Question

Find the square root of a given number by prime factorization method.
900

Answer 1

Hint: First using the prime factorization method we’ll find the prime factors of the given number.
Then using the property of taking the square root of a number we’ll find the square root of the given number using some of the factors obtained by prime factorization.

Complete step-by-step answer:
Given data: the given number is 900
Using the prime factorization method i.e. writing a number as the product of its factors as prime numbers
Let the number say N
Then prime factorization of $N = {P_1}^{e1} \times {P_2}^{e2} \times ...... \times {P_n}^{en}$ where ${P_1},{P_2},......,{P_n}$are prime numbers.
Therefore, prime factorizing 900
$\begin{array}{*{20}{c}}
  2 \\
  2 \\
  3 \\
  3 \\
  5 \\
  5 \\
  1
\end{array}\left| {\begin{array}{*{20}{c}}
  {900} \\
  {450} \\
  {225} \\
  {75} \\
  {25} \\
  5 \\
  1
\end{array}} \right.$
We can write 900 in the form of the product of its factors,
i.e. $900 = 2 \times 2 \times 3 \times 3 \times 5 \times 5$
$\therefore 900 = {2^2} \times {2^2} \times {3^2}.........(i)$
Now we know that if $x = {a^2}$
$\sqrt x = \pm a$and also ${a^2} \times {b^2} = {\left( {ab} \right)^2}$
Now, taking square root on both sides of the equation(i), we get,
$ \Rightarrow \sqrt {900} = \pm (2 \times 3 \times 5)$
Simplifying the terms of bracket, we get,
$ \Rightarrow \sqrt {900} = \pm 30$
Therefore, from the above equation, we can say that the square root of the 900 will +30 and -30.

Note: Mostly we just give the solution of the equation ${x^2} = {a^2}$ as the $x = a$ but this solution is not the only real solution of the equation and the solutions are $x = \pm a$, we just ignore the second the negative case. In this question, we have gone through the same situation so remember to take the negative case also as it will affect our answer.
We can also verify that the solution of the equation ${x^2} = {a^2}$is given by $x = \pm a$
Let's take the equation
i.e. ${x^2} = {a^2}$
subtracting both sides by ${a^2}$
$ \Rightarrow {x^2} - {a^2} = 0$
Using ${y^2} - {z^2} = (y + z)(y - z)$
$ \Rightarrow (x - a)(x + a) = 0$
i.e. $(x - a) = 0$ or $x + a = 0$
$\therefore x = a$or $x = - a$
$\therefore x = \pm a$