Problem 36 - A Graphical Approach to College Algebra

Question

Solve each equation. For equations with real solutions, support your answers graphically. $$(x+4)(x-1)=-5 x-4$$

Step-by-Step Solution

Answer

The solutions are \( x = 0 \) and \( x = -8 \).

Step 1: Expand the left side of the equation

Start by expanding the expression on the left-hand side of the equation. We have \((x+4)(x-1)\). Use the distributive property (FOIL method) to expand:\[ (x+4)(x-1) = x^2 - x + 4x - 4 = x^2 + 3x - 4 \]

Step 2: Set the equation to zero

Combine the expanded left side with the right side of the equation to form a quadratic equation. The original equation is:\[ x^2 + 3x - 4 = -5x - 4 \] Add \(5x + 4\) to both sides to set the equation to zero:\[ x^2 + 3x - 4 + 5x + 4 = 0 \] Simplify the equation:\[ x^2 + 8x = 0 \]

Step 3: Solve the quadratic equation

Factor the quadratic equation \(x^2 + 8x = 0\). Factor out the common term \(x\) from both terms:\[ x(x + 8) = 0 \]Now set each factor to zero and solve for \(x\):1. \( x = 0 \)2. \( x + 8 = 0 \rightarrow x = -8 \)

Step 4: Verify solutions graphically

Graph the expressions: The left side is \( (x+4)(x-1) \) and the right side is \( -5x - 4 \).Plot both functions: Let \( y_1 = (x+4)(x-1) \)Let \( y_2 = -5x - 4 \)The solutions are the \(x\)-coordinates where the graphs intersect. You will see intersections at \( x = 0 \) and \( x = -8 \).

Step 5: Write final solutions

State the solutions you've found:The solutions to the equation \((x+4)(x-1) = -5x - 4\) are \(x = 0\) and \(x = -8\).

Key Concepts

Distributive Property

The distributive property is a fundamental tool in algebra. It allows us to simplify expressions by distributing one operation over another within parentheses. In this exercise, we use the distributive property to expand

• the expression \((x+4)(x-1)\).

The process is often referred to as "FOIL" (First, Outer, Inner, Last), which helps in multiplying two binomials.
Here’s how it works:

• **First**: Multiply the first terms in each binomial: \(x \times x = x^2\).
• **Outer**: Multiply the outer terms: \(x \times -1 = -x\).
• **Inner**: Multiply the inner terms: \(4 \times x = 4x\).
• **Last**: Multiply the last terms: \(4 \times -1 = -4\).

After applying these steps, we combine all the terms to get

• \(x^2 - x + 4x - 4\).

Simplifying this gives us

• \(x^2 + 3x - 4\).

We used the distributive property to expand and tidy up our polynomial expression!

Factoring

Factoring is essentially about breaking down an expression into products of simpler terms that, when multiplied together, return the original expression. Once we set the quadratic equation

• \(x^2 + 8x = 0\),

we used factoring to find its solutions.
To factor, look for common terms in the equation. Here, the common term in

• \(x^2 + 8x\)

is

• \(x\).

Factoring it out gives

• \(x(x + 8) = 0\).

According to the zero product property, if multiple factors equal zero, then at least one of the factors must be zero. This principle leads us directly to the solutions:

• \(x = 0\) and \(x + 8 = 0\).

Solving

• \(x + 8 = 0\)

gives us another solution:

• \(x = -8\).

Using factoring, we efficiently found the values of \(x\) where the original equation holds true.

Graphical Solutions

Graphical solutions involve representing equations on a graph to visually identify solutions. In the exercise, we had two expressions: the left side,

• \((x+4)(x-1)\)

and the right side,

• \(-5x - 4\).

By plotting these on a graph as

• \(y_1 = (x+4)(x-1)\)
and
• \(y_2 = -5x - 4\),

we can easily see where the two equations intersect.
These intersection points are where the values of \(x\) satisfy both sides of the equation.
The graphical approach makes it clear; you can observe the

• meeting points at \(x = 0\) and \(x = -8\),

confirming our solutions. This visual method is an excellent way to verify algebraic solutions, giving a clear picture of where equations equalize on a coordinate plane.

Intersection Points

Intersection points are where two graphs meet, representing solutions to an equation where two expressions are equal. In our exercise, plotting

• \(y_1 = (x+4)(x-1)\)
and
• \(y_2 = -5x - 4\)

helps find the intersection points. The goal is to discover where

• both lines or curves cross each other.

These crossing points are our solutions, \(x = 0\) and \(x = -8\). These values of \(x\) make both equations equivalent.
Intersection points provide a tangible way to confirm our analytical solutions, ensuring the steps in algebra resonate visually.
By checking these points, we gain confidence that our calculations correctly solve the equation. They help bridge the gap between algebraic manipulation and visual comprehension.