt {44} + \sqrt 3 $.

A is the correct option.

Note: While solving a quadratic equation, if we face any difficulty in factoring it using the middle term, we can also use a direct formula to find its roots.
Let we have a quadratic equation, $a{x^2} + bx + c = 0$, then its roots can be determined using the formula:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$