Problem 52 - Elementary Algebra

Question

Simplify each expression by performing the indicated operation. $$ (1+\sqrt{3})^{2} $$

Step-by-Step Solution

Answer

Answer: The simplified form of the expression \((1+\sqrt{3})^{2}\) is \(4 + 2\sqrt{3}\).

Step 1: Use the binomial theorem

In our case, we have \((a+b)^2\) where \(a=1\) and \(b=\sqrt{3}\). Using the binomial theorem, we can write this as: $$(1+\sqrt{3})^2 = 1^2 + 2(1)(\sqrt{3}) + (\sqrt{3})^2$$

Step 2: Calculate each term

Calculate the square of each term and the product of the middle term: $$1^2 = 1$$ $$2(1)(\sqrt{3}) = 2\sqrt{3}$$ $$(\sqrt{3})^2 = 3$$

Step 3: Combine the terms

Now, add the terms together to get the simplified expression: $$1 + 2\sqrt{3} + 3$$

Step 4: Simplify the expression

Combine the like terms (1 and 3) to get the final simplified expression: $$1 + 3 + 2\sqrt{3} = 4 + 2\sqrt{3}$$ The simplified expression is \((1+\sqrt{3})^{2} = 4 + 2\sqrt{3}\).

Key Concepts

Binomial Theorem

The Binomial Theorem is a quick way of expanding (or multiplying out) a binomial expression that is raised to a power, such as \( (a+b)^n \). The theorem provides a formula that tells us how to expand an expression of the form \( (a+b)^n \) into a sum involving terms of the form \( a^k b^{n-k} \), where the coefficient of each term is a specific positive integer known as a binomial coefficient. These coefficients can be found in Pascal's triangle or calculated using the formula \( \frac{n!}{k!(n-k)!} \) where \( n! \) denotes the factorial of \( n \).

When simplifying an expression like \( (1+\sqrt{3})^2 \), applying the binomial theorem allows us to expand it systematically into \( 1^2 + 2(1)(\sqrt{3}) + (\sqrt{3})^2 \). This provides a clear structure that helps avoid errors and ensures every term is accounted for in the expansion.

Simplifying Square Roots

Simplifying square roots is about expressing the square root in its simplest radical form. A square root, denoted by \( \sqrt{x} \), asks the question: 'Which number, when multiplied by itself, gives \( x \)?' When a square root of a number is not a perfect square (meaning it doesn't result in a whole number), we simplify the expression by factoring out perfect squares if possible.

For example, \( \sqrt{3} \) cannot be simplified further because 3 is a prime number. However, if we have \( \sqrt{8} \) you can simplify it to \( 2\sqrt{2} \) because \( 8 = 4 \times 2 \) and \( \sqrt{4} = 2 \. In the given problem, the term \( (\sqrt{3})^2 \) is simplified as 3.

Exponentiation

Exponentiation is a mathematical operation involving two numbers, the base \( a \) and the exponent \( n \) in the operation \( a^n \). This operation tells us to multiply the base \( a \) by itself \( n \) times. When \( n \) is a positive integer, the operation is straightforward: for instance, \( 3^2 = 3 \times 3 \.

It becomes slightly more complex with non-integer exponents, such as square roots which can be considered exponents of 1/2, meaning \( x^{1/2} = \sqrt{x} \). In our example, exponentiation comes into play in computing the square of \( \sqrt{3} \) which involves squaring the square root, effectively canceling out the root and leaving us with just the original number under the root, 3.

Basic Algebra

Basic algebra is the branch of mathematics that deals with symbols and rules for manipulating those symbols. It is a unifying thread of almost all of mathematics and includes everything from solving elementary equations to studying abstractions such as groups, rings, and fields.

In the context of our problem, algebra comes into play when we simplify the expression \( (1+\sqrt{3})^{2} \) by first expanding it using the binomial theorem and then combining like terms. Combining like terms is a fundamental concept in algebra that involves adding or subtracting terms with the same variable to a power. For example, the expression \( 1 + 2\sqrt{3} + 3 \) has like terms (the numbers) which we can add to get \( 4 + 2\sqrt{3} \. This step is crucial after expansion, as it leads to the expression in its most simplified form.