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.

\(4 \times 4= 16\)

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.

\(\sqrt{a}= \pm b\)^{2}= a. b can either be negative or positive, either way, the squared value will be the same.

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

\(\sqrt{1}= \pm 1\)

### Simplifying Square Roots

Square roots are the most common type of radical used. A square root “un-squares” a number. For example, because 5^{2 }= 25, we say the square root of 25 is \(\pm{5}\)

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, \(\sqrt{-81}\)

For example, if we found \(\sqrt{8}\)

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, \(\sqrt{36} \times \sqrt{5}\)

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

\(\sqrt{25 \times 3}\)

\(\sqrt{25} \times \sqrt{3}\)

\( 5 \sqrt{3}\)

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 chapter-wise 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.

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