Problem 61 - Intermediate Algebra

Question

Multiply and then simplify if possible. $$ (2 \sqrt{7}+3 \sqrt{5})(\sqrt{7}-2 \sqrt{5}) $$

Step-by-Step Solution

Answer

The simplified form is \(-16 - \sqrt{35}\).

Step 1: Apply the Distribution

We need to apply the distributive property (also known as the FOIL method for binomials) to the expression \((2 \sqrt{7}+3 \sqrt{5})(\sqrt{7}-2 \sqrt{5})\). This means multiplying each term in the first binomial by each term in the second binomial.

Step 2: Multiply First Terms

Multiply the first terms of each binomial: \(2 \sqrt{7} \times \sqrt{7} = 2 \times 7 = 14\).

Step 3: Multiply Outer Terms

Multiply the outer terms: \(2 \sqrt{7} \times (-2 \sqrt{5}) = -4 \times \sqrt{35}\).

Step 4: Multiply Inner Terms

Multiply the inner terms: \(3 \sqrt{5} \times \sqrt{7} = 3 \times \sqrt{35}\).

Step 5: Multiply Last Terms

Multiply the last terms: \(3 \sqrt{5} \times (-2 \sqrt{5}) = -6 \times 5 = -30\).

Step 6: Combine Like Terms

Add all the terms obtained from the distributions:\[ 14 - 4 \sqrt{35} + 3 \sqrt{35} - 30 \].Combine the like terms:\(-4 \sqrt{35} + 3 \sqrt{35} = -\sqrt{35}\).The simplified expression is:\[ 14 - 30 - \sqrt{35} = -16 - \sqrt{35} \].

Key Concepts

Binomials

Binomials are algebraic expressions that contain two terms. Each term can be a number, a variable, or the product of numbers and variables. In the context of our exercise, the binomials are \(2 \sqrt{7} + 3 \sqrt{5}\) and \(\sqrt{7} - 2 \sqrt{5}\).
These binomials are expressions with radical components. Understanding the structure of binomials is crucial when applying algebraic operations, such as multiplication, because they dictate how terms pair with each other during the process.

• The first term of the first binomial is \(2 \sqrt{7}\), and the first term of the second binomial is \(\sqrt{7}\).
• The second term of the first binomial is \(3 \sqrt{5}\), and the second term of the second binomial is \(-2 \sqrt{5}\).

Recognizing these pairings is important when using methods like the distributive property to expand and simplify expressions. Knowing the definition and structure of binomials helps tremendously in algebraic manipulations.

Distributive Property

The distributive property is a fundamental concept in algebra that allows you to expand or distribute factors over addition or subtraction. In algebraic terms, it states that for any numbers or expressions \(a\), \(b\), and \(c\), the expression \(a(b + c)\) is equivalent to \(ab + ac\).
This concept is particularly useful in manipulating binomials. In our exercise, we applied the distributive property to solve \((2 \sqrt{7} + 3 \sqrt{5})(\sqrt{7} - 2 \sqrt{5})\).

• First, distribute \(2 \sqrt{7}\) to both terms in the second binomial, \(\sqrt{7}\) and \(-2 \sqrt{5}\).
• Then, distribute \(3 \sqrt{5}\) in the same manner.

This process, often referred to as the FOIL method when dealing with two binomials, ensures that each term in the first binomial is multiplied by each term in the second binomial. After distribution, sum the products and combine like terms to simplify the expression further.

Simplifying Radicals

Simplifying radicals involves reducing expressions that contain square roots (or other root types) to their simplest form. This means rewriting the radicals so that no perfect square factors remain under the root sign, if possible. However, in some cases like our exercise, simplifying refers primarily to combining like terms that involve radicals.
After multiplying the binomials and using the distributive property, the result was:\[ 14 - 4 \sqrt{35} + 3 \sqrt{35} - 30 \]
This expression contains terms with radicals, specifically \(-4 \sqrt{35}\) and \(3 \sqrt{35}\). When simplifying radicals in an expression:

• Look for common radical components to combine. Here, \(-4 \sqrt{35}\) and \(3 \sqrt{35}\) combine to \(-\sqrt{35}\).
• Simplify the numerical parts separately, combining \(14\) and \(-30\) to get \(-16\).

The final simplified expression becomes \(-16 - \sqrt{35}\). This step ensures the expression is as straightforward as possible, making further algebraic operations easier.