Problem 21 - Algebra 1: Concepts and Skills

Question

Simplify the expression. $$ \sqrt{80}-\sqrt{45} $$

Step-by-Step Solution

Answer

The simplified expression is \(\sqrt{5}\).

Step 1: Breaking Down the Numbers into Their Prime Factors

First, break down 80 and 45 into their prime factors. 80 can be expressed as \(2^4 \times 5\) and 45 as \(3^2 \times 5\).

Step 2: Simplifying the Square Roots

Then, simplify each square root: \(\sqrt{80}\) can be simplified to \(4\sqrt{5}\) (since \(2^2\) can come out of the square root) and likewise, \(\sqrt{45}\) simplifies to \(3\sqrt{5}\).

Step 3: Performing the Subtraction

Finally, subtract \(3\sqrt{5}\) from \(4\sqrt{5}\) giving the answer as \(\sqrt{5}\) (as 4-3=1).

Key Concepts

Prime Factorization

Prime factorization is the process of breaking down a whole number into a product of prime numbers. Prime numbers are numbers greater than 1, whose only divisors are 1 and themselves. For example, the number 80 can be broken down into prime factors as follows:

• 80 is divisible by 2, the smallest prime number, so 80 ÷ 2 = 40.
• 40 is also divisible by 2, hence 40 ÷ 2 = 20.
• 20 ÷ 2 = 10, and again 10 ÷ 2 = 5.
• Finally, 5 is a prime number.

So, the prime factorization of 80 is expressed as \(2^4 \times 5\).
Similarly, for 45:

• 45 is divisible by 3, resulting in 45 ÷ 3 = 15.
• 15 is again divisible by 3, thus 15 ÷ 3 = 5.
• 5 is a prime number.

Thus, the prime factorization of 45 is \(3^2 \times 5\). Breaking down numbers into their prime factors simplifies the calculation of square roots and helps in various algebraic manipulations.

Square Roots

A square root of a number is a value that, when multiplied by itself, gives the original number. When simplifying expressions involving square roots, like \(\sqrt{80}\) and \(\sqrt{45}\), we use their prime factorizations.
From the prime factors, we determine which numbers can be "taken out" of the square root. For example:

• From \(80 = 2^4 \times 5\), since \(2^4 = (2^2)^2\), we can take \(2^2\) out of the square root as one 4, leaving behind \(\sqrt{5}\). So, \(\sqrt{80} = 4\sqrt{5}\).
• For \(45 = 3^2 \times 5\), the \(3^2\) can be taken outside the root as one 3, leaving \(\sqrt{5}\) inside. So, \(\sqrt{45} = 3\sqrt{5}\).

The simplification of square roots using prime factorization makes them easier to work with in algebraic expressions.

Algebraic Manipulation

Algebraic manipulation involves performing operations to simplify or solve expressions and equations. In the context of this problem, once we have simplified the individual square roots, we can combine like terms.

• The expression \(\sqrt{80} - \sqrt{45}\) is simplified to \(4\sqrt{5} - 3\sqrt{5}\).
• These terms can be treated like similar "like terms" in algebra because both are in terms of \(\sqrt{5}\).
• Subtracting these, we have \(4\sqrt{5} - 3\sqrt{5} = (4-3)\sqrt{5} = \sqrt{5}\).

The final answer shows how algebraic manipulation and simplification can yield a simplified outcome. This process helps in reducing complexity in algebraic expressions and making calculations much more manageable.