Problem 68 - Algebra 2

Question

Simplify each expression. $$ (3-\sqrt{6})^{2} $$

Step-by-Step Solution

Answer

The simplified form of \((3-\sqrt{6})^{2}\) is \(15 - 6 \sqrt{6}\).

Step 1: Understanding the definition of square of a binomial

The formula for square of a binomial \((a-b)^{2}\) is \(a^{2}-2ab+b^{2}\). The given expression is \((3-\sqrt{6})^{2}\) and here, \(a=3\) and \(b=\sqrt{6}\).

Step 2: Apply the formula

Replace \(a\) and \(b\) by 3 and \(\sqrt{6}\) respectively in the formula \(a^{2}-2ab+b^{2}\) to get the new expression \(3^{2}-2(3)(\sqrt{6})+(\sqrt{6})^{2}\).

Step 3: Simplify the math expression

Perform the arithmetic operations to simplify the expression: \(3^{2}\) becomes 9, \(\sqrt{6}^{2}\) becomes 6 and \(2(3)(\sqrt{6})\) becomes \(6\sqrt{6}\). Substituting these we get \(9-6\sqrt{6}+6\). Add the constants together to get a simplified expression of \(15-6\sqrt{6}\).

Key Concepts

Square of a Binomial

Squaring a binomial is a common task when dealing with algebraic expressions. It's like squaring a tiny math puzzle—just a bit more structured! Each binomial consists of two parts, say \(a\) and \(b\). To square a binomial like \((a-b)^2\), you apply the formula:

• Square of the first term \(a^2\)
• Minus double the product of both terms \(-2ab\)
• Plus the square of the second term \(b^2\)

In the case of \((3-\sqrt{6})^2\), \(a\) is 3 and \(b\) is \(\sqrt{6}\). By simply plugging these values into the formula, you can break the expression down into bite-sized manageable pieces:

• \(3^2 = 9\)
• \(-2 \times 3 \times \sqrt{6} = -6\sqrt{6}\)
• \((\sqrt{6})^2 = 6\)

Resulting in the simplified expression of \(9 - 6\sqrt{6} + 6\). This approach eases the workload and guides you to the answer!

Algebraic Expressions

An algebraic expression is like a crafting kit, mixing numbers, variables, and operations to construct equations. Understanding them is crucial for solving different types of math problems. When you deal with expressions such as \((3-\sqrt{6})^2\), you're interacting with numbers and operations that can be simplified using algebraic rules.
These rules are:

• Applying operations like addition, subtraction, multiplication, and division.
• Using algebraic identities to simplify expressions more effectively.

When tackling algebraic expressions, break them down:- Substitute the known parts—like in \((3-\sqrt{6})^2\) to find \(a\) and \(b\). - Use distributive properties, especially in expanding expressions involving multiplication. - Then solve and simplify by performing arithmetic operations step-by-step. Mastering these basic operations makes handling more complex expressions a whole lot simpler.

Radicals Simplification

Simplifying radicals comes into play once you deal with square roots or any root expressions. It’s a way to make things neat and tidy in algebra, helping in reducing expressions to more manageable forms. A radical, like \(\sqrt{6}\), signifies the square root. To simplify a radical, you’ll aim to:

• Express the number under the root with its prime factors.
• Identify perfect squares and extract them from inside the root.

In expressions like \((3-\sqrt{6})^2\), simplification involves understanding that when you square the radical \((\sqrt{6})^2\), it simplifies to 6.
Similarly, terms such as \(-6\sqrt{6}\) are simplified using arithmetic operations with other constants. Breaking it down eases our understanding and manipulation of the expression, letting us reach the final simplified form of \(15-6\sqrt{6}\). Radicals may seem complex, but focus on the number structure and practice makes them clearer.