Problem 35 - Elementary and Intermediate Algebra

Question

Multiply and simplify. All variables represent positive real numbers. $$ (\sqrt{2}+1)(\sqrt{2}-3) $$

Step-by-Step Solution

Answer

The product simplifies to \(-1 - 2\sqrt{2}\).

Step 1: Apply the Distributive Property

Use the distributive property to expand the expression \((\sqrt{2}+1)(\sqrt{2}-3)\). This means multiplying each term in the first binomial by each term in the second binomial.\[(\sqrt{2}+1)(\sqrt{2}-3) = \sqrt{2} \cdot \sqrt{2} + \sqrt{2} \cdot (-3) + 1 \cdot \sqrt{2} + 1 \cdot (-3)\]

Step 2: Simplify the Expression

Calculate each of the multiplications and simplify:1. \(\sqrt{2} \cdot \sqrt{2} = (\sqrt{2})^2 = 2\)2. \(\sqrt{2} \cdot (-3) = -3\sqrt{2}\)3. \(1 \cdot \sqrt{2} = \sqrt{2}\)4. \(1 \cdot (-3) = -3\)Now combine these results:\[2 - 3\sqrt{2} + \sqrt{2} - 3\]

Step 3: Combine Like Terms

Combine the like terms from the simplified expression:There are two constant terms: \(2\) and \(-3\), which add up to \(-1\).There are also two \(\sqrt{2}\) terms: \(-3\sqrt{2}\) and \(\sqrt{2}\), which add to \(-2\sqrt{2}\).Thus, the final expression is:\[-1 - 2\sqrt{2}\]

Key Concepts

Distributive Property

The distributive property is a fundamental tool in algebra that helps in simplifying expressions, especially when dealing with multiplication over addition or subtraction. The formula is generally expressed as \( a(b+c) = ab + ac \). This means you multiply each term inside the parenthesis by the term outside. In the context of binomials, like in our original exercise, you distribute each term in the first binomial across every term in the second.
For example, the problem \((\sqrt{2}+1)(\sqrt{2}-3)\) requires us to apply the distributive property to each term:

• Multiply \(\sqrt{2}\) by \(\sqrt{2}\), resulting in \((\sqrt{2})^2 = 2\).
• Multiply \(\sqrt{2}\) by \(-3\), resulting in \(-3\sqrt{2}\).
• Multiply \(1\) by \(\sqrt{2}\), resulting in \(\sqrt{2}\).
• Multiply \(1\) by \(-3\), resulting in \(-3\).

This step shows how the distributive property expands a simple multiplication into multiple simpler operations, making the problem easier to handle.

Simplifying Expressions

Simplifying expressions is crucial for making mathematical results easier to understand and work with. Once you've applied the distributive property and expanded the expression, the next step is simplification.
In our problem, after expanding we got:

• \(2\) from \((\sqrt{2})^2\)
• \(-3\sqrt{2}\) from \(\sqrt{2} \cdot (-3)\)
• \(\sqrt{2}\) from \(1 \cdot \sqrt{2}\)
• \(-3\) from \(1 \cdot (-3)\)

Now, simplification involves combining these results to further reduce the expression:

• Combine like terms, meaning adding or subtracting similar terms. For instance, the constants \(2\) and \(-3\) are combined to \(-1\).
• The radical terms \(-3\sqrt{2}\) and \(\sqrt{2}\) are "like terms" because they both involve \(\sqrt{2}\). Combined, they result in \(-2\sqrt{2}\).

The simplified expression we arrive at is \(-1 - 2\sqrt{2}\). This process of combining like terms streamlines the expression, making it easier to interpret or manipulate further.

Radicals in Algebra

Radicals, often containing square roots, are commonly encountered in algebra and can sometimes appear intimidating. However, manipulating them follows some straightforward rules.
A radical like \(\sqrt{n}\) is a number that, when squared, gives \(n\). Keep in mind that the distribution and combination of radicals need to adhere to standard arithmetic rules:

• When you multiply \(\sqrt{a}\) by \(\sqrt{a}\), it simplifies to \(a\). This is due to the property that \((\sqrt{a})^2 = a\).
• Like terms involving radicals (e.g., \(3\sqrt{2}\) and \(2\sqrt{2}\)) must have the same radical part to be combined through addition or subtraction, treating them like any other variable term such as \(x\).

When combining radicals, ensure they have identical radicands (the number inside the root symbol). In our problem, the terms \(-3\sqrt{2}\) and \(\sqrt{2}\) can be combined because they share the same radicand, \(\sqrt{2}\), resulting in \(-2\sqrt{2}\). Mastery of these concepts is essential for tackling more complex algebra involving radicals.