Problem 14
Question
For the following problems, simplify each of the square root expressions. $$ (\sqrt{10}-\sqrt{3})(\sqrt{5}+\sqrt{2}) $$
Step-by-Step Solution
Verified Answer
Question: Simplify the following square root expression: $(\sqrt{10} - \sqrt{3})(\sqrt{5} + \sqrt{2})$
Answer: The simplified expression is $5\sqrt{2} + 2\sqrt{5} - \sqrt{15} - \sqrt{6}$.
1Step 1: Distribute the terms
First, we use the distributive property (FOIL) to multiply the two binomials.
$$
(\sqrt{10} - \sqrt{3})(\sqrt{5} + \sqrt{2}) = (\sqrt{10}\times\sqrt{5}) + (\sqrt{10}\times\sqrt{2}) - (\sqrt{3}\times\sqrt{5}) - (\sqrt{3}\times\sqrt{2})
$$
2Step 2: Simplify the square root terms
Now, we simplify the square root terms by multiplying the square root values together.
$$
= \sqrt{50} + \sqrt{20} - \sqrt{15} - \sqrt{6}
$$
3Step 3: Simplify further
Finally, we can simplify a few of the square root terms further by breaking them down into their prime factors.
$$
= \sqrt{2^1\cdot5^2} + \sqrt{2^2\cdot5^1} - \sqrt{3\cdot5} - \sqrt{6}
$$
Now, we can pull out any factors that have a pair:
$$
= 5\sqrt{2} + 2\sqrt{5} - \sqrt{15} - \sqrt{6}
$$
4Step 4: Write the final simplified expression
The given square root expression simplifies to:
$$
5\sqrt{2} + 2\sqrt{5} - \sqrt{15} - \sqrt{6}
$$
Key Concepts
Distributive PropertyFOIL MethodPrime FactorizationRadical Expressions
Distributive Property
The distributive property is a rule in mathematics that allows us to multiply a single term by a sum or difference inside parentheses. Think of distribution as the act of spreading something out or handing it out evenly. In math, if we have a term outside the parentheses, say 'a', and a sum inside like '(b + c)', then distributing 'a' across 'b' and 'c' would give us 'ab + ac'.
Using the distributive property makes it easier to handle complex expressions, such as those involving square roots. For instance, when simplifying \(\sqrt{10}-\sqrt{3})(\sqrt{5}+\sqrt{2})\), you'll use this principle to multiply each term from the first binomial \(\sqrt{10}-\sqrt{3}\) by each term in the second \(\sqrt{5}+\sqrt{2}\), which eventually results in four unique products.
Using the distributive property makes it easier to handle complex expressions, such as those involving square roots. For instance, when simplifying \(\sqrt{10}-\sqrt{3})(\sqrt{5}+\sqrt{2})\), you'll use this principle to multiply each term from the first binomial \(\sqrt{10}-\sqrt{3}\) by each term in the second \(\sqrt{5}+\sqrt{2}\), which eventually results in four unique products.
FOIL Method
The FOIL method is a mnemonic that stands for First, Outer, Inner, Last, which are the steps to multiplying two binomials together. It's essentially an application of the distributive property. To utilize the FOIL method, multiply the First terms in each binomial, then the Outer terms, followed by the Inner terms, and finally the Last terms.
In the exercise \(\sqrt{10}-\sqrt{3})(\sqrt{5}+\sqrt{2})\), you apply FOIL to expand this into \(\sqrt{10}\times\sqrt{5}) + (\sqrt{10}\times\sqrt{2}) - (\sqrt{3}\times\sqrt{5}) - (\sqrt{3}\times\sqrt{2})\). Following FOIL ensures you won't miss any terms and each multiplication step is accounted for.
In the exercise \(\sqrt{10}-\sqrt{3})(\sqrt{5}+\sqrt{2})\), you apply FOIL to expand this into \(\sqrt{10}\times\sqrt{5}) + (\sqrt{10}\times\sqrt{2}) - (\sqrt{3}\times\sqrt{5}) - (\sqrt{3}\times\sqrt{2})\). Following FOIL ensures you won't miss any terms and each multiplication step is accounted for.
Prime Factorization
Prime factorization is the process of breaking down a composite number into the product of its prime factors. It's like finding the basic building blocks for numbers. Prime numbers are those that have only two distinct positive factors: 1 and the number itself.
When factorizing numbers under a square root, like when you have \(\sqrt{50}\), \(\sqrt{20}\), \(\sqrt{15}\), and \(\sqrt{6}\), you search for prime numbers that when multiplied together will give you the original number. For \(\sqrt{50}\), this would be \(2^1 \cdot 5^2\) because 2 and 5 are prime and their product is 50. Finding the prime factors helps in simplifying square root expressions by allowing you to 'pull out' squares from under the radical, simplifying the expression.
When factorizing numbers under a square root, like when you have \(\sqrt{50}\), \(\sqrt{20}\), \(\sqrt{15}\), and \(\sqrt{6}\), you search for prime numbers that when multiplied together will give you the original number. For \(\sqrt{50}\), this would be \(2^1 \cdot 5^2\) because 2 and 5 are prime and their product is 50. Finding the prime factors helps in simplifying square root expressions by allowing you to 'pull out' squares from under the radical, simplifying the expression.
Radical Expressions
Radical expressions involve roots, such as square roots, cube roots, and so on. The square root symbol \(\sqrt{}\) is called a radical, and the number inside is called the radicand. Simplifying radical expressions means breaking them down into the simplest form possible.
For instance, when you encounter \(\sqrt{50} + \sqrt{20} - \sqrt{15} - \sqrt{6}\), you want to express each term as simply as you can. By using prime factorization on the radicands, you identify any perfect squares you can extract outside the radical, which simplifies the expression. This process can lead to the solution \(5\sqrt{2} + 2\sqrt{5} - \sqrt{15} - \sqrt{6}\), which is the simplified form of the original expression and can't be reduced further using elementary operations.
For instance, when you encounter \(\sqrt{50} + \sqrt{20} - \sqrt{15} - \sqrt{6}\), you want to express each term as simply as you can. By using prime factorization on the radicands, you identify any perfect squares you can extract outside the radical, which simplifies the expression. This process can lead to the solution \(5\sqrt{2} + 2\sqrt{5} - \sqrt{15} - \sqrt{6}\), which is the simplified form of the original expression and can't be reduced further using elementary operations.
Other exercises in this chapter
Problem 13
Write the principal and secondary square roots of each number. Use a calculator to obtain a decimal approximation for the two square roots of 35 . Round to two
View solution Problem 13
Simplify each square root. $$ \sqrt{\frac{9}{4}} $$
View solution Problem 14
Simplify each expression by performing the indicated operation. $$ -\sqrt{10}-2 \sqrt{10} $$
View solution Problem 14
For the following problems, simplify the expressions. $$ \frac{\sqrt{75}}{5 \sqrt{3}} $$
View solution