Question

The value of \((\sqrt{5}+2)^{6}+(\sqrt{5}-2)^{6}\) is (1) a positive integer (2) a negative integer (3) an irrational number (4) a rational number but not an integer

Short answer

Based on the step-by-step solution, find the value of \((\sqrt{5}+2)^{6}+(\sqrt{5}-2)^{6}\) and determine whether it is a positive integer, a negative integer, an irrational number, or a rational number but not an integer. Answer: The value of \((\sqrt{5}+2)^{6}+(\sqrt{5}-2)^{6}\) is \(2(3189+615\sqrt{5})\), which is an irrational number.

Step-by-step solution

Recognize the format

This exercise consists of two parts. In the first part, we need to find the value of \((\sqrt{5}+2)^{6}+(\sqrt{5}-2)^{6}\). In the second part, we need to determine the type of the value obtained.

Expand using the binomial theorem

To find the value of \((\sqrt{5}+2)^{6}+(\sqrt{5}-2)^{6}\), we can use the binomial theorem which states that \((a+b)^{n}=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}\) where \(\binom{n}{k}=\dfrac{n!}{k!(n-k)!}\). Applying the binomial theorem for both terms: \((\sqrt{5}+2)^6 = \sum_{k=0}^6\binom{6}{k}(\sqrt{5})^{6-k}(2)^{k}\) and \((\sqrt{5}-2)^6 = \sum_{k=0}^6\binom{6}{k}(\sqrt{5})^{6-k}(-2)^{k}\)

Combine the sums and simplify

By adding the two sums, we obtain: \((\sqrt{5}+2)^{6}+(\sqrt{5}-2)^{6} = \sum_{k=0}^6\binom{6}{k}(\sqrt{5})^{6-k}(2)^{k} + \sum_{k=0}^6\binom{6}{k}(\sqrt{5})^{6-k}(-2)^{k}\) Combine the two sums: \(\sum_{k=0}^6\binom{6}{k}(\sqrt{5})^{6-k}(2)^{k}+\binom{6}{k}(\sqrt{5})^{6-k}(-2)^{k} = \sum_{k=0}^6\binom{6}{k}(\sqrt{5})^{6-k}[(2)^{k}+(-2)^{k}]\) Now, notice that when k is even, \((2)^{k}+(-2)^{k} = 2(2)^k\), and when k is odd, \((2)^{k}+(-2)^{k} = 0\). So, we can simplify the sum: \(\sum_{k=0}^6\binom{6}{k}(\sqrt{5})^{6-k}[(2)^{k}+(-2)^{k}] = 2\sum_{\substack{k=0\\ k\, even}}^6\binom{6}{k}(\sqrt{5})^{6-k}(2)^{k}\)

Calculate the sum

Now, calculate the sum by substituting the even values of k and the corresponding binomial coefficients: \(2\left[\binom{6}{0}(\sqrt{5})^{6}(2)^0+\binom{6}{2}(\sqrt{5})^{4}(2)^2+\binom{6}{4}(\sqrt{5})^{2}(2)^4+\binom{6}{6}(\sqrt{5})^{0}(2)^6\right]\) \(=2\left[1(\sqrt{5})^{6}+15(\sqrt{5})^{4}(4)+15(\sqrt{5})^{2}(16)+1(64)\right]\) \(=2\left[3125+375\sqrt{5}+240\sqrt{5}+64\right]\) \(=2(3189+615\sqrt{5})\)

Determine the type

Now, we can see that the value of \((\sqrt{5}+2)^{6}+(\sqrt{5}-2)^{6}\) is \(2(3189+615\sqrt{5})\), which is a rational multiple of an irrational number. Thus, the result is an: (3) irrational number

Key concepts

Irrational Numbers

Irrational numbers are numbers that cannot be expressed as the quotient of two integers. They have non-repeating, non-terminating decimal expansions, making them impossible to represent precisely using a fraction. A classic example of an irrational number is the square root of a non-perfect square, such as \(\sqrt{5}\). Since 5 is not a perfect square, \(\sqrt{5}\) cannot be simplified into a fraction, thus classified as irrational.

In the given exercise, the expression \(\sqrt{5}+2\) and \(\sqrt{5}-2\) raise to the power of 6, highlight an essential concept about irrational numbers. When we perform operations involving them, the result might still contain irrational numbers, especially when you apply functions like squaring or other arithmetic operations. Despite converting some parts of these expressions to rational values, like simplifying using binomial theorem, the presence of \(\sqrt{5}\) maintains its irrational nature.

Understanding irrational numbers and their properties helps to recognize why certain types of number results are classified as irrational, not merely because they contain complex operations but due to the inherent characteristics of the numbers involved.

Rational Numbers

Rational numbers are numbers that can be expressed as a fraction \(\frac{a}{b}\), where \(a\) and \(b\) are integers, and \(b eq 0\). These include all integers, fractions, and finite or repeating decimals. Rational numbers are significant because they include values that can be precisely measured and calculated.

In the context of the exercise, understanding rational numbers is crucial when determining the nature of the expression \((\sqrt{5}+2)^6+(\sqrt{5}-2)^6\). Despite the individual components, \(\sqrt{5}+2\) and \(\sqrt{5}-2\), does not make the expression itself a rational number. The expression simplifies to contain multiplying factors involving irrational numbers, affecting the overall rationality.

Hence, even if you multiply a rational number by an irrational number, the outcome remains irrational. This principle is essential when examining various algebraic expressions to deduce their core nature as rational or irrational.

Integer Operations

Integer operations involve basic arithmetic calculations such as addition, subtraction, multiplication, and division, where the operands are integers. Integers are more than just whole numbers; they include negative values and zero as well. Operations with integers follow standard arithmetic rules but can sometimes result in values outside the integer domain, such as when dividing one integer by another that doesn't yield a whole number.

In our exercise, although we are dealing primarily with exponents and binomial expansion, integer operations come into play significantly. Consider the binomial coefficients calculated during the expansion process using the binomial theorem. These coefficients are pure integers, and they guide the simplification stages as multipliers for radical terms, like \((\sqrt{5})^{6-k}\).

Integer operations are essential for reading and predicting results in large polynomial expansions and ensure we preserve the order of operations correctly. While dealing with irrational or rational numbers, maintaining integer operations' accuracy guarantees the calculated expressions' validity and helps identify fundamental number type distinctions.