Problem 100
Question
Solving a Quadratic Equation Find all real solutions of the equation. $$x(x-20)+3(x-20)=0$$
Step-by-Step Solution
Verified Answer
The solutions to the equation are \(x = 20\) and \(x = -3\).
1Step 1: Simplify the equation
Start by distributing the terms in brackets to simplify the equation. This yields: \(x^2 - 20x + 3x - 60 = 0\) which simplifies to \(x^2 - 17x - 60 = 0\).
2Step 2: Factor the quadratic equation
Next, the quadratic equation is factored. The factored form of the equation is \((x - 20)(x + 3) = 0\). This is found by looking for two numbers that add to -17 and multiply to -60.
3Step 3: Solve for \(x\)
Finally, use the zero product property \((ab = 0 \Rightarrow a = 0 \vee b = 0)\), which states that if the product of multiple factors is zero, at least one of the factors must be zero. Here, there are two possible solutions: \(x - 20 = 0 \Rightarrow x = 20\) and \(x + 3 = 0 \Rightarrow x = -3\).
Key Concepts
Factoring QuadraticsZero Product PropertyReal Solutions
Factoring Quadratics
Factoring quadratics is a critical skill when attempting to solve quadratic equations. It involves rewriting the equation in a product form, where the equation is expressed as the product of two or more factors. For example, when faced with a quadratic equation like \(x^2 - 17x - 60 = 0\), the goal is to identify two binomial expressions that multiply to give the original quadratic equation.
To factor a quadratic equation, look for two numbers that add up to the coefficient of the \(x\) term, which is -17 in our case, and multiply to give the constant term, here being -60. This process can be likened to solving a puzzle, where these two numbers are the missing pieces. Once you find these numbers, you can write the factored form as \(x - a\) and \(x - b\), where \(a\) and \(b\) are the solutions to the puzzle. For the equation at hand, the factored form is \(x - 20)(x + 3) = 0\), showcasing that -20 and 3 are the numbers we were looking for. Factoring is a powerful tool as it breaks down the equation into simpler parts, making it easier to solve for the variable \(x\).
To factor a quadratic equation, look for two numbers that add up to the coefficient of the \(x\) term, which is -17 in our case, and multiply to give the constant term, here being -60. This process can be likened to solving a puzzle, where these two numbers are the missing pieces. Once you find these numbers, you can write the factored form as \(x - a\) and \(x - b\), where \(a\) and \(b\) are the solutions to the puzzle. For the equation at hand, the factored form is \(x - 20)(x + 3) = 0\), showcasing that -20 and 3 are the numbers we were looking for. Factoring is a powerful tool as it breaks down the equation into simpler parts, making it easier to solve for the variable \(x\).
Zero Product Property
Understanding the zero product property is crucial when solving factored quadratic equations. This fundamental algebraic property indicates that if a product of two or more factors equals zero, at least one of the factors must be zero. It is symbolically represented as \(ab = 0 \Rightarrow a = 0 \vee b = 0\).
Applying this property to the equation \(x - 20)(x + 3) = 0\), we infer that for the equation to hold true, either \(x - 20\) or \(x + 3\) must be equal to zero. These separate conditions provide us the potential solutions for the quadratic equation. By setting each factor equal to zero and solving for \(x\), we can determine the values that make the original quadratic equation true. This simple property transforms the problem of solving a quadratic equation into an easier task of solving two linear equations.
Applying this property to the equation \(x - 20)(x + 3) = 0\), we infer that for the equation to hold true, either \(x - 20\) or \(x + 3\) must be equal to zero. These separate conditions provide us the potential solutions for the quadratic equation. By setting each factor equal to zero and solving for \(x\), we can determine the values that make the original quadratic equation true. This simple property transforms the problem of solving a quadratic equation into an easier task of solving two linear equations.
Real Solutions
Real solutions are the possible values of \(x\) that satisfy a quadratic equation. When solved correctly, quadratic equations can have two real solutions, one real solution, or none at all. In our example equation, \(x - 20)(x + 3) = 0\), the zero product property tells us that either \(x - 20 = 0\) or \(x + 3 = 0\).
Solving these equations, we find that \(x = 20\) and \(x = -3\) are the real solutions. It's important to note that these solutions are the points at which the graph of the quadratic equation— a parabola—intersects the \(x\)-axis. In cases where no real solutions exist, the parabola does not cross the \(x\)-axis at all, indicating that the solutions are complex or imaginary numbers. Understanding real solutions is not only essential for solving quadratic equations but also for interpreting their graphical representations.
Solving these equations, we find that \(x = 20\) and \(x = -3\) are the real solutions. It's important to note that these solutions are the points at which the graph of the quadratic equation— a parabola—intersects the \(x\)-axis. In cases where no real solutions exist, the parabola does not cross the \(x\)-axis at all, indicating that the solutions are complex or imaginary numbers. Understanding real solutions is not only essential for solving quadratic equations but also for interpreting their graphical representations.
Other exercises in this chapter
Problem 99
The average salaries \(S\) (in thousands of dollars) of secondary classroom teachers in the United States from 2005 through 2013 can be approximated by the mode
View solution Problem 99
Find the product. $$(4 x+1)^{2}$$
View solution Problem 100
use the models below, which approximate the numbers of Bed Bath \(\&\) Beyond stores \(B\) and Williams-Sonoma stores \(W\) for the years 2000 through \(2013,\)
View solution Problem 100
The total public debt \(D\) (in trillions of dollars) in the United States from 2005 through 2014 can be approximated by the model $$D=0.051 t^{2}+0.20 t+5.0,5
View solution