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Add or subtract terms whenever possible. $$3 \sqrt{54}-2 \sqrt{24}-\sqrt{96}+4 \sqrt{63}$$

Short Answer

Expert verified
\(\sqrt{6} + 12\sqrt{7}\)

Step by step solution

01

Simplify square roots

Begin by simplify square roots involved in the expression. Find the perfect square factors of each number under the root to simplify the square roots. The perfect squares factors for the numbers are as follows:\(54 = 9 \times 6\)\(24 = 4 \times 6\)\(96 = 16 \times 6\)\(63 = 9 \times 7\)Now, apply the square root to each factor:\(\sqrt{54} = \sqrt{9 \times 6} = 3\sqrt{6}\)\(\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}\)\(\sqrt{96} = \sqrt{16 \times 6} = 4\sqrt{6}\)\(\sqrt{63} = \sqrt{9 \times 7} = 3\sqrt{7}\)Substitute these values back into the expression.
02

Substitute and Combine Like Terms

In the original expression \(3\sqrt{54}-2\sqrt{24}-\sqrt{96}+4\sqrt{63}\), replace the roots:\(3\sqrt{54}-2\sqrt{24}-\sqrt{96}+4\sqrt{63} = 3\cdot3\sqrt{6} - 2\cdot2\sqrt{6} - 4\sqrt{6} + 4\cdot3\sqrt{7} = 9\sqrt{6} - 4\sqrt{6} - 4\sqrt{6} + 12\sqrt{7} = 1\sqrt{6} + 12\sqrt{7}\)
03

Writing the final answer

Rewrite \(1\sqrt{6} + 12\sqrt{7}\) as \(\sqrt{6} + 12\sqrt{7}\), as the 1 coefficient is understood and does not need to be written explicitly. This is the final simplified expression.

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Key Concepts

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

Radicals
Radicals are mathematical expressions that involve root symbols, commonly square roots or higher-degree roots such as cube roots. Simplifying radicals involves rewriting them in their simplest form. In precalculus, students learn to identify and extract perfect square factors from under the square root to simplify the expression. For example, the square root of 54 can be simplified because 54 has a factor of 9, which is a perfect square. By knowing that \( \sqrt{9} = 3 \) and applying the property that \( \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} \) for positive numbers, the radical \( \sqrt{54} \) can be simplified to \( 3\sqrt{6} \). Simplifying radicals is crucial for various algebraic operations, such as adding, subtracting, or even multiplying radicals.
Perfect Square Factors
A perfect square is an integer that can be expressed as the square of another integer. For instance, 16 is a perfect square since it is the result of \( 4^2 \). Identifying perfect square factors within a number underneath a radical symbol allows us to simplify the radical expression. When simplifying the square root of a number, you should factor the number under the root and look for perfect square factors. For instance, when breaking down \( \sqrt{96} \), we recognize that 96 can be expressed as \( 16 \times 6 \), where 16 is a perfect square. We can then simplify \( \sqrt{96} \) to \( 4\sqrt{6} \) because \( \sqrt{16} = 4 \) and \( \sqrt{6} \) remains under the radical. It's an essential technique in precalculus, making larger, more complex radicals far more manageable.
Combine Like Terms
Combining like terms is a fundamental technique used in algebra to simplify expressions. Like terms are terms in an expression that have the same variable raised to the same power, although they may have different coefficients. With radicals, 'like terms' have the same radical part, such as \( \sqrt{6} \) in different terms. When combining like terms in our example \( 9\sqrt{6} - 4\sqrt{6} - 4\sqrt{6} \), we add or subtract the coefficients of the like radical terms: \( 9 - 4 - 4 \) results in 1, so we combine them to get \( \sqrt{6} \). Similarly, if there were another term with \( \sqrt{7} \) instead of \( \sqrt{6} \) as in our example \( 12\sqrt{7} \) , it would remain unchanged because there are no other \( \sqrt{7} \) terms to combine with. This process reduces the expression to its simplest form and makes it easier to handle in further operations.
Precalculus
Precalculus is a branch of mathematics that prepares students for calculus, but it's much more than just an antechamber to more advanced study. It includes a thorough exploration of algebraic concepts like the ones we've discussed: simplifying radicals, factoring to identify perfect squares, and combining like terms. By the time students complete precalculus, they have a strong foundation in these algebraic processes, enabling them to tackle calculus problems with greater confidence. Moreover, understanding these concepts allows students to solve a variety of practical problems in science, engineering, and economics, where simplification of expressions is often required for analysis or solving equations.

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Most popular questions from this chapter

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