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Chapter 12: Problem 30

Without solving the equation, decide how many solutions it has. $$ (x-1)(x-2)=0 $$

Short Answer

Expert verified
Answer: The equation has 2 solutions.

Step by step solution

01

Identify the Factors

In the given equation \((x-1)(x-2)=0\), there are two factors: \((x-1)\) and \((x-2)\).
02

Consider the Zero Product Property

The Zero Product Property states that if the product of two or more factors is equal to zero, then at least one of the factors must be equal to zero. In this case, we have \((x-1)(x-2)=0\). So, either \((x-1)=0\) or \((x-2)=0\).
03

Count the Distinct Solutions

Since there are two distinct factors, we have two possible solutions for \(x\). So, the given equation has 2 solutions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Zero Product Property
The zero product property is a fundamental concept in algebra. It states that if the product of two or more factors equals zero, then at least one of these factors must be zero. This property forms the basis for solving quadratic equations by factoring.
In an equation like
  • \((a)(b) = 0\)
we can conclude that either
  • \(a = 0\)
  • or \(b = 0\)
This is because multiplying any number by zero results in zero. Understanding this property helps in identifying solutions to equations quickly.
Factoring
Factoring involves expressing a polynomial as the product of its simplest components or factors. For quadratic equations, this often means breaking down the expression into binomials.
Take the example:
  • \((x-1)(x-2) = 0\)
This expression is already factored, showcasing two binomials: \((x-1)\) and \((x-2)\).
This step prepares the equation for applying the zero product property. Factoring transforms complex expressions into simpler forms, making it easier to find solutions.
Solutions of Equations
Solutions of an equation are the values that satisfy the equation, making it true. For the equation
  • \((x-1)(x-2) = 0\),
you determine solutions using the zero product property. Here, the solutions are where each factor equals zero:
  • When \((x-1) = 0\), solving gives \(x = 1\).
  • When \((x-2) = 0\), solving gives \(x = 2\).
Thus, the equation has two solutions: \(x = 1\) and \(x = 2\). Each solution corresponds to one factor. This method shows how factoring and the zero product property help find the exact solutions of quadratic equations.

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Most popular questions from this chapter

Evaluate the expressions in Problems \(51-54\) given that \(f(x)=2 x^{3}+3 x-3, \quad g(x)=3 x^{2}-2 x-4\) \(h(x)=f(x) g(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}\) $$ a_{0} $$

Without solving the equation, decide how many solutions it has. $$ \left(x^{2}-4\right)(x+5)=0 $$

For what values of \(a\) does the equation have a solution in \(x\) ? $$ \left(a x^{2}+1\right)(x-a)=0 $$

For what values of \(a\) does the equation have a solution in \(x\) ? $$ a-x^{5}=0 $$

Give the value of \(a\) that makes the statement true. The constant term of \((t+2)^{2}(t-a)^{2}\) is 9 .
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