Problem 68
Question
Solve each of the equations. $$x^{2}-14 x=0$$
Step-by-Step Solution
Verified Answer
The solutions are \( x = 0 \) and \( x = 14 \).
1Step 1: Factor the Equation
The given equation is a quadratic equation: \[ x^2 - 14x = 0 \] The first step is to factor the equation. Notice that you can factor an \(x\) from each term: \[ x(x - 14) = 0 \]
2Step 2: Apply the Zero-Product Property
The Zero-Product Property states that if a product of two factors is zero, then at least one of the factors must be zero. We apply this property to our factored equation:- Set each factor equal to zero: \[ x = 0 \] \[ x - 14 = 0 \]
3Step 3: Solve Each Equation
Solve each of the simple equations obtained from the factored form:- The first equation \( x = 0 \) is already solved.- Solve the second equation for \( x \): \[ x - 14 = 0 \] Add 14 to both sides: \[ x = 14 \]
4Step 4: State the Solutions
The solutions to the original equation \( x^2 - 14x = 0 \) are the values of \( x \) that satisfy both simple equations. Therefore, the solutions are:- \( x = 0 \)- \( x = 14 \)
Key Concepts
Factoring MethodZero-Product PropertySolving Quadratic Equations
Factoring Method
The factoring method is a fundamental technique used to solve quadratic equations, typically expressed in the form \( ax^2 + bx + c = 0 \). This method involves rewriting the quadratic equation as a product of its factors. In other words, you transform a polynomial into a product of simpler expressions. This transformation is crucial because it lays the groundwork for applying other solving methods, such as the zero-product property.
Here's how it typically works:
Here's how it typically works:
- Identify common factors in each term of the quadratic equation. In equations like \( x^2 - 14x = 0 \), you can see that an \( x \) is common in both terms.
- Factor out the common term and rewrite the expression as a product. For our equation, factoring yields \( x(x - 14) = 0 \).
Zero-Product Property
The zero-product property is a simple but powerful concept that states: if the product of two numbers is zero, then at least one of those numbers must be zero. In mathematical terms, if \( a \times b = 0 \), then either \( a = 0 \) or \( b = 0 \). This property allows us to break down complex equations into simpler pieces that are easier to solve.
When we apply this property to a factored quadratic equation such as \( x(x - 14) = 0 \), we obtain two straightforward equations to solve:
When we apply this property to a factored quadratic equation such as \( x(x - 14) = 0 \), we obtain two straightforward equations to solve:
- \( x = 0 \)
- \( x - 14 = 0 \)
Solving Quadratic Equations
Solving quadratic equations is an essential skill in algebra, involving finding the values of the variable that satisfy the equation. Once your quadratic equation is in a factored form, as seen in equations like \( x(x - 14) = 0 \), you can easily apply the zero-product property to find the solutions.
Let's go through the example:
Let's go through the example:
- Start by solving each equation from the factored form separately. The equation \( x = 0 \) is already solved.
- For the equation \( x - 14 = 0 \), add 14 to both sides: \( x = 14 \).
Other exercises in this chapter
Problem 68
Should help you pull together all of the factoring techniques of this chapter. Factor completely each polynomial, and indicate any that are not factorable using
View solution Problem 68
Find all real number solutions for each equation. $$x^{5}-x=0$$
View solution Problem 68
Find each quotient. $$\frac{-a^{4} b^{5} c}{a^{2} b^{4} c}$$
View solution Problem 68
Simplify by removing the inner parentheses first and working outward. $$[9 x y-(4 x+x y-y)]-[4 y-(2 x-x y+6 y)]$$
View solution