Problem 47 - Elementary Algebra

Question

For the following problems, select a number (or numbers) to show that For what real number is \((6 a)^{2}\) equal to \(6 a^{2} ?\)

Step-by-Step Solution

Answer

Answer: The equation holds true for a = 0.

Step 1: Simplify the given equation

To solve the equation, we first need to simplify it. We can do that by expanding the left side of the equation: \((6a)^2 = (6a)(6a) = 36a^2\) So, the equation simplifies to: \(36a^2 = 6a^2\)

Step 2: Rearrange and simplify the equation

Move all the terms to one side of the equation: \(36a^2 - 6a^2 = 0\) Now, combine the like terms on the left side: \(30a^2 = 0\)

Step 3: Solve for a

Now, we need to find the value(s) of a for which the equation holds true. Divide both sides of the equation by 30: \(a^2 = 0\) Since the only number whose square is 0 is 0 itself, we can deduce that: \(a = 0\) So, the real number a for which \((6a)^2 = 6a^2\) is a = 0.

Key Concepts

Algebraic Expressions

In algebra, an algebraic expression is a combination of constants, variables, and operations (such as addition, subtraction, multiplication, and division). For example, in the equation \( (6a)^2 = 6a^2 \) that we're examining, \( (6a)^2 \) and \( 6a^2 \) are both algebraic expressions involving the variable \('a'\).

Understanding how to manipulate these expressions is critical when solving equations. Specifically, knowing the properties of exponents and operations allows us to realize that \( (6a)^2 \) symbolizes the square of the expression in the parenthesis—meaning, it must be multiplied by itself—leading to a different outcome than simply squaring the \( 'a' \) in \( 6a^2 \) without first distributing the exponent over the 6. This distinction is central in understanding how algebraic expressions are structured and interpreted.

Simplify Equations

To simplify an equation, we look to reduce it to its most basic form without changing its meaning in order to find the solution. Simplifying includes combining like terms and reducing expressions to make solving easier. For instance, in our example, we simplify the expression \( (6a)^2 \) to \( 36a^2 \) by expanding the square of the binomial (as detailed later).

Simplification often involves rearranging the equation, like we did by moving all terms to one side to get \( 36a^2 - 6a^2 = 0 \) and then combining the like terms to end up at \( 30a^2 = 0\). By methodically simplifying, we reduce the chances of errors, and the equation becomes more straightforward to solve. It's also easier to see that dividing by \( 30 \) reduces the equation to \( a^2 = 0 \)—a clear path to the solution.

Square of a Binomial

The square of a binomial is a special product where a two-term expression (a binomial) is multiplied by itself. It follows a specific pattern: \[ (a + b)^2 = a^2 + 2ab + b^2 \] When dealing with the square of a binomial, we utilize this pattern. However, an often misunderstood aspect is when a coefficient is located outside the binomial, as seen in \( (6a)^2 \).

In our exercise, the expression \( (6a)^2 \) is the square of a binomial where \( 6 \) is the coefficient and \( a \) is the variable. Since no addition or subtraction is within the parentheses, the coefficient itself is also squared in the process—leading to \( 36a^2\). This step is pivotal in realizing that squaring the entire term \( 6a \) yields a different result than individually squaring \( 6 \) and \( a \) as in \( 6a^2\). Understanding this strong distinction allows us to correctly manipulate algebraic expressions.