Suppose we know the area of a square and we want to know the length of the side. What do we do?
The area of the square is 16 square inches. Since the area of a rectangle is length x width, it follows that the area of the square is x times x. (Note that x.x can be written as x^{2})
Therefore, x^{2} = 16.
\(4 \times 4= 16\) , it follows that each side is 4 inches long. (Note that it is also true that\(\left ( 4 \right ) \times \left ( 4 \right )\) , but a negative number is impossible in this problem)
The relationship between 16 and 4 can be expressed in two different ways;
16 is the square of 4 which is written as 16 = 4^{2}
4 is the square root of 16 which is written as \( 4=\sqrt{16} \)
But the value of (4)^{2} is 16 as well. Thus the square root of 16 can be stated as +4 and 4 as well.
We can now give the following formal definition.
\(\sqrt{a}= \pm b\) (where a is a nonnegative number) means that b is the number such that b^{2}= a. b can either be negative or positive, either way, the squared value will be the same.
Examples : \(\sqrt{9}= \pm 3\) Since \(3^{2}= 9\)
\(\sqrt{1}= \pm 1\) Since \(1^{2}= 1\)
Simplifying Square Roots
Square roots are the most common type of radical used. A square root “unsquares” a number. For example, because 5^{2 }= 25, we say the square root of 25 is \(\pm{5}\). The square root of 25 is written as \(\sqrt{25}\)
The radical sign, when first used was an R with a line through the tail, similar to our prescription symbol today. The R came from the Latin, “radix”, which can be translated as “source” or “foundation”. It wasn’t until the 1500’s that our current symbol was first used in Germany (but even then it was just a check mark with no bar over the numbers.
Sample questions:
Example 1.
\(\sqrt{1}= \pm 1\)

\(\sqrt{121}= \pm 11\)

\(\sqrt{4}= \pm 2\)

\(\sqrt{625}= \pm 25\)

\(\sqrt{9}= \pm 3\)

\(\sqrt{81}\)
= undefined 
The final example, \(\sqrt{81}\) is currently undefined as negatives have no square root. This is because if we square a positive or a negative, the answer will be positive. Thus we can only take square roots of positive numbers. In another lesson, we will define a method we can use to work with and evaluate negative square roots, but for now, we will simply say they are undefined. Not all numbers have a nice even square root.
For example, if we found \(\sqrt{8}\) on our calculator, the answer would be 2.828427124746190097603377448419… and even this number is a rounded approximation of the square root. To be as accurate as possible, we will never use the calculator to find decimal approximations of square roots. Instead, we will express roots in the simplest radical form. We will do this using a property known as the product rule of radicals
Product Rule of Square Roots: \(\sqrt{a \times b}= \sqrt{a} \times \sqrt{b}\)
We can use the product rule to simplify an expression such as \(\sqrt{36 \times 5}\)
by splitting it into two roots, \(\sqrt{36} \times \sqrt{5}\) and simplifying the first root, \(6 \sqrt{5}\)
The trick in this 1 process is being able to translate a problem like \(\sqrt{180}\)
Into \(\sqrt{36 \times 5}\)
There are several ways this can be done. The most common and, with a bit of practice, the fastest method, is to find perfect squares that divide evenly into the radicand or number under the radical. This is shown in the next example.
Example 2:\(\sqrt{75}\):
75 is divisible by 25 which is a perfect square
\(\sqrt{25 \times 3}\); Split into factors
\(\sqrt{25} \times \sqrt{3}\): Product rule, take the square root of 25
\( 5 \sqrt{3}\) is our Solution.
If there is a coefficient in front of the radical to begin with, the problem merely becomes a big multiplication problem.
Byju’s help students by delivering chapterwise and detailed solutions to the questions of NCERT books so that they can compare their answers with the sample answers given here – NCERT Solutions for Square Roots.