Problem 43 - Algebra A Combined Function

Question

Solve. See Examples 1 through 5. $$ (t+3)^{2}-2(t+3)-8=0 $$

Step-by-Step Solution

Answer

The solutions are \(t = 1\) and \(t = -5\).

Step 1: Identify Substitution

Notice that the expression \((t + 3)\) is common in the equation. To simplify, let \(u = t + 3\). This substitution will simplify the expression to \(u^2 - 2u - 8 = 0\).

Step 2: Solve the Quadratic Equation

Solve the quadratic equation \(u^2 - 2u - 8 = 0\). We can factor this equation: \((u - 4)(u + 2) = 0\).

Step 3: Find Values of u

From the factors \((u - 4)(u + 2) = 0\), we get two possible values for \(u\): \(u = 4\) and \(u = -2\).

Step 4: Reverse Substitution

Substitute back \(t + 3 = u\). So we have two equations: \(t + 3 = 4\) and \(t + 3 = -2\).

Step 5: Solve for t

For \(t + 3 = 4\), solve for \(t\): \[t = 4 - 3 = 1\]For \(t + 3 = -2\), solve for \(t\): \[t = -2 - 3 = -5\].

Step 6: Verify Solutions

Substitute \(t = 1\) and \(t = -5\) back into the original equation to verify solutions:- For \(t=1\): \((1+3)^2 - 2(1+3) - 8 = 16 - 8 - 8 = 0\).- For \(t=-5\): \((-5+3)^2 - 2(-5+3) - 8 = 4 + 4 - 8 = 0\). Both solutions satisfy the original equation.

Key Concepts

Substitution Method

The substitution method is a helpful technique, especially when dealing with quadratic equations. In the given exercise, you noticed that the expression \((t + 3)\) repeated in the equation. To simplify solving it, we can set \(u = t + 3\). By substituting \(u\) for \((t + 3)\), we convert the equation into a simpler form: \(u^2 - 2u - 8 = 0\). This makes the equation easier to handle because it looks like a standard quadratic equation.
This clever substitution can reduce complexity and reveal the structure of the equation, enabling easier manipulation and solution finding.
The key is to spot repeating patterns or isolated terms that simplify the equation when replaced with a single variable.

Factoring

Factoring is the method used to break down complicated equations into simpler ones by finding expressions that multiply to give the original equation.
For the equation \(u^2 - 2u - 8 = 0\), we factor it into \((u - 4)(u + 2) = 0\).
What we are essentially doing here is finding two numbers that multiply to \(-8\) and add to \(-2\), which are \(-4\) and \(2\).
This factorization technique helps to find the roots or solutions of quadratic equations efficiently.

• Check for common factors.
• Look for product-sum combinations that work for the equation.
This step simplifies solving the equation because it converts a quadratic expression into linear factors that are much easier to manage.

Solving Equations

Once you have factored the quadratic equation into \((u - 4)(u + 2) = 0\), solving is straightforward.
The equation is zero when either \(u - 4 = 0\) or \(u + 2 = 0\). Solving these gives us \(u = 4\) and \(u = -2\).
This is an application of the zero product property which states that if a product of factors equals zero, at least one of the factors must be zero.
Solving quadratic equations by factoring is a widely used method because it breaks the original complex equation into simpler components, making it easier to isolate and calculate the unknown variable values.

• Set each factor equal to zero.
• Solve each equation to find possible solutions.

This is a powerful tool in algebra that simplifies the problem-solving process significantly.

Reverse Substitution

Reverse substitution is the final step where you go back from the substituted variable. You substitute back to express the solution in terms of the original variable.
In this problem, we replace back \(u\) with \(t + 3\), meaning we solve for \(t\) using the two found \(u\) values:

• For \(u = 4\), substitute to get \(t + 3 = 4\), thus \(t = 1\).
• For \(u = -2\), substitute to get \(t + 3 = -2\), thus \(t = -5\).

In reverse substitution, you ensure the solutions you found actually correspond to the original variable.
Always check that the solutions satisfy the original equation after reverse substitution to confirm they are correct. This step completes the journey from simplifying the equation with substitution back to solving it for the exact variable of interest.