Problem 6 - Precalculus: Mathematics for Calculus

Question

The equation \((x+1)^{2}-5(x+1)+6=0\) is of ______ type. To solve the equation, we set \(W=\) ______ The resulting quadratic equation is ______ .

Step-by-Step Solution

Answer

The equation is quadratic type; set \(W=x+1\). The quadratic equation is \(W^2 - 5W + 6 = 0\).

Step 1: Identify the Type of Equation

The equation given is \[(x+1)^2 - 5(x+1) + 6 = 0.\]This equation is in terms of the expression \((x + 1)\). Notice that it is a quadratic expression with respect to \((x + 1)\), because the highest power of \((x + 1)\) is 2. Therefore, the equation is of quadratic type.

Step 2: Substitute a New Variable

To simplify solving this quadratic equation, we use a substitution. Let \(W = x + 1\). This substitution means we will express everything in terms of \(W\) instead of \(x+1\). This reduces the complexity of the original problem.

Step 3: Rewrite the Equation Using the New Variable

We substitute \(W = x + 1\) into the original equation, rewriting it as:\[W^2 - 5W + 6 = 0.\]This new equation is a standard quadratic equation in terms of \(W\).

Step 4: Solve the Quadratic Equation

To solve the equation \(W^2 - 5W + 6 = 0\), we can factor it.This factors as:\[(W - 2)(W - 3) = 0.\]Setting each factor equal to zero gives us the solutions:\[W - 2 = 0 \quad \Rightarrow \quad W = 2,\]\[W - 3 = 0 \quad \Rightarrow \quad W = 3.\]

Step 5: Substitute Back to Find x

Recall that \(W = x + 1\). We substitute back to find the corresponding \(x\) values:\[W = 2 \quad \Rightarrow \quad x + 1 = 2 \quad \Rightarrow \quad x = 1,\]\[W = 3 \quad \Rightarrow \quad x + 1 = 3 \quad \Rightarrow \quad x = 2.\]

Key Concepts

The Substitution Method

The substitution method is a powerful tool used in solving equations, especially when handling quadratic equations. It essentially simplifies complex expressions by substituting one part of the equation with a variable. This process makes it much easier to solve the equation step-by-step.

In the context of quadratic equations, the substitution method revolves around the idea of reducing a quadratic expression into something simpler. By using a new variable to replace a more complicated part of the equation, we can transform it into a standard form that we are more familiar with.

For example, given the equation \[(x+1)^2 - 5(x+1) + 6 = 0,\]we substitute \(W = x + 1\). This substitution helps simplify the equation into a standard quadratic form \(W^2 - 5W + 6 = 0\).

• This method is highly beneficial when the equation has repetitive expressions.
• It allows one to deal with a simpler equation, making it easier to solve.
• Once the simpler equation is solved, the original variable is reintroduced for the final solution.

Factoring Quadratics

Factoring quadratics is a fundamental technique used to find the roots or solutions of a quadratic equation. When an equation is in the form of \(ax^2 + bx + c = 0\), factoring allows us to express it as a product of two linear factors.

For instance, consider the quadratic equation \(W^2 - 5W + 6 = 0\). To factor this equation, we look for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3. Hence, the quadratic can be factored as \((W - 2)(W - 3) = 0\).

The steps involved in factoring a quadratic equation include:

• Finding two numbers whose product is equal to the constant term (in this case, 6) and whose sum equals the coefficient of the linear term (in this case, -5).
• Rewriting the quadratic as a product of two binomials based on these numbers.
• Setting each factor equal to zero and solving for the variable.

Variables Substitution

Variables substitution is a versatile method that offers a new perspective on solving equations. By substituting a variable, we transform the problem into a new equation that is often easier to manage. This technique is especially useful in simplifying equations where a specific expression repeats or complicates the form of the equation.

In our exercise, we set \(W = x + 1\), transforming the original equation \((x+1)^2 - 5(x+1) + 6 = 0\) into \(W^2 - 5W + 6 = 0\). By doing so, we're simplifying the equation, making it a straightforward task to find the solution.

Here are some advantages of using variable substitution:

• It simplifies complex expressions within the equation.
• It provides clarity by breaking down the problem into more manageable parts.
• It allows for a tailored approach to solving specific types of equations, like quadratics.

Once the solution to the new equation is found, the variable substitution is reversed to find the value of the original variable.