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Chapter 6: Problem 45

Solve each quadratic equation using the method that seems most appropriate. $$(x+2)(x-7)=10$$

Short Answer

Expert verified
The solutions to the equation are \(x = 8\) and \(x = -3\).

Step by step solution

01

Expand the Equation

First, expand the left side of the equation to remove the parentheses. Use the distributive property to multiply \((x+2)(x-7)\): \[(x+2)(x-7) = x^2 - 7x + 2x - 14\]Simplify this to:\[x^2 - 5x - 14\]
02

Move All Terms to One Side

Subtract 10 from both sides of the equation to set the equation to zero:\[x^2 - 5x - 14 = 10 \rightarrow x^2 - 5x - 14 - 10 = 0 \]\[x^2 - 5x - 24 = 0\]
03

Factor the Quadratic Equation

Look for two numbers that multiply to \(-24\) and add to \(-5\). The numbers \(-8\) and \(3\) work:\[x^2 - 5x - 24 = (x - 8)(x + 3) = 0\]
04

Solve for the Roots

Set each factor equal to zero and solve for \(x\):1. \(x - 8 = 0\) \(x = 8\)2. \(x + 3 = 0\) \(x = -3\)
05

Verify the Solutions

Plug the solutions back into the original equation to verify: - For \(x = 8\): \((8+2)(8-7) = 10 \rightarrow 10(1) = 10\).- For \(x = -3\): \((-3+2)(-3-7) = 10 \rightarrow (-1)(-10) = 10\).Both solutions satisfy the original equation, confirming they are correct.

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Key Concepts

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

Distributive Property
The distributive property is a fundamental algebraic principle used to eliminate parentheses by distributing a multiplier to terms within the parentheses. It is particularly helpful when expanding expressions such as \((x+2)(x-7)\). The property states that \(a(b + c) = ab + ac\).

Applying to Quadratics:
Consider \( (x + 2)(x - 7) \). By applying the distributive property:
  • First, multiply \(x \) by every term in the second parenthesis: \(x(x - 7) = x^2 - 7x\).
  • Next, distribute \(2 \) the same way: \(2(x - 7) = 2x - 14\).
  • Combine these results: \(x^2 - 7x + 2x - 14\).
This method allows you to simplify and align similar terms, leading to \(x^2 - 5x - 14\).

Such simplification is essential to form a standard quadratic equation, which is key for further analysis and solving.
Factoring Quadratics
Factoring is a strategy for solving quadratic equations by expressing them as a product of simpler binomials. For example, the equation \(x^2 - 5x - 24 = 0\) can be factored.

Steps to Factor:
  • Identify the Constants: We need two numbers that multiply to \(-24\) (the constant term) and add to \(-5\) (the linear coefficient).
  • Finding the Numbers: Consider the numbers \(-8\) and \(+3\). They satisfy both required conditions: \(-8 \times 3 = -24\) and \(-8 + 3 = -5\).
  • Formulate the Factors: Use these numbers to factor the quadratic: \(x^2 - 5x - 24 = (x - 8)(x + 3)\).
This factorization lets us quickly solve the equation by setting each factor equal to zero. Factoring is a practical and powerful tool for finding solutions, known as the roots of the equation.
Roots of Equations
The solutions to a quadratic equation, often referred to as the roots, are the values of \(x\) that satisfy the equation \(ax^2+bx+c=0\).

Finding Roots:
  • Once the quadratic is factored, use each factor to set up an equation: \(x - 8 = 0\) and \(x + 3 = 0\).
  • Solve for \(x\):
    • For \(x - 8 = 0\), you find \(x = 8\).
    • For \(x + 3 = 0\), you find \(x = -3\).
The values \(x = 8\) and \(x = -3\) are the roots, meaning they make the original equation true.

Verification:
Verifying the roots by substituting back into the original equation is crucial to ensure they are correct. For instance:
  • For \(x = 8\), substitute to get \((8 + 2)(8 - 7) = 10 \).
  • For \(x = -3\), substitute to check \((-3 + 2)(-3 - 7) = 10 \).
Both substitutions confirm that \(x = 8\) and \(x = -3\) are indeed the solutions. This verification process assures the reliability of the factored results.

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