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Chapter 7: Problem 42

Simplify. \((5+\sqrt{6})(5-\sqrt{2})\)

Short Answer

Expert verified
The simplified expression is 19.

Step by step solution

01

Recognize Formula

Identify that this is a difference of squares problem. The formula for the difference of squares is \((a+b)(a-b) = a^2 - b^2\). Here, \(a = 5\) and \(b = \sqrt{6}\).
02

Apply the Formula

Substitute \(a = 5\) and \(b = \sqrt{6}\) into the difference of squares formula: \[(5)^2 - (\sqrt{6})^2\]
03

Calculate Each Square

Calculate each part of the expression separately: \(5^2 = 25\) and \((\sqrt{6})^2 = 6\).
04

Simplify the Expression

Subtract the results from each other: \(25 - 6 = 19\).

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Key Concepts

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

Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and arithmetic operations such as addition, subtraction, multiplication, and division. They form the building blocks of algebra and help express mathematical ideas concisely. In the expression \((5+\sqrt{6})(5-\sqrt{6})\), the terms are separated by arithmetic operations, and each term comprises constants and variables. Algebraic expressions do not produce direct results but rather represent mathematical relationships and can be simplified or manipulated to solve problems.
Simplifying Expressions
Simplifying expressions involves reducing them to their simplest form, making them easier to work with while retaining their original value. The process often includes combining like terms, factoring, or using special formulas, such as the difference of squares.

For example, in our exercise, \((5+\sqrt{6})(5-\sqrt{6})\) represents the difference of squares formula \((a^2 - b^2)\), where \(a = 5\) and \(b = \sqrt{6}\). By applying the formula, simplifying this expression becomes straightforward. This not only makes it easier to handle but also to understand the mathematical relationship depicted.
Square Roots
The square root of a number is a value that, when multiplied by itself, results in the original number. For example, the square root of 6 is expressed as \(\sqrt{6}\), and multiplying \(\sqrt{6}\) by itself yields 6. Square roots are crucial in algebra, often appearing in formulas, expressions, and equations.

In the given exercise, understanding square roots plays a key role in simplifying the expression. The terms \(5+\sqrt{6}\) and \(5-\sqrt{6}\) each include a square root, which is integral to applying the difference of squares formula and simplifying the expression to its simplest form.

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Most popular questions from this chapter

Find \((f+g)(x),(f-g)(x),(f \cdot g)(x),\) and \(\left(\frac{f}{g}\right)(x)\) for each \(f(x)\) and \(g(x)\) $$ \begin{array}{l}{f(x)=10 x-20} \\ {g(x)=x-2}\end{array} $$

Find each power. $$ (2 \sqrt{x}-3)^{2} $$

OPEN ENDED Determine a value of \(b\) for which \(b^{\frac{1}{6}}\) is an integer.

Simplify each expression. $$ \left(b^{\frac{1}{3}}\right)^{\frac{3}{5}} $$

$$ \sqrt{2}-\sqrt{x+6} \leq-\sqrt{x} $$
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