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Evaluate each expression in Exercises \(1-12,\) or indicate that the root is not a real number. $$-\sqrt{36}$$

Short Answer

Expert verified
\(-\sqrt{36} = -6\).

Step by step solution

01

Identify the operation

In the given expression, \(-\sqrt{36}\), the operation to be conducted is a square root operation on 36, followed by a negative sign operation.
02

Conduct the square root operation

Taking the square root of the number 36, it results in the number 6. So, \(\sqrt{36} = 6\).
03

Applying the negative sign

The negative sign outside the square root applies to the result of the square root operation. Therefore, \(-\sqrt{36} = -6\).

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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 those containing a root, such as a square root, cube root, or any higher-order root. The symbol \(\sqrt{}\) represents the square root. In radical expressions, the value inside the radical sign is called the radicand. Understanding radical expressions is crucial as they are commonly found in many mathematical calculations. For instance, to find \(\sqrt{36}\), 36 is the radicand, and the output of this operation is the positive number 6, because 6 multiplied by itself gives 36.

Often, there is a need to simplify radical expressions to make calculations more manageable. Radical expressions can also include variables and are simplified using similar principles to those with numeric radicands. When you encounter a negative sign outside a radical, such as in \(-\sqrt{36}\), it affects the final result by making it negative after you've found the square root.
Real Numbers
Real numbers include both rational and irrational numbers, encompassing all numbers that can be located on the number line. This means all whole numbers, fractions, terminating or repeating decimals, and non-repeating, non-terminating decimals like \(\pi\) fall under real numbers. Square roots also generally fall within real numbers, provided the number inside the square root (the radicand) is non-negative.

However, there are certain scenarios where the root might not be considered a 'real' number. When you attempt to take a square root of a negative number, the result is not a real number. For example, \(\sqrt{-36}\) is not a real number, because you cannot find any real number that multiplies by itself to give a negative product. Positive radicands like 36, however, always provide real number roots, ensuring results like 6 sit comfortably within the set of real numbers.
Negative Numbers
Negative numbers are less than zero, and appear on the left side of the number line. They are used to describe a loss, decrease, or decay in various mathematical contexts. When dealing with square roots, it's vital to discern the placement of negatives.

A common source of confusion is when a negative sign is placed outside a square root, such as \(-\sqrt{36}\). Here, the negative is not under the square root but affects the positive result derived from \(\sqrt{36}\). This operation provides \(-6\), as the negative sign alters the final calculation. On the other hand, attempting to find \(\sqrt{-36}\) is a situation involving imaginary numbers, as square rooting a negative doesn't yield a result within the real number set.
  • Negative numbers are the result of a subtraction that exceeds the original number.
  • In square root operations, external negatives adjust the final value post-calculation.
  • Understanding the placement of negatives prevents misunderstandings in both real and complex numbers.

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Most popular questions from this chapter

Use the strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality. To earn an A in a course, you must have a final average of at least \(90 \% .\) On the first four examinations, you have grades of \(86 \%, 88 \%, 92 \%,\) and \(84 \% .\) If the final examination counts as two grades, what must you get on the final to earn an A in the course?

Use the strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality. On two examinations, you have grades of 86 and \(88 .\) There is an optional final examination, which counts as one grade. You decide to take the final in order to get a course grade of A, meaning a final average of at least 90 . a. What must you get on the final to earn an A in the course? b. By taking the final, if you do poorly, you might risk the B that you have in the course based on the first two exam grades. If your final average is less than \(80,\) you will lose your \(\mathrm{B}\) in the course. Describe the grades on the final that will cause this to happen.

Perform the indicated operations. Simplify the result, if possible. $$\frac{a b}{a^{2}+a b+b^{2}}+\left(\frac{a c-a d-b c+b d}{a c-a d+b c-b d} \div \frac{a^{3}-b^{3}}{a^{3}+b^{3}}\right)$$

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. The equation \((2 x-3)^{2}=25\) is equivalent to \(2 x-3=5\).

Will help you prepare for the material covered in the first section of the next chapter. If \(y=4-x^{2},\) find the value of \(y\) that corresponds to values of \(x\) for each integer starting with -3 and ending with 3

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