Problem 86 - Algebra A Combined Function

Question

Solve each equation. $$ (x+6)(x-6)=(2 x-9)(x+4) $$

Step-by-Step Solution

Answer

The solutions are \(x = 0\) and \(x = 1\).

Step 1: Expand the Left Side

First, expand the left side of the equation \((x+6)(x-6)\) using the difference of squares formula. This gives:\[(x+6)(x-6) = x^2 - 36\].

Step 2: Expand the Right Side

Next, expand the right side \((2x-9)(x+4)\) using the distributive property:\[(2x-9)(x+4) = 2x(x+4) - 9(x+4) = 2x^2 + 8x - 9x - 36 = 2x^2 - x - 36\].

Step 3: Equate and Form simple Equation

Set the expanded left side equal to the expanded right side:\[x^2 - 36 = 2x^2 - x - 36\].Subtract \(x^2\) and add 36 from both sides to simplify the equation.

Step 4: Simplify the Equation

Perform the simplification:\[0 = 2x^2 - x - 36 - x^2 + 36\], which combines to:\[0 = x^2 - x\].

Step 5: Factor the Equation

Factor the equation \(x^2 - x = 0\) by taking out the common factor \(x\):\[x(x - 1) = 0\].

Step 6: Solve for x

Set each factor equal to zero and solve for \(x\):\[x = 0\] \[x - 1 = 0 \Rightarrow x = 1\].

Key Concepts

Difference of Squares

When dealing with quadratic expressions like \((x+6)(x-6)\), the difference of squares formula is a valuable tool. The formula \(a^2 - b^2 = (a + b)(a - b)\) is used to simplify expressions where one perfect square is subtracted from another. In our problem, \(x+6\) and \(x-6\) are the components that form the difference of squares:

• Here, \(a = x\) and \(b = 6\).
• When applied, \((x+6)(x-6)\) results in \(x^2 - 36\).

This transformation enables us to deal with a simpler polynomial expression, making subsequent operations like solving equations much more manageable. It's important to remember this shortcut, as it saves time and reduces complexity when expanding expressions formed by the product of two binomials structured this way. Always look for opportunities to apply the difference of squares in algebraic expressions for quicker simplification.

Distributive Property

The distributive property is a fundamental technique in algebra, allowing us to expand expressions like \((2x-9)(x+4)\). In essence, the distributive property involves multiplying each term inside a set of parentheses by a term outside of it.

In our exercise, we involved the following steps:- Distribute \(2x\) to both \(x\) and \(4\), leading to \(2x^2 + 8x\).- Distribute \(-9\) to both \(x\) and \(4\), leading to \(-9x - 36\).

After distributing and combining like terms, the result is \(2x^2 - x - 36\). This property is crucial in simplifying expressions and setting the stage for solving equations, as it allows the expansion and combination of terms leading to a more straightforward equation to solve. Remember, applying the distributive property correctly is key to unpacking more complex algebraic expressions.

Factoring

Factoring is the process of breaking down an expression into a product of its factors, and it's integral for simplifying equations. In our exercise, after simplifying both sides of the equation, we arrived at \(x^2 - x = 0\). Recognizing common factors in each term of the equation is the first step in factoring.

• Here, both terms share a common factor of \(x\).
• By pulling out the \(x\), we rewrite the equation as \(x(x - 1) = 0\).

Factoring simplifies the process of solving the equation, as it reduces multi-term polynomials into their linear components. It reveals the roots of the equation easily by setting each factor equal to zero, leading directly to the solutions. In practice, recognizing and applying common factoring techniques will often streamline finding solutions to polynomial equations.

Solving Equations

Solving equations requires putting all the learned algebraic techniques into practice to find the variable values that satisfy the equation. In this exercise, once we had \(x(x - 1) = 0\) through factoring, solving it involved setting each factor equal to zero:

• Setting \(x = 0\) gives one solution.
• For \(x - 1 = 0\), solving gives \(x = 1\).

This method, often referred to as the **zero-product property**, is crucial because it tells us that if a product of factors equals zero, at least one factor must also be zero. Therefore, solving an equation through factoring uncovers all possible solutions, simplifying the equation-solving process into manageable steps. Always remember, once the equation is factored, solving becomes almost straightforward as you reduce the problem to simple algebraic equations.