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,
$\begin{matrix}
2|\underline{900} \\
2|\underline{450} \\
3|\underline{225} \\
3|\underline{75} \\
5|\underline{25} \\
5|\underline{5} \\
1
\end{matrix}$
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 $\dfrac{1}{2}$ . So, we now modify the equation as,
$\Rightarrow \sqrt{900}={{\left( {{2}^{2}}\times {{3}^{2}}\times {{5}^{2}} \right)}^{\dfrac{1}{2}}}$
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,
$\Rightarrow \sqrt{900}={{\left( {{2}^{2}} \right)}^{\dfrac{1}{2}}}\times {{\left( {{3}^{2}} \right)}^{\dfrac{1}{2}}}\times {{\left( {{5}^{2}} \right)}^{\dfrac{1}{2}}}$
We also know another property of the indices that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ . Modifying the equation accordingly, we get,
\[\begin{align}
  & \Rightarrow \sqrt{900}=\left( {{2}^{2\times \dfrac{1}{2}}} \right)\times \left( {{3}^{2\times \dfrac{1}{2}}} \right)\times \left( {{5}^{2\times \dfrac{1}{2}}} \right) \\
 & \Rightarrow \sqrt{900}=\left( 2 \right)\times \left( 3 \right)\times \left( 5 \right) \\
 & \Rightarrow \sqrt{900}=30 \\
\end{align}\]
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.