Square Root of 512

The square root of 512 can be determined by the prime factorisation and long division method, as 512 is not a perfect square number there is no rational root. A perfect square number is a number that can be written as a square of a particular number and whose square root is a rational number. The square root of a number is just the opposite of squaring. In radical form, the square root of 512 is denoted as √512, and in the exponential form, it is written as (512)^½. In this article, we shall learn how to find the square root of 512.

Learn more about square and square roots.

What is Square Root of 512?

The square root of 512 is a number whose square gives the result 512. Now, 512 is not a perfect square number because we cannot find any whole number which could be multiplied twice to get 512. Thus, we find an approximate value of the square root of 512 as it is an irrational number.

We can also say that the square root of 512 is the root of the quadratic equation x² – 512

x² – 512 = 0

⇒ x² = 512

Taking square roots on both sides, we get

⇒ x = √512

⇒ x = ±16√2.

How to Find the Square Root of 512?

Let us find the square root of 512 using the prime factorisation and long division method. The repeated subtraction method will not work as 512 is not a perfect square, and its square root is an irrational number.

Prime Factorisation Method

To find the square root of 512 by the prime factorisation method, we first prime factorise 512 and then make pairs of two to get the square root.

Prime factorisation of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Square root of 512 = √[2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2] = 2 × 2 × 2 × 2 × √2 = 16√2.

Long Division Method

To find the square root of 512 by the long division method, we shall write 512 as a dividend and pair its digits from right to left. Now, we shall calculate the square root as follows:

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

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Solved Examples on Square Root of 512

Example 1:

Find the diameter of the circle whose area is 512𝜋 m².

Solution:

Let r be the radius of the circle.

The area of the circle = 𝜋r² = 512𝜋 m²

⇒ r² = 512 (taking square root on both the sides)

⇒ r = √512

⇒ r = 16√2 m (taking the positive root as length cannot be negative)

∴ the diameter of the circle = 2 × 16√2 = 32√2 m

Example 2:

Find the length of the base of a triangle whose area is 128 m² and the height is half its base.

Solution:

Let b be the base of the triangle.

Height of the triangle = b/2

Area of the triangle = ½ × b × b/2 = 128 m²

⇒ ¼ × b² = 128

⇒ b² = 512 (taking square root on both the side)

⇒ b = √512 = 16√2 m

∴ the length of the base of the triangle is 16√2 m.

Example 3:

What is the smallest number that should be multiplied with 512 to make it a perfect square number? Also, find the square root of that number obtained.

Solution:

Now, the square root of 512 = 16√2.

Then 512 × 2 = 1024 will be the perfect square number. Thus, 2 is the smallest number that should be multiplied.

√1024 = 32.