Problem 29
Question
Simplify the expression. $$ \sqrt{125} $$
Step-by-Step Solution
Verified Answer
The simplified expression for \( \sqrt{125}\) is \(5\sqrt{5}\).
1Step 1: Factorize the Radicand
We begin by factorizing the radicand \(\sqrt{125}\). Break down the number \(125\) into its prime factors, which are \(5 \times 5 \times 5\). So, \( \sqrt{125} = \sqrt{5 \times 5 \times 5} \).
2Step 2: Understand the Property of Square Root
The square root of a number is such that when it is multiplied by itself gives the original number. That means the square root of \(x^2\) is \(x\). Now applying this to our simplified root, we can pull out pairs of same numbers as they are perfect squares. Therefore, \(\sqrt{5 \times 5 \times 5}= 5\sqrt{5}\).
3Step 3: Final Simplification
The final simplified version of the square root is \(5\sqrt{5}\). There are no other simplifications that can be done.
Key Concepts
FactorizationSquare Root PropertiesPrime Factors
Factorization
Factorization is the process of breaking down a number or expression into a product of its factors. In our problem, we are dealing with the number 125, which needs to be factorized. Factorization helps simplify complex expressions and is particularly useful in simplifying square roots. By expressing 125 as a product of its prime factors, we gain insight into its structure and pave the way toward simplification.
To factorize 125, find the smallest prime number that divides it. Since 125 is an odd number, it is not divisible by 2, the smallest prime number. So, we move to the next prime number, which is 3, but 125 is not divisible by 3 either. Then, we try with 5. Dividing 125 by 5, we get 25, which can also be divided by 5, resulting in another factor of 5 and leaving us with 5. Thus, 125 can be expressed as:
To factorize 125, find the smallest prime number that divides it. Since 125 is an odd number, it is not divisible by 2, the smallest prime number. So, we move to the next prime number, which is 3, but 125 is not divisible by 3 either. Then, we try with 5. Dividing 125 by 5, we get 25, which can also be divided by 5, resulting in another factor of 5 and leaving us with 5. Thus, 125 can be expressed as:
- 5 × 5 × 5
Square Root Properties
Understanding the properties of square roots is key to simplifying them effectively. A square root is essentially the inverse operation of squaring a number. The square root of a number is the value that, when squared, gives the original number. For example, the square root of 9 is 3 because 3 squared is 9. This basic property helps in handling square roots in mathematical problems.
For simplification, one of the most crucial properties is the ability to "take out" pairs of identical factors from under the square root. For instance, if the radicand has pairs of identical factors, each pair can be taken out as a single factor. In our example of \(\sqrt{125} \\), after factorization, we have:
For simplification, one of the most crucial properties is the ability to "take out" pairs of identical factors from under the square root. For instance, if the radicand has pairs of identical factors, each pair can be taken out as a single factor. In our example of \(\sqrt{125} \\), after factorization, we have:
- \( \sqrt{5 \times 5 \times 5} \)
- Since \( \sqrt{5 \times 5} \) equals 5, we can take out one 5.
- The remaining expression inside the square root is simply 5.
Prime Factors
Prime factors are the building blocks of numbers. A prime factor is a factor that is a prime number, meaning it can only be divided by 1 and itself. The use of prime factors in simplification processes like finding square roots is fundamental because they allow the identification of pairs that can be simplified further. Knowing how to find prime factors and apply them in operations is a critical skill in math.
To find the prime factors of any given number, divide the number by the smallest possible prime numbers until you reach 1. In the case of 125, as we broke it down in the factorization section:
To find the prime factors of any given number, divide the number by the smallest possible prime numbers until you reach 1. In the case of 125, as we broke it down in the factorization section:
- We identified that 125 comprises three 5s.
- So the prime factors of 125 are 5, 5, and 5.
Other exercises in this chapter
Problem 29
Find the value of \(b^{2}\)- 4ac for the equation. $$4 x^{2}+5 x+1=0$$
View solution Problem 29
Determine whether the equation has two solutions, one solution, or no real solution. \(4 x^{2}-5 x+1=0\)
View solution Problem 29
Evaluate the expression. Check the results by squaring each root. $$ \pm \sqrt{49} $$
View solution Problem 29
Solve the equation or write no real solution. Write the solutions as integers if possible. Otherwise, write them as radical expressions. $$ n^{2}=49 $$
View solution