Question

c.√16
f.√108

Explanation:

Part c: $\boldsymbol{\sqrt{16}}$

Step1: Recall square root of perfect square

We know that $4\times4 = 16$, so $\sqrt{16}$ is the number which when multiplied by itself gives 16.
$\sqrt{16}=4$

Step1: Factor the radicand

First, we factor 108 into its prime factors. We know that $108 = 36\times3$, and 36 is a perfect square ($6\times6 = 36$).

Step2: Apply square root property

Using the property $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ (where $a\geq0$ and $b\geq0$), we can write $\sqrt{108}=\sqrt{36\times3}=\sqrt{36}\times\sqrt{3}$.
Since $\sqrt{36} = 6$, we have $\sqrt{108}=6\sqrt{3}$.

Answer:

4

Part f: $\boldsymbol{\sqrt{108}}$