Problem 6 - Algebra 2 and Trigonometry

Question

Solve and check each of the equations. \(x^{2}+6 x+5=0\)

Step-by-Step Solution

Answer

The solutions to the equation are \(x = -1\) and \(x = -5\).

Step 1: Identify the Problem Type

We are given a quadratic equation of the form \(x^2 + 6x + 5 = 0\). This equation is in standard form \(ax^2 + bx + c = 0\) where \(a=1\), \(b=6\), and \(c=5\). Our task is to find the values of \(x\) that solve this equation.

Step 2: Set Up to Factor the Quadratic

To factor the quadratic equation, we need to find two numbers whose product is \(a \times c = 5\) and whose sum is \(b = 6\). These numbers will help us break down the middle term and factor the quadratic.

Step 3: Factor the Quadratic

The numbers that multiply to 5 and add to 6 are 5 and 1. Rewrite the quadratic equation using these numbers: \[x^2 + 5x + x + 5 = 0\]Now, factor by grouping: \[(x^2 + 5x) + (x + 5) = 0\]Factor out the common factors in each group: \[x(x + 5) + 1(x + 5) = 0\]Notice \((x + 5)\) is a common factor: \[(x + 1)(x + 5) = 0\]

Step 4: Solve the Factored Equations

Now set each factor equal to zero and solve for \(x\):1. \(x + 1 = 0\) Solve for \(x\): \(x = -1\)2. \(x + 5 = 0\) Solve for \(x\): \(x = -5\)

Step 5: Verify the Solutions

Substitute \(x = -1\) and \(x = -5\) back into the original equation to ensure they satisfy the equation.1. For \(x = -1\): \((-1)^2 + 6(-1) + 5 = 1 - 6 + 5 = 0\) The equation is satisfied.2. For \(x = -5\): \((-5)^2 + 6(-5) + 5 = 25 - 30 + 5 = 0\) The equation is satisfied.

Key Concepts

Factoring Quadratics

Factoring quadratics is a method used to simplify and solve equations that can be expressed in the form \( ax^2 + bx + c = 0 \). It involves breaking down the middle term \( bx \) by identifying two numbers that multiply to give \( a \times c \), the product of the first and last coefficients, and add up to \( b \). For our quadratic equation \( x^2 + 6x + 5 = 0 \), the task was to find two numbers that fulfill these conditions.
These numbers, 5 and 1, allow us to rewrite the middle term, \( 6x \), in terms of those numbers, transforming the equation into \( x^2 + 5x + x + 5 = 0 \).
By grouping terms, \( (x^2 + 5x) + (x + 5) \), and factoring out common components in each group, \( x(x + 5) + 1(x + 5) \), we identify a shared factor, \( x + 5 \). Thus, we can express it as the product \((x + 1)(x + 5) = 0\).
This process of breaking down and grouping terms allows us to transform the quadratic into a factorable expression, setting the stage for solving.

Solving Equations

Once a quadratic is factored, solving the equation becomes straightforward. We use the property that if a product of factors equals zero, at least one of the factors must be zero. In our example, after factoring, we have \((x + 1)(x + 5) = 0\).
This gives us two separate equations to solve:

• \( x + 1 = 0 \)
• \( x + 5 = 0 \)

Solve these simple equations by isolating \( x \).
For \( x + 1 = 0 \), subtract 1 from both sides to find \( x = -1 \).
For \( x + 5 = 0 \), subtract 5 from both sides to find \( x = -5 \).
The solutions \( x = -1 \) and \( x = -5 \) indicate the points where the quadratic equation touches or crosses the x-axis. This illustrates why quadratics often have two solutions.

Verifying Solutions

When we solve a quadratic equation, it's crucial to verify that our solutions satisfy the original equation. This ensures accuracy and confirms that our factoring and solving processes were done correctly.
To verify, substitute each solution back into the original equation \( x^2 + 6x + 5 = 0 \).
For \( x = -1 \), replace \( x \) in the equation to get \((-1)^2 + 6(-1) + 5 = 1 - 6 + 5 = 0\). The left-hand side simplifies to zero, satisfying the equation.
For \( x = -5 \), substitute to obtain \((-5)^2 + 6(-5) + 5 = 25 - 30 + 5 = 0\). This also simplifies to zero.
The successful reduction to zero for both cases proves that \( x = -1 \) and \( x = -5 \) are true solutions. Verifying solutions is an essential final step in problem-solving, confirming the accuracy of the solutions obtained.