SOLUTION
Now, 18 is not a perfect square, because its square root is a recurring decimal
![\sqrt[]{18}=4.2426406\ldots.](https://img.qammunity.org/2023/formulas/mathematics/college/jdw5r84rztl5v14s7hs0039jz0lk9dyz69.png)
18 lies between two perfect squares, which are 16 and 25
![\begin{gathered} \sqrt[]{16}=4 \\ \sqrt[]{25}=5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5m4mk8wd09jyhmvisnfr6yzvbm73az0jph.png)
So the squre root of 18 will be greater than 4, but less than 5.
That is 4 is less than the square root of 18, and square root of 18 is less than 5. Which is
![4<\sqrt[]{18}<5](https://img.qammunity.org/2023/formulas/mathematics/college/r2s6zpuuayb2npe5xe0f3uo1tpckp7emyx.png)
Option A is the correct answer