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In Exercises \(1-38,\) multiply as indicated. If possible, simplify any radical expressions that appear in the product. $$(\sqrt{6}+2)(\sqrt{6}-2)$$

Short Answer

Expert verified
The solution of the given exercise (\( \sqrt{6} \) + 2)(\( \sqrt{6} \) - 2) is 2

Step by step solution

01

Recognize the pattern

Recognize the pattern that the given expressions are in the form of difference of squares, i.e., (a+b)(a-b). Here, a is \( \sqrt{6} \) and b is 2.
02

Apply the formula of the difference of squares

The difference of squares formula is a^2 - b^2. Applying this formula for given expressions, (\( \sqrt{6} \) + 2)(\( \sqrt{6} \) - 2) becomes (\( \sqrt{6} \))^2 - 2^2.
03

Simplify the expression

Now, calculate the squares. (\( \sqrt{6} \))^2 is 6 and 2^2 is 4. Subtract them to get 6-4.
04

Final Simplification

The result of 6-4 is 2 which is the final solution

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radical Expressions
Radical expressions are mathematical expressions that involve roots, usually square roots, cube roots, or higher-order roots. In the given exercise, you see a square root, \( \sqrt{6} \), which is a common type of radical expression.
A radical expression typically consists of a radicand (the number inside the radical sign) and the degree of the root indicated by the index, though in the case of square roots, the index is often omitted.
  • Example: In \( \sqrt{6} \), 6 is the radicand.
Understanding how to handle and simplify radical expressions is key in many areas of algebra.
When you multiply or add radical expressions, it's important to simplify them wherever possible. This often involves rationalizing denominators or reducing square roots by factoring out perfect squares.
In our exercise, instead of simplifying directly by manipulating radicals, we apply a more strategic approach through recognizing patterns like the Difference of Squares, which allows us to simplify the expression more efficiently.
Simplification
Simplification is the process of reducing a mathematical expression to its most basic form. In mathematics, especially in algebra, it's important to simplify your expressions to make them easier to understand and work with.
Simplification includes many different operations, such as combining like terms, reducing fractions, and, like in our exercise, applying algebraic identities.
  • For example, the expression \((\sqrt{6}+2)(\sqrt{6}-2)\) can be simplified using the Difference of Squares identity.
  • This powerful algebraic identity allows us to move from a more complex product to a simple subtraction \( a^2 - b^2 \).
By transforming the original expression with the Difference of Squares, we turned it into \((\sqrt{6})^2 - 2^2\), which simplifies further to 6 - 4. This yields a simple result, 2, which is much more straightforward and easier to interpret.
Simplifying expressions often allows for more clarity in problem-solving and helps avoid errors that can arise from dealing with unnecessarily complicated expressions.
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and arithmetic operations. They form the basis of algebra and are used to represent relationships and solve problems involving unknown quantities.
  • An algebraic expression can be as simple as \( x + 2 \) or as complex as \((\sqrt{6} + 2)(\sqrt{6} - 2)\).
In the exercise provided, the algebraic expression \((\sqrt{6}+2)(\sqrt{6}-2)\) takes advantage of the Difference of Squares to simplify the product of two binomials.

When working with algebraic expressions, identifying common patterns, like difference of squares, can significantly streamline the solving process. Algebraic techniques such as factoring, expanding, and simplifying are essential tools for tackling these problems efficiently.
Applying the formulas correctly, such as \( a^2 - b^2 \) in this case, not only helps in reaching the correct solution but also enhances understanding of underlying algebraic structures and relationships. Thus, mastering algebraic expressions and their simplification remains a fundamental skill throughout mathematics.

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Most popular questions from this chapter

Use a graphing utility to solve each radical equation. Graph each side of the equation in the given viewing rectangle. The equation's solution set is given by the \(x\) -coordinate(s) of the point (s) of intersection. Check by substitution. $$\begin{aligned} &\sqrt{x}+3=5\\\ &[-1,6,1] \text { by }[-1,6,1] \end{aligned}$$

Use a graphing utility to solve each radical equation. Graph each side of the equation in the given viewing rectangle. The equation's solution set is given by the \(x\) -coordinate(s) of the point (s) of intersection. Check by substitution. $$\begin{aligned} &\sqrt{x^{2}+3}=x+1\\\ &[-1,6,1] \text { by }[-1,6,1] \end{aligned}$$

Use a graphing utility to solve each radical equation. Graph each side of the equation in the given viewing rectangle. The equation's solution set is given by the \(x\) -coordinate(s) of the point (s) of intersection. Check by substitution. $$\begin{aligned} &4 \sqrt{x}=x+3\\\ &[-1,10,1] \text { by }[-1,14,1] \end{aligned}$$

The early Greeks believed that the most pleasing of all rectangles were golden rectangles, whose ratio of width to height is \(\frac{w}{h}=\frac{2}{\sqrt{5}-1}\) The Parthenon at Athens fits into a golden rectangle once the triangular pediment is reconstructed. (PICTURE NOT COPY) Rationalize the denominator of the golden ratio. Then use a calculator and find the ratio of width to height, correct to the nearest hundredth, in golden rectangles.

In Exercises \(129-132\), determine if each operation is performed correctly by graphing the function on each side of the equation with your graphing utility. Use the given viewing rectangle. The graphs should be the same. If they are not, correct the right side of the equation and then use your graphing utility to verify the correction. $$\begin{aligned} &(\sqrt{x}+1)^{2}=x+1\\\ &[0,8,1] \text { by }[0,15,1] \end{aligned}$$

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