[tex]\sqrt[]{2025}\text{ = 45}[/tex]Explanation:[tex]\begin{gathered} \text{Given:} \\ \sqrt[]{2025} \end{gathered}[/tex]
To find the square root of the number above, first let's break the number down into smaller factors to determine if it is a perfect square
[tex]\begin{gathered} \text{First divide by 5 as it is divisible by it (last digit is 5)} \\ \frac{2025}{5}\text{ = }405 \\ 2025\text{ = 5 }\times\text{ 405} \\ \frac{405}{5}\text{ = 81 (divisible by 5 as last digit is 5)} \\ 405\text{ = 5 }\times81 \\ 2025\text{ = 5 }\times5\text{ }\times\text{ 81} \end{gathered}[/tex]
[tex]\begin{gathered} We\text{ can break down 81 if we don't know its roots}\colon \\ 81\text{ divisible by 3 = }\frac{81}{3}\text{ = 27} \\ 27\text{ divided by 3 = }\frac{27}{3}\text{ = 9} \\ 9\text{ divided by 3 = }\frac{9}{3}\text{ = 3} \\ 3\text{ divided by 3 = 3/3 = 1} \\ \\ We\text{ divided 3 four times} \\ 81\text{ = 3}\times3\times3\times3 \end{gathered}[/tex][tex]\begin{gathered} 2025\text{ = 5}\times5\times81 \\ 2025\text{ = 5}\times5\times3\times3\times3\times3 \\ \text{for it to be p}\operatorname{erf}ect\text{ square, we would have the same number multiplied twice} \\ So\text{ we n}eed\text{ to group our factors in groups of 2 of same number} \\ \\ 2025\text{ = (5}\times5)\times(3\times3)\times(3\times3) \\ \end{gathered}[/tex][tex]\begin{gathered} \sqrt[]{2025}\text{ = }\sqrt[]{\text{(5}\times5)\times(3\times3)\times(3\times3)} \\ we\text{ will take only one of the numbers appearing twice:} \\ =\text{ 5 }\times\text{ 3 }\times\text{ 3} \\ =\text{ 45} \\ \\ \text{Hence, we can say 45 }\times\text{ 45 = 2025} \\ \text{and }\sqrt[]{2025}\text{ = 45} \end{gathered}[/tex]
