Square Root of 2601

Let us find the square root of 2601 by the prime factorisation and long division method. As 2601 is a perfect square number, we will get a rational root of 2601. A perfect square number is a number which can be written as a square of a particular number. The square root of a number is just the opposite of squaring. If a² = b, then b^1/2 = a. In radical form, the square root of 2601 is denoted as √2601, and in the exponential form, it is written as (2601)^½. In this article, we shall learn how to find the square root of 2601.

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What Is the Square Root of 2601?

The square root of 2601 is a number whose square gives the result 2601. For 2601, if we square ±51, the answer will be 2061. Thus, the square root of 2601 is 51. We can also say that the solutions of the quadratic equation x² – 2601 = 0 are the square roots of 2601.

x² – 2601 = 0

⇒ x² = 2601 (taking square roots on both sides)

⇒ x = √2601 = ±51

How to Find the Square Root of 2601?

Let us find the square root of 2601 using the prime factorisation and long division method. The repeated subtraction method is very efficient in finding the square root of 2601, as 2601 is a large number.

Prime Factorisation Method

To find the square root of 2601 by the prime factorisation method, we first prime factorise 2601 and then make pairs of two to get the square root.

Prime factorisation of 2601 = 3 × 3 × 17 × 17

Square root of 2601 = √[3 × 3 × 17 × 17] = 3 × 17 = 51

Long Division Method

To find the square root of 2601 by the long division method, we shall write 2601 as a dividend and pair its digits from right to left. Now, we have to calculate the square root of 2601:

To learn how to find the square root of any number long division method, click here.

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Solved Examples on Square Root of 2601

Example 1:

Find the radius of the circle whose area is 2601𝜋 m².

Solution:

Let r be the radius of the circle.

The area of the circle = 𝜋r² = 2601𝜋 m²

⇒ r² = 2601 (taking square root on both sides)

⇒ r = √2601

⇒ r = 51 m (taking the positive root as length cannot be negative)

∴ the radius of the circle = 51 m

Example 2:

Find the roots of the equation (x² – 3x)/9 = (687 – x)/3.

Solution:

We have (x² – 3x)/9 = (687 – x)/3

⇒ (x² – 3x)/3 = (687 – x)

⇒ x² – 3x = 2601 – 3x

⇒ x² – 2601 = 0

⇒ x = √2601 = ±51

∴ x = 51 and –51

Example 3:

What is the smallest number that should be subtracted from 2701 to make it a perfect square number? Also, find the square root of the number obtained.

Solution:

Now, 51² = 2601 < 2701 < 2704 = 52²

∴ 2701 – 100 = 2601

∴ 100 is the number that should be subtracted from 2701.

√2601 = 51