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Chapter 4: Problem 5

Solve the equation \((x+2)(x-3)=14\) by constructing a table of values. Use integer values of \(x\) between \(-6\) and \(6 .\)

Short Answer

Expert verified
The solutions are x = -4 and x = 5.

Step by step solution

01

Set up the equation

Given the equation is (x+2)(x-3)=14. Our goal is to check different integer values for x between -6 and 6 and find the ones satisfying the equation.
02

Construct a table

Create a table with three columns: one for the value of x, the second for (x+2)(x-3), and the third for comparing it with 14.
03

Fill the table with values

Substitute each integer value of x from -6 to 6 into the expression (x+2)(x-3) and record the results in the table.
04

Evaluate each value of x

Calculate as follows for each integer value: x = -6: (-6+2)(-6-3) = (-4)(-9) = 36 (not equal to 14) x = -5: (-5+2)(-5-3) = (-3)(-8) = 24 (not equal to 14) x = -4: (-4+2)(-4-3) = (-2)(-7) = 14 (equal to 14, so one solution is x = -4) x = -3: (-3+2)(-3-3) = (-1)(-6) = 6 (not equal to 14) x = -2: (-2+2)(-2-3) = (0)(-5) = 0 (not equal to 14) x = -1: (-1+2)(-1-3) = (1)(-4) = -4 (not equal to 14) x = 0: (0+2)(0-3) = (2)(-3) = -6 (not equal to 14) x = 1: (1+2)(1-3) = (3)(-2) = -6 (not equal to 14) x = 2: (2+2)(2-3) = (4)(-1) = -4 (not equal to 14) x = 3: (3+2)(3-3) = (5)(0) = 0 (not equal to 14) x = 4: (4+2)(4-3) = (6)(1) = 6 (not equal to 14) x = 5: (5+2)(5-3) = (7)(2) = 14 (equal to 14, so another solution is x = 5) x = 6: (6+2)(6-3) = (8)(3) = 24 (not equal to 14)
05

Identify solutions

From the table, the values of x that make the equation true are x = -4 and x = 5.

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Key Concepts

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

Quadratic Equations
Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, with \(a\) not equal to zero. The equation \( (x+2)(x-3) = 14\) is a quadratic equation.
Quadratic equations often appear in various real-life contexts such as physics, engineering, and finance.
To solve a quadratic equation, you usually have a few main methods: factoring, completing the square, using the quadratic formula, and graphing.
In this specific exercise, rather than using these traditional methods, we use a table of values to find integer solutions. This method has its unique benefits, such as providing students a clear visual pathway on how different solutions are derived.
Table of Values
A table of values is a helpful tool for solving equations, especially when seeking integer solutions. It helps you visualize the relationship between two variables. In this exercise, we create a table with three columns, and we test different values of \(x\) to see which ones satisfy the equation \( (x+2)(x-3) = 14\).
Here's a step-by-step approach:
  • Set up the columns: The first column lists the integer values of \(x\) from \(-6\) to \(6\).
  • The second column shows the result of the expression \( (x+2)(x-3)\) for each corresponding value of \(x\).
  • The third column is for comparison: we check if the result equals the target value \(14\).
By systematically substituting each value of \(x\) in the range and recording the results, you can clearly identify which \(x\) values make the equation true.
Integer Solutions
Integer solutions are values of \(x\) that are whole numbers, without fractions or decimals. When solving the equation \( (x+2)(x-3) = 14\) using the table of values, we searched specifically for integer solutions within the range \(-6\) to \(6\).
The completed table demonstrates that:
  • When \(x = -4\), the expression \( (x+2)(x-3)\) equals \(14\).
  • When \(x = 5\), the expression \( (x+2)(x-3)\) also equals \(14\).
This means \(-4\) and \(5\) are the integer solutions for the given quadratic equation.
Finding integer solutions can sometimes be easier than finding non-integer solutions, especially when working within a specific range. This method provides a clear and systematic way to check each possible value.

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