Problem 29 - Contemporary Precalculus

Question

Expand and (where possible) simplify the expression. $$(\sqrt{x}+1)^{6}$$

Step-by-Step Solution

Answer

Question: Expand and simplify the expression \((\sqrt{x}+1)^{6}\). Answer: The expanded and simplified expression is \(x^3 + 6x^{\frac{5}{2}} + 15x^2 + 20x^{\frac{3}{2}} + 15x + 6x^{\frac{1}{2}} + 1\).

Step 1: Use the binomial theorem

The binomial theorem states that for any positive integer n, and any real numbers a and b: $$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$ Here, we have n = 6, a = \(\sqrt{x}\), and b = 1. We will substitute these values into the formula.

Step 2: Substitute the values and expand

Now, we will substitute n = 6, a = \(\sqrt{x}\), and b = 1 into the binomial theorem: $$(\sqrt{x}+1)^6 = \sum_{k=0}^{6} \binom{6}{k} (\sqrt{x})^{6-k} (1)^k$$ By expanding the expression, we will get: $$(\sqrt{x}+1)^6 = \binom{6}{0} (\sqrt{x})^{6} (1)^0 + \binom{6}{1} (\sqrt{x})^{5} (1)^1 + \binom{6}{2} (\sqrt{x})^{4} (1)^2 + \binom{6}{3} (\sqrt{x})^{3} (1)^3 + \binom{6}{4} (\sqrt{x})^{2} (1)^4 + \binom{6}{5} (\sqrt{x})^{1}(1)^5 + \binom{6}{6} (\sqrt{x})^{0} (1)^6$$

Step 3: Simplify the expressions

Now, we will simplify the expressions: $$(\sqrt{x}+1)^6 = 1(x^3) + 6(x^{\frac{5}{2}}) + 15(x^2) + 20(x^{\frac{3}{2}}) + 15(x) + 6(x^{\frac{1}{2}}) + 1$$ By putting all the terms together, the final simplified expression is: $$(\sqrt{x}+1)^6 = x^3 + 6x^{\frac{5}{2}} + 15x^2 + 20x^{\frac{3}{2}} + 15x + 6x^{\frac{1}{2}} + 1$$

Key Concepts

Binomial Expansion

The binomial expansion is a powerful tool in algebra that allows us to expand expressions raised to a power in the form of \begin{align*}(a+b)^n\text{, where\}a\text{\and\}b\text{\are constants\and\}n\text{\is a positive integer. The expansion is based on the binomial theorem, which is given by the formula:}\text{$$}(a+ b)^n= \text{$$}\text{sum}_{k=0}^{n}\text{binom}{n}{k}a^{n-k}b^k\text{$$}To expand the expression }\text{$$}(\text{sqrt}{x}+1)^{6}\text{$$}, we apply this theorem by identifying }\text{a} = \text{sqrt}{x}\text{\and\}\text{b} = 1. This results in a series of terms that are combinations of }\text{$$}\text{binom}{6}{k}(\text{sqrt}{x})^{6-k}(1)^k\text{$$}, which then need to be simplified by doing the indicated operations. This process allows the original compact expression to be written out in an expanded form, making it easier to understand and work with in later calculations.

Simplifying Expressions

Once the expression has been expanded using the binomial theorem, the next step is simplifying expressions to make them more manageable and to identify like terms. Simplifying can involve several steps:

• Combining like terms, which are terms that have the same variables raised to the same power.
• Reducing coefficients, which involves basic arithmetic to combine constants.
• Applying exponent rules, such as recognizing that \text{$$}(\text{sqrt}{x})^2= x\text{ and }\text{sqrt}{x}^{n}= x^{n/2}\text{$$}.

With the given exercise, simplification involves expressing powers of the square root of x\text{\as powers of\}x\text{\ to a fractional power, which leverages the property that\}\text{sqrt}{x} = x^{1/2}. These simpler forms allow us to see the structure of the polynomial we've created through binomial expansion more clearly. This simplification is vital for further algebraic manipulations and understanding the behavior of the function described by the polynomial.

Radical Equations

Radical equations involve roots, such as square roots or cube roots, within the equation. The key to solving these equations lies in eliminating the radical, which is often done by raising both sides of the equation to the power of the radical−in the case of square roots, we square both sides. A critical point to remember when working with radical equations is to check for extraneous solutions that might be introduced when we perform operations to eliminate the radials.For example, if we have an equation with \text{sqrt}{x}, to isolate x, we would square both sides. However, after finding the value of x, we need to substitute back into the original equation to ensure it is a valid solution since squaring is not a reversible operation over the set of real numbers.It's also worth noting that radical expressions, like those we encounter during binomial expansion, emphasize the need to understand simplifying radicals since the terms we work with often involve fractional exponents as seen with terms like \text{$$}x^{\frac{5}{2}}\text{ or }x^{\frac{3}{2}}\text{$$}. Understanding the properties of radicals and how to manipulate them is crucial to both solving radical equations and simplifying expressions derived from binomial expansions.