Problem 11 - College Algebra

Question

For the following exercises, simplify each expression. $$ \sqrt{98} $$

Step-by-Step Solution

Answer

\(7\sqrt{2}\)

Step 1: Identify the Prime Factors

First, let's start by finding the prime factors of the number under the square root, which is 98. The prime factors of 98 can be found by dividing it by the smallest prime numbers until we reach 1. \( 98 \div 2 = 49 \), and \( 49 \div 7 = 7 \). Thus, 98 can be expressed as the product \( 2 \times 7 \times 7 \).

Step 2: Group the Prime Factors into Pairs

Next, we find pairs of the same prime numbers. In \( 2 \times 7 \times 7 \), the number 7 appears as a pair (since we have two 7s). This will help us simplify the square root by taking one number from each pair outside the square root.

Step 3: Simplify the Square Root

Now that we have identified the pair, we can simplify the expression. The square root of a pair of 7s is 7, because \( \sqrt{7^2} = 7 \). Therefore, we can rewrite the expression as \( 7\sqrt{2} \). This simplification is possible because only one '7' survives from the paired 7s under the square root.

Key Concepts

Prime Factorization

Prime factorization involves breaking down a composite number into the product of its prime numbers. Prime numbers are those greater than 1 that are only divisible by 1 and themselves. For instance, in the example of 98, we start by dividing by the smallest prime number 2, and then proceed to 7, the next smallest prime number that can divide the quotient without a remainder.

• Start with the smallest prime, which is often 2. If 2 doesn't work, move to the next prime which is 3, and so on.
• Continue dividing until you end up with all prime numbers, which can no longer be divided further.

Breaking the number 98, for example, starts as:

• 98 divided by 2 gives 49.
• Next, divide 49 by 7 to result in 7.
• Thus, the prime factors of 98 are 2, 7, and 7 (i.e., \(2 \times 7 \times 7\)).

Using prime factorization allows us to see the numbers that build up the original number, laying the groundwork for simplification in various mathematical processes, including simplifying square roots.

Square Roots

Simplifying square roots might seem complex, but it becomes manageable with the help of prime factorization. Square roots essentially ask what number multiplied by itself will give the original number. In our example of simplifying \(\sqrt{98}\), the process looks into whether part of the number's factorization can form a perfect square.

• Identify the pairs of numbers or numbers that repeat themselves within the factorization.
• Each complete pair will come out from under the square root as a single number.

For our square root:

• The prime factorization gives us \(2 \times 7 \times 7\).
• The numbers 7 and 7 partner to form \(7^2\), earning one 7 out of the root: \(\sqrt{7^2} = 7\).
• This simplifies the expression \(\sqrt{98}\) to \(7\sqrt{2}\), where 2 remains inside the root as it does not form a complete pair.

These steps highlight how simplification turns seemingly complex numbers into more streamlined forms suitable for further mathematical operations.

Algebraic Expressions

Algebraic expressions involve operations like addition, multiplication, and division applied to numbers and variables. Simplifying these expressions often requires similar steps to simplifying numerical expressions, like reducing square roots.

When working with square roots in algebraic expressions:

• Factorize any numbers under the root sign, just as seen with numerical square roots.
• Simplify by taking out pairs as singular entities outside the root.
• Consider variables that might involve the same process, treating them like numbers.

In equations and functions, simplifying terms like \(7\sqrt{2}\) makes expressions easier to work with.

• This form can readily combine with other terms, aiding in finding solutions or simplifying further.
• Understanding this process enhances your ability to manipulate and solve more complex algebraic expressions and equations seamlessly.

By integrating these processes, one can turn complex problems into simpler tasks, reinforcing the importance of mastering square root simplification within algebraic contexts.