Problem 34
Question
The equation \(x(x-9)(x-9)=0\) has three factors. However, it has only two solutions. Explain why.
Step-by-Step Solution
Verified Answer
The equation has only two unique solutions, 0 and 9, because the factor \((x-9)\) appears twice but represents the same solution.
1Step 1: Identify the Factors
The given equation is \(x(x-9)(x-9)=0\). There are three factors: \(x\), \((x-9)\), and \((x-9)\).
2Step 2: Use the Zero Product Property
According to the zero product property, for the product of factors to be zero, at least one of the factors must be zero. Therefore, set each factor equal to zero: \(x = 0\) and \(x - 9 = 0\).
3Step 3: Solve for the Variable
Solve the equations obtained from each factor: \(x = 0\) and \(x - 9 = 0\). The solution to the second equation is \(x = 9\).
4Step 4: List the Solutions
The solutions found are \(x = 0\) and \(x = 9\).
5Step 5: Explain the Duplicate Factor
The factor \((x-9)\) appears twice, but it still represents the same solution \(x = 9\). Therefore, only two unique solutions exist: 0 and 9.
Key Concepts
Factoring EquationsUnique SolutionsDuplicate Factors
Factoring Equations
When solving quadratic equations, one powerful strategy is factoring.
Factoring transforms the equation into a product of simpler expressions, or 'factors.' For example, the equation in our exercise is given as: \(x(x-9)(x-9) = 0\).
Here, we identified three factors: \(x\), \(x - 9\), and \(x - 9\).
Factoring helps break down the equation into smaller parts, making it easier to solve. To solve the equation, we use the Zero Product Property.
Factoring transforms the equation into a product of simpler expressions, or 'factors.' For example, the equation in our exercise is given as: \(x(x-9)(x-9) = 0\).
Here, we identified three factors: \(x\), \(x - 9\), and \(x - 9\).
Factoring helps break down the equation into smaller parts, making it easier to solve. To solve the equation, we use the Zero Product Property.
Unique Solutions
Using the Zero Product Property, we say that if a product of factors equals zero, at least one of the factors must be zero.
This means we set each factor equal to zero: \(x = 0\) and \(x - 9 = 0\).
Solving these equations gives us the solutions: \(x = 0\) and \(x = 9\).
Although the original problem had three factors, two of them were the same (\(x - 9\)).
This leads to only two unique solutions: 0 and 9.
This means we set each factor equal to zero: \(x = 0\) and \(x - 9 = 0\).
Solving these equations gives us the solutions: \(x = 0\) and \(x = 9\).
Although the original problem had three factors, two of them were the same (\(x - 9\)).
This leads to only two unique solutions: 0 and 9.
Duplicate Factors
In algebra, sometimes equations have repeated factors, known as duplicate factors.
In our exercise, \((x - 9)\) appeared twice.
Even though it shows up more than once, it represents the same solution.
Here, \(x - 9 = 0\) yields the solution \(x = 9\). This happens twice due to the duplicate factor, but it counts as only one unique solution.
In summary, duplicate factors do not add new solutions; they reinforce an existing one.
In our exercise, \((x - 9)\) appeared twice.
Even though it shows up more than once, it represents the same solution.
Here, \(x - 9 = 0\) yields the solution \(x = 9\). This happens twice due to the duplicate factor, but it counts as only one unique solution.
In summary, duplicate factors do not add new solutions; they reinforce an existing one.
Other exercises in this chapter
Problem 33
Use a pattern to factor. Check. Identify any prime polynomials. $$ x^{18}-y^{50} $$
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Use the \(a c\) method to factor. Check the factoring. Identify any prime polynomials. $$ 3 u^{2}+14 u+8 $$
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Use a pattern to factor. Check. Identify any prime polynomials. $$ x^{32}-y^{90} $$
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Use the \(a c\) method to factor. Check the factoring. Identify any prime polynomials. $$ 7 b^{2}+b-6 $$
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