Question

Find the square root of a given number $900$ by prime factorisation method?

Answer 1

Hint: We start dividing the number $900$ by $2$ and then by $3$ and then $5$ . We then express $900$ as a product of the prime factors raised to some powers, like $900=2^2\times 3^2\times 5^2$ . Then, we apply the properties of indices and find out the answer.

Complete step by step solution:
Prime factorisation of a number means dividing a number continuously by prime numbers, starting with the lowest possible prime factor of the number, till the number reduces to a prime number. The given number is then written in the form of a product of the prime factors raised to some power. Let us now prime factorise the given number $900$ . Clearly, $900$ being an even number is divisible by the prime number $2$ .
$900$ upon division by $2$ gives $450$ . $450$ upon division by $2$ gives $225$ . $225$ upon division by $3$ gives $75$ . $75$ upon division by $3$ gives $25$ . $25$ upon division by $5$ gives $5$ . This entire thing can be shown schematically as,
$\begin{array}{c}2|\underset{―}{900}\\ 2|\underset{―}{450}\\ 3|\underset{―}{225}\\ 3|\underset{―}{75}\\ 5|\underset{―}{25}\\ 5|\underset{―}{5}\\ 1\end{array}$
Thus, $900$ upon prime factorisation gives $900=2^2\times 3^2\times 5^2$ .
Now, if we want to find out the square root of $900$ , we can simply do so by taking square root on both sides of the equation $900=2^2\times 3^2\times 5^2$ , which gives,
$\Rightarrow \sqrt{900}=\sqrt{2^2\times 3^2\times 5^2}$
Now, the square root is nothing but the power ${\displaystyle \frac{1}{2}}$ . So, we now modify the equation as,
$\Rightarrow \sqrt{900}={(2^2\times 3^2\times 5^2)}^{{\displaystyle \frac{1}{2}}}$
We know the property of indices that, ${\left(a^m{b}^n\right)}^p={\left(a^m\right)}^p\times {\left(b^n\right)}^p$ . So, if we transform the above equation according to this property we get,
$\Rightarrow \sqrt{900}={\left(2^2\right)}^{{\displaystyle \frac{1}{2}}}\times {\left(3^2\right)}^{{\displaystyle \frac{1}{2}}}\times {\left(5^2\right)}^{{\displaystyle \frac{1}{2}}}$
We also know another property of the indices that ${\left(a^m\right)}^n=a^{mn}$ . Modifying the equation accordingly, we get,
$$\begin{array}{rl}& \Rightarrow \sqrt{900}=\left(2^{2\times {\displaystyle \frac{1}{2}}}\right)\times \left(3^{2\times {\displaystyle \frac{1}{2}}}\right)\times \left(5^{2\times {\displaystyle \frac{1}{2}}}\right)\\ & \Rightarrow \sqrt{900}=\left(2\right)\times \left(3\right)\times \left(5\right)\\ & \Rightarrow \sqrt{900}=30\end{array}$$
Therefore, we can conclude that the square root of $900$ is $30$.

Note: Applying Square root after prime factorisation is often useful especially for some numbers, for which do not seem to be a perfect square. But, for executing this entire process successfully and without errors, we should apply the prime factorisation correctly, and should carefully carry out the property of indices.