Problem 40 - College Algebra with Corequisite Support

Question

For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring. $$ (x+1)^{2}-8(x+1)-9=0 $$

Step-by-Step Solution

Answer

The real solutions are \(x = 8\) and \(x = -2\).

Step 1: Identify the Substitution Variable

Recognize the expression \((x+1)\) appearing repeatedly in the equation. Let \(u = x+1\). Substitute \(u\) into the equation to rewrite \((x+1)^2 - 8(x+1) - 9 = 0\) as \(u^2 - 8u - 9 = 0\).

Step 2: Rewrite the Equation in Terms of the New Variable

The equation \((x+1)^2 - 8(x+1) - 9 = 0\) now simplifies to the quadratic equation \(u^2 - 8u - 9 = 0\). This equation is in quadratic form and ready for factoring.

Step 3: Factor the Quadratic Equation

Factor the quadratic equation \(u^2 - 8u - 9\). Find two numbers that multiply to \(-9\) and add up to \(-8\). These numbers are \(-9\) and \(+1\). Thus, the equation factors as \((u - 9)(u + 1) = 0\).

Step 4: Solve for the Substitute Variable

Set each factor from \((u - 9)(u + 1) = 0\) equal to zero and solve for \(u\).- For \(u - 9 = 0\), we get \(u = 9\).- For \(u + 1 = 0\), we get \(u = -1\).

Step 5: Substitute Back to Original Variable

Recall that \(u = x + 1\). Substitute back to find the values of \(x\).- If \(u = 9\), then \(x + 1 = 9\), giving \(x = 8\).- If \(u = -1\), then \(x + 1 = -1\), giving \(x = -2\).

Step 6: Verify the Real Solutions

Check each solution in the original equation to ensure they satisfy it.- For \(x = 8\), substitute into the original equation: \((8 + 1)^2 - 8(8 + 1) - 9 = 0\), which simplifies to \(81 - 72 - 9 = 0\).- For \(x = -2\), substitute into the original equation: \((-2 + 1)^2 - 8(-2 + 1) - 9 = 0\), which simplifies to \(1 + 16 - 9 = 0\). Both calculations confirm the solutions are correct.

Key Concepts

Factoring

Factoring is a fundamental technique used to solve quadratic equations. Essentially, it involves rewriting the equation as a product of two smaller expressions set to zero. Here are the basics of how it works:

• Identify a quadratic equation of the form \( ax^2 + bx + c = 0 \).
• Find two numbers that multiply to the constant term \( c \) and add up to the coefficient \( b \).
• Use these numbers to break down the middle term (the one with \( x \)) and rewrite the equation.
• Factor by grouping and simplify the expression to its factored form.

This approach breaks down complex expressions into simpler, solvable components. In our exercise, the equation \( u^2 - 8u - 9 = 0 \) was factored as \((u - 9)(u + 1) = 0\), which simplifies finding solutions.

Substitution Method

The substitution method simplifies solving complex equations by introducing a new variable. This method is especially useful when the equation contains a recurring expression. Here’s how it works:

• Identify the part of the equation that repeats. Assign it a substitute variable.
• Rewrite the equation using this new variable and solve it as a standard quadratic equation.
• Once solved, substitute back the original terms for your new variable to find the original variable's solutions.

In the given problem, the repeated expression \( (x+1) \) was replaced with \( u \), turning the equation into \( u^2 - 8u - 9 = 0 \). This transformation allowed us to factor and solve the equation more easily.

Real Solutions

Finding real solutions involves determining values that satisfy the original equation without any imaginary numbers. For a quadratic equation, these solutions are often the x-values where the function crosses the x-axis on a graph.To verify real solutions from a factored equation:

• Set each factor equal to zero and solve for the variable.
• Ensure these solutions satisfy the original equation when substituting back.

In the exercise, the factors \((u - 9)\) and \((u + 1)\) provided \( u = 9 \) and \( u = -1 \). Substituting \( u \) back as \( x+1 \) gave the solutions \( x = 8 \) and \( x = -2 \). Testing these in the original equation confirmed they are indeed real solutions.