Problem 30
Question
Solve each equation. You will need to use the factoring techniques that we discussed throughout this chapter. $$x^{2}+x-2=0$$
Step-by-Step Solution
Verified Answer
The solutions are \(x = -2\) and \(x = 1\).
1Step 1: Identify the Equation and Goal
The given equation is a quadratic equation: \(x^2 + x - 2 = 0\). The goal is to factor the quadratic equation and solve for the values of \(x\) that satisfy the equation.
2Step 2: Factor the Quadratic
To factor the quadratic equation \(x^2 + x - 2\), we need to find two numbers whose product is \(-2\) (the constant term) and whose sum is \(1\) (the coefficient of the linear term). The numbers that satisfy these conditions are \(2\) and \(-1\). We can rewrite the quadratic equation as \((x + 2)(x - 1) = 0\).
3Step 3: Apply the Zero Product Property
The Zero Product Property states that if the product of two expressions is zero, then at least one of the expressions must be zero. This gives us two equations to solve: \(x + 2 = 0\) and \(x - 1 = 0\).
4Step 4: Solve Each Equation
Solving \(x + 2 = 0\), we subtract 2 from both sides to find \(x = -2\). Solving \(x - 1 = 0\), we add 1 to both sides to find \(x = 1\).
Key Concepts
Zero Product PropertyQuadratic Equation SolutionsPolynomial FactoringSolving Algebraic Equations
Zero Product Property
The Zero Product Property is a fundamental concept in algebra that simplifies the task of solving quadratic equations. It states that if the product of two or more terms is zero, then at least one of the terms must be zero. In mathematical terms, if \(a \cdot b = 0\), then \(a = 0\) or \(b = 0\). This property is particularly useful when dealing with factored forms of quadratic equations.
When a quadratic equation is expressed as the product of two binomials, such as \((x + 2)(x - 1) = 0\), each binomial becomes a candidate for being zero. Thus, you solve two simpler equations, \(x + 2 = 0\) and \(x - 1 = 0\), to find the solutions for \(x\). By applying the Zero Product Property, you not only simplify the process, but also ensure a systematic approach to finding solutions.
So remember, whenever you have a quadratic equation that can be factored into a product of two terms, use this property to break down the task into smaller, manageable parts.
When a quadratic equation is expressed as the product of two binomials, such as \((x + 2)(x - 1) = 0\), each binomial becomes a candidate for being zero. Thus, you solve two simpler equations, \(x + 2 = 0\) and \(x - 1 = 0\), to find the solutions for \(x\). By applying the Zero Product Property, you not only simplify the process, but also ensure a systematic approach to finding solutions.
So remember, whenever you have a quadratic equation that can be factored into a product of two terms, use this property to break down the task into smaller, manageable parts.
Quadratic Equation Solutions
Quadratic equations are polynomial equations of degree two, generally written in the form \(ax^2 + bx + c = 0\). The solutions to these equations can often be found using multiple methods, including factoring, the quadratic formula, or completing the square. In this example, you are working with the equation \(x^2 + x - 2 = 0\).
After factoring this specific quadratic into \((x + 2)(x - 1) = 0\), you apply the Zero Product Property to find the solutions. Solving the resulting equations, \(x + 2 = 0\) and \(x - 1 = 0\), gives the solutions \(x = -2\) and \(x = 1\), respectively. These solutions indicate the points where the parabola, represented by the equation \(x^2 + x - 2 = 0\), intersects the x-axis.
Finding solutions to quadratic equations helps in graphing the corresponding parabola and understanding the behavior of quadratic functions in various applied contexts.
After factoring this specific quadratic into \((x + 2)(x - 1) = 0\), you apply the Zero Product Property to find the solutions. Solving the resulting equations, \(x + 2 = 0\) and \(x - 1 = 0\), gives the solutions \(x = -2\) and \(x = 1\), respectively. These solutions indicate the points where the parabola, represented by the equation \(x^2 + x - 2 = 0\), intersects the x-axis.
Finding solutions to quadratic equations helps in graphing the corresponding parabola and understanding the behavior of quadratic functions in various applied contexts.
Polynomial Factoring
Polynomial factoring breaks down larger expressions into simpler multiplied factors. In the context of solving equations, it’s a way to simplify the problem so it can be solved more easily. Here, the quadratic \(x^2 + x - 2\) has been factored into \((x+2)(x-1)\).
To factor quadratics, you look for two numbers that multiply to give the constant term (here, \(-2\)) and add to give the coefficient of the linear term (here, \(1\)). For \(x^2 + x - 2\), the numbers are \(2\) and \(-1\). You then use these numbers to rewrite the middle term, allowing you to group and factor by grouping if needed.
Factoring polynomials is a skill that needs practice, as it is a key step in solving algebraic equations. Understanding how to efficiently factor will make solving quadratic equations much quicker and intuitive.
To factor quadratics, you look for two numbers that multiply to give the constant term (here, \(-2\)) and add to give the coefficient of the linear term (here, \(1\)). For \(x^2 + x - 2\), the numbers are \(2\) and \(-1\). You then use these numbers to rewrite the middle term, allowing you to group and factor by grouping if needed.
Factoring polynomials is a skill that needs practice, as it is a key step in solving algebraic equations. Understanding how to efficiently factor will make solving quadratic equations much quicker and intuitive.
Solving Algebraic Equations
Algebraic equations involve finding the value of the variable that makes the equation true. Solving these equations often requires isolating the variable using algebraic principles and operations. In the case of quadratic equations such as \(x^2 + x - 2 = 0\), this means breaking down the equation into simpler parts, often by factoring.
Once factored, you use the Zero Product Property to solve the simpler equations \(x + 2 = 0\) and \(x - 1 = 0\). Solving these gives the roots or solutions of the original quadratic equation.
Once factored, you use the Zero Product Property to solve the simpler equations \(x + 2 = 0\) and \(x - 1 = 0\). Solving these gives the roots or solutions of the original quadratic equation.
- The solution \(x = -2\) means that if you replace \(x\) by \(-2\), the equation becomes true.
- Similarly, \(x = 1\) achieves the same for the second equation.
Other exercises in this chapter
Problem 29
Find each product. $$(4 a b)\left(-2 a^{2} b\right)(7 a)$$
View solution Problem 29
Subtract the polynomials using the horizontal format. \(2 x^{3}+x^{2}-7 x-2\) from \(5 x^{3}+2 x^{2}+6 x-13\)
View solution Problem 30
Factor completely each of the polynomials and indicate any that are not factorable using integers. $$4 x^{2}-x+6$$
View solution Problem 30
Factor each of the following polynomials completely. Indicate any that are not factorable using integers. Don't forget to look first for a common monomial facto
View solution