Problem 1
Question
Solve the following equations, if possible. $$ (x-7)(x+4)=0 $$
Step-by-Step Solution
Verified Answer
Answer: The solutions are x = 7 and x = -4.
1Step 1: Set each factor equal to zero
To find the solutions of the given equation, set each factor equal to zero:
$$
(x-7) = 0
$$
and
$$
(x+4) = 0
$$
2Step 2: Solve for x in each equation
Now, solve each equation for x to find the two possible solutions.
For the first equation:
$$
x - 7 = 0 \Rightarrow x = 7
$$
For the second equation:
$$
x + 4 = 0 \Rightarrow x = -4
$$
3Step 3: Write the final solution
The solutions to the given equation are \(x = 7\) and \(x = -4\). These are the values that make the equation true.
Key Concepts
Factoring MethodRoots of the EquationZero Product Property
Factoring Method
Understanding the factoring method is key to solving quadratic equations like the one given in the exercise: \((x-7)(x+4)=0\). But what exactly does factoring mean in this context? Factoring allows us to represent a quadratic equation as a product of its linear factors. This is useful because it simplifies the equation into parts that are easier to work with. In this case, the equation is already factored, meaning it’s written as the product of two binomials: \((x-7)\) and \((x+4)\).
This method transforms complex expressions into manageable pieces. By using this approach, one can quickly apply the next fundamental principle in solving the equation.
This method transforms complex expressions into manageable pieces. By using this approach, one can quickly apply the next fundamental principle in solving the equation.
Roots of the Equation
Finding the roots of the equation is like solving a puzzle. Roots, often called solutions or zeros, are the values of \(x\) that satisfy the equation \((x-7)(x+4)=0\). To find these values, we exploit the equation's factored form. By setting each factor equal to zero, we begin our journey towards the roots.
For instance, when \(x-7=0\), solving this gives \(x=7\). Similarly, solving \(x+4=0\) yields \(x=-4\). Thus, the roots of the equation are \(x=7\) and \(x=-4\). These values solve the equation because substituting them back renders the original equation true. Identifying these roots explains why this method is so effective.
For instance, when \(x-7=0\), solving this gives \(x=7\). Similarly, solving \(x+4=0\) yields \(x=-4\). Thus, the roots of the equation are \(x=7\) and \(x=-4\). These values solve the equation because substituting them back renders the original equation true. Identifying these roots explains why this method is so effective.
Zero Product Property
The Zero Product Property is a fundamental rule in algebra that underpins much of solving quadratics through factoring. This property states that if a product of two numbers is zero, then at least one of the multiplicands must be zero. In the context of our equation, \((x-7)(x+4)=0\), it's applied to find the solutions.
Using this property, one can confidently set each factor of the equation to zero: \((x-7)=0\) and \((x+4)=0\). Solving each one gives the potential solutions, \(x=7\) and \(x=-4\). Thanks to the Zero Product Property, this method not only simplifies finding solutions but also ensures accuracy, making this approach reliable for students tackling similar quadratic equations.
Using this property, one can confidently set each factor of the equation to zero: \((x-7)=0\) and \((x+4)=0\). Solving each one gives the potential solutions, \(x=7\) and \(x=-4\). Thanks to the Zero Product Property, this method not only simplifies finding solutions but also ensures accuracy, making this approach reliable for students tackling similar quadratic equations.
Other exercises in this chapter
Problem 1
Determine the values of \(a, b,\) and \(c\) in the following quadratic equations. $$ 4 x^{2}-3 x+5=0 $$
View solution Problem 1
Solve each of the following quadratic equations using the method of extraction of roots. $$ x^{2}-144=0 $$
View solution Problem 1
Which of the following equations are quadratic equations? Answer "yes" or "no" to each equation. $$ 6 x^{2}-4 x+9=0 $$
View solution Problem 2
For the following problems, solve the equations. $$ (b+1)(b-6)=0 $$
View solution