Question

Find the square root of a number 7744 by Prime Factorisation Method.

Answer 1

Hint: Here, we will proceed by representing the number whose square root is to be found as the product of its prime factors according to the Prime Factorization Method and then using the formulas like ${a^b} \times {c^b} = {\left( {a \times c} \right)^b}$ and $\sqrt {{a^2}} = a$ to get the result.

Complete step-by-step answer:

In order to find the square root of the number 7744, we will represent this number as the product of prime factors of this number.
According to Prime Factorisation Method,
$7744 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 \times 11$
Let us make pairs of two same prime factors from the above product of prime factors, we will get
 $
  7744 = \left( {2 \times 2} \right) \times \left( {2 \times 2} \right) \times \left( {2 \times 2} \right) \times \left( {11 \times 11} \right) \\
   \Rightarrow 7744 = {2^2} \times {2^2} \times {2^2} \times {11^2} \\
 $
By taking the square root on both sides of the above equation, we get
$\sqrt {7744} = \sqrt {{2^2} \times {2^2} \times {2^2} \times {{11}^2}} $
Using the formula ${a^b} \times {c^b} = {\left( {a \times c} \right)^b}$ in the above equation, we get
$\sqrt {7744} = \sqrt {{{\left( {2 \times 2 \times 2 \times 11} \right)}^2}} $
Since, we know that $\sqrt {{a^2}} = a$, the above equation becomes
$
  \sqrt {7744} = 2 \times 2 \times 2 \times 11 \\
   \Rightarrow \sqrt {7744} = 88 \\
 $
Therefore, the square root of the number 7744 is 88.

Note: In this particular problem, we are asked for the square root of the number 7744 (i.e., $\sqrt {7744} = {\left( {7744} \right)^{\dfrac{1}{2}}}$) that’s why we have taken pairs of two same prime factors at a time. But if we would have been asked for a cube root of any number then we would have taken pairs of three same prime factors at a time.