Chapter 1: Problem 6
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 ______ .
Short Answer
Expert verified
The equation is quadratic type; set \(W=x+1\). The quadratic equation is \(W^2 - 5W + 6 = 0\).
Step by step solution
01
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.
02
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.
03
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 \(W\).
04
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:\[W - 2 = 0 \quad \Rightarrow \quad W = 2,\]\[W - 3 = 0 \quad \Rightarrow \quad W = 3.\]
05
Substitute Back to Find x
Recall that \(W = x + 1\). We substitute back to find the corresponding \(x\) values:\[W = 2 \quad \Rightarrow \quad x + 1 = 2 \quad \Rightarrow \quad x = 1,\]\[W = 3 \quad \Rightarrow \quad x + 1 = 3 \quad \Rightarrow \quad x = 2.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \(W^2 - 5W + 6 = 0\).
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:
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:
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.