121 LearnersLast updated on May 26th, 2025
Square Root of 433
If a number is multiplied by itself, the result is a square. The inverse operation is finding the square root. Square roots are used in various fields such as engineering, physics, and finance. Here, we will discuss the square root of 433.
What is the Square Root of 433?
The square root is the inverse operation of squaring a number. The number 433 is not a perfect square. The square root of 433 can be expressed in both radical and exponential forms. In radical form, it is expressed as √433, whereas in exponential form it is (433)^(1/2). √433 ≈ 20.809, which is an irrational number because it cannot be expressed as a fraction of two integers.
Finding the Square Root of 433
The prime factorization method is typically used for perfect square numbers. However, for non-perfect squares like 433, we use the long division method and the approximation method. Let us now learn the following methods:
- Long division method
- Approximation method
Square Root of 433 by Long Division Method
The long division method is particularly useful for finding the square roots of non-perfect squares. Here are the steps to find the square root using this method:
Step 1: Group the digits of 433 from right to left. Since 433 has only three digits, we consider it as 4 | 33.
Step 2: Find the largest number whose square is less than or equal to 4. This number is 2, because 2^2 = 4. Place 2 as the first digit of the quotient. Subtract 4 from 4 to get a remainder of 0.
Step 3: Bring down 33 to make it the new dividend. Double the divisor (2), which is now 4.
Step 4: Find a digit 'n' such that 4n × n is less than or equal to 33. In this case, n=8, because 48 × 8 = 384, which is greater than 33. Thus, n=7, because 47 × 7 = 329, which is closer.
Step 5: Subtract 329 from 330 (33 with a decimal point added) to get 1. Let the quotient be 20.7.
Step 6: Continue the process to find more decimals. Add two zeros to the remainder to get 100. Repeat the process to get a more precise result.
So, the approximate square root of √433 is 20.809.
Square Root of 433 by Approximation Method
The approximation method is another way to find square roots, which is simpler for a quick estimate. Here is how to approximate the square root of 433:
Step 1: Identify the closest perfect squares around 433. The perfect squares are 400 (20^2) and 441 (21^2). Therefore, √433 lies between 20 and 21.
Step 2: Use interpolation to approximate: (433 - 400) / (441 - 400) = 33 / 41 ≈ 0.805 Adding this to the lower bound gives 20 + 0.805 = 20.805. Thus, √433 ≈ 20.809.
Common Mistakes and How to Avoid Them in the Square Root of 433
Common errors occur when finding square roots, such as ignoring the negative root or making mistakes in long division. Let's explore some common mistakes students make.
Mistake 1
Forgetting about the negative square root
It's important to remember that every positive number has both positive and negative square roots. However, in most practical applications, we use only the positive square root.
For example, √433 ≈ 20.809, but there is also -20.809.
Square Root of 433 Examples
Problem 1
Can you help Alex find the area of a square box if its side length is given as √433?
The area of the square is 433 square units.
Explanation
The area of the square = side^2.
The side length is given as √433.
Area of the square = (√433)^2 = 433.
Therefore, the area of the square box is 433 square units.
Problem 2
A square-shaped patio measuring 433 square feet is built. If each of the sides is √433, what will be the square feet of half of the patio?
216.5 square feet
Explanation
Divide the given area by 2, as the patio is square-shaped.
Dividing 433 by 2 = 216.5 So, half of the patio measures 216.5 square feet.
Problem 3
Calculate √433 × 4.
About 83.236
Explanation
First, find the square root of 433 which is approximately 20.809, then multiply 20.809 by 4. So, 20.809 × 4 ≈ 83.236.
Problem 4
What will be the square root of (216 + 217)?
The square root is 21
Explanation
To find the square root, first calculate the sum of (216 + 217).
216 + 217 = 433, and √433 ≈ 20.809, which rounds to 21.
Therefore, the square root of (216 + 217) is approximately ±21.
Problem 5
Find the perimeter of a rectangle if its length 'l' is √433 units and the width 'w' is 43 units.
The perimeter of the rectangle is approximately 127.618 units.
Explanation
Perimeter of the rectangle = 2 × (length + width). Perimeter = 2 × (√433 + 43) ≈ 2 × (20.809 + 43) = 2 × 63.809 = 127.618 units.
FAQ on Square Root of 433
1.What is √433 in its simplest form?
2.Mention the factors of 433.
3.Calculate the square of 433.
4.Is 433 a prime number?
5.433 is divisible by?
6.How does learning Algebra help students in Vietnam make better decisions in daily life?
7.How can cultural or local activities in Vietnam support learning Algebra topics such as Square Root of 433?
8.How do technology and digital tools in Vietnam support learning Algebra and Square Root of 433?
9.Does learning Algebra support future career opportunities for students in Vietnam?
Important Glossaries for the Square Root of 433
- Square root: The square root of a number is a value that, when multiplied by itself, gives the original number. Example: √16 = 4.
- Irrational number: An irrational number cannot be expressed as a fraction of two integers, where the denominator is not zero. Example: √2.
- Approximation: The process of finding a value that is close to, but not exactly equal to, a certain number. Example: π ≈ 3.14.
- Long division method: A method used to find the square root of a number by dividing and averaging in a step-by-step manner.
- Prime number: A number greater than 1 that has no positive divisors other than 1 and itself. Example: 433.
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About BrightChamps in Vietnam
Jaskaran Singh Saluja
About the Author
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
Fun Fact
: He loves to play the quiz with kids through algebra to make kids love it.
