How do you find the square root \[289?\]

Hint: The given question is to find out the square root of the given number. Since \[4\] is the square of \[2\]. Therefore the square root of \[4{\text{ is }} \pm 2\]. Hence we can say that square root is nothing but those number if we square the square root number then we get same number back or we can say that square of \[2 = 4\]
Square root of \[4 = \pm 2, - 2\]

Complete step-by-step answer:
The given question is to find out the square root of the given number. Square of any number means to multiply that number with itself and hence the number obtained is known as square of the number. On the other hand if we do the square root of the obtained number we get the same number twice but with opposite signs, One is positive and other is negative. Herein the given question, we have to find out the square root of the given number which is \[289\]
Here\[,\] we have to find out the square root of \[289.\]
We will find out the square root of \[289\] by prime factorization method. Prime factorization method is the method in which we make the possible smallest factors of the given number and from them\[,\]and when we multiply the factors \[,\] we get the same number again.
Here we have to find out the square root of \[289.\].
Since \[289\] is divisible by \[17\] only means from \[2{\text{ to }}16,\]\[289\] is not divisible by any number and \[289,\] on divided by \[17,\] left with \[17\]. Also \[17\] is divisible by \[17,\] therefore we get the \[2\] factors of \[289\] which are \[17{\text{ and }}17.\]
Now \[,\] square root of \[289{\text{ is }}\sqrt {289} \]
And we can write \[289{\text{ as }}17 \times 17\]
Because these are the factors of \[289\]
\[{\text{So, }}\sqrt {289} = \sqrt {17 \times 17} \]
Inside the square root\[,\] twice \[17\] becomes \[17\] if we take \[17\] out of the square root.
Therefore we get
\[\sqrt {289} = 17\]
Hence square root of \[289{\text{ is }}17\]

Note: If we take square root on both sides we get
\[{\left( {\sqrt {289} } \right)^2} = {\left( {17} \right)^2}\]
In left side square root and square gets cancel and we get
\[289 = 289\]
Because square of \[17{\text{ is }}289\] and we also know square of \[2{\text{ is }}4\] and square root of \[4{\text{ is }}2{\text{ and }} - 2\] because if we take square of \[2{\text{ and }}\left( { - 2} \right),\] we get \[4\]. Therefore square root of \[4{\text{ is }}2\] and \[ - 2.\]