Problem 32 - Algebra for College Students

Question

Find the following products and express answers in simplest radical form. All variables represent nonnegative real numbers. \((\sqrt{2}+6)(\sqrt{2}-2)\)

Step-by-Step Solution

Answer

The product is \(-2\).

Step 1: Set Up the Expression

We are given the expression \((\sqrt{2}+6)(\sqrt{2}-2)\). Notice that this expression appears to be a difference of squares, which follows the form \((a+b)(a-b) = a^2 - b^2\). Here, \(a = \sqrt{2}\) and \(b = 2\).

Step 2: Apply the Difference of Squares Formula

Using the formula \((a+b)(a-b) = a^2 - b^2\), substitute \(a\) and \(b\) with \(\sqrt{2}\) and \(2\) respectively: \((\sqrt{2})^2 - (2)^2\).

Step 3: Simplify the Squared Terms

Calculate \((\sqrt{2})^2\): this simplifies to \(2\).Calculate \((2)^2\): this simplifies to \(4\).So, we now have the expression \(2 - 4\).

Step 4: Perform the Final Calculation

Subtract 4 from 2: \(2 - 4 = -2\).

Key Concepts

Difference of Squares

The concept of the difference of squares is a powerful identity in algebra.
It provides a nifty way to handle expressions like \((a+b)(a-b)\).
By its very nature, the formula \((a+b)(a-b) = a^2 - b^2\) helps simplify expressions into a more manageable form.
In essence, the difference of squares involves two binomials: one with a sum, the other with a difference.
When multiplied, these cancel out the middle terms, leaving us with just a difference.
Here in the problem, we've identified that \(a = \sqrt{2}\) and \(b = 2\).
This detection allows us to apply the formula directly. Such detection is not just about recognition but understanding the structure. The simplicity of the expression \(a^2 - b^2\) helps streamline calculations. Whether \(a\) or \(b\) are numbers, radicals, or variables, the principle remains the same.
Recognizing this structure can save you huge amounts of time and effort in complex algebraic manipulations.

Simplification of Radicals

Simplifying radicals is a key skill when working with square roots. A radical is in its simplest form when there are no square roots left to find, and no fractions inside the radical.
In our step-by-step solution, we reached the step \( (\sqrt{2})^2 \). When simplifying, knowing that \((\sqrt{n})^2\) equals \(n\) is crucial.
Here, \(\sqrt{2}\) times itself results in \(2\) because the square and the square root cancel out.
It is also important to simplify any resulting expression completely. For instance, when you reach an expression like \(2 - 4\), ensuring that this reduction yields the simplest form, \(-2\), is critical.
Practicing this simplification will help reinforce understanding and ensure accuracy in future problems.

Multiplying Binomials

Multiplying binomials is a foundational skill in algebra, often utilizing patterns or identities like the difference of squares.
Here, we dealt with the multiplication of \((\sqrt{2}+6)(\sqrt{2}-2)\), which seems complex, but simplifies through recognizing patterns.
Typically, multiplying binomials involves combining terms step by step. In formulas like \((a+b)(a-b)=a^2-b^2\), direct multiplication through distribution or employing identities becomes straightforward.
Identifying the pattern and leveraging it reduces effort significantly, providing quick insight into the problem.
Thus, mastering the multiplication of different binomials can greatly aid in transforming strenuous algebraic tasks into manageable calculations.
This knowledge frequently clarifies the operations and deepens understanding of further algebraic manipulations.