Example  3
Determine four consecutive odd integers such that if the first integer is increased by two, the second integer is decreased by two, the third integer is multiplied by two and the fourth integer is squared, the sum of the four resulting integers is seven.

Step 1: Write what you know.

To help organize the information, use a table.

Number Integer  Operation    Result
\(1\)   \(2x + 1\)
\( +2\) \(2x + 1 + 2 = 2x + 3\)
\(2\)   \(2x + 3\) \( -2\) \(2x + 3 - 2 = 2x + 1\)
\(3\)   \(2x + 5\) \(\times 2\)
\(2\left( {2x + 5} \right) = 4x + 10\)
\(4\)   \(2x + 7\)
squared \(\left( {2x + 7} \right)^2 = 4x^2 + 28x + 49\)

The sum of the four resulting values is seven.


Step 2: Analyze how to solve the problem.

Add the four expressions from the last column together.  Their sum will equal seven.


Step 3: Solve the problem.

\(\begin{align}
 \left( {2x + 3} \right) + \left( {2x + 1} \right) + \left( {4x + 10} \right) + \left( {4x^2 + 28x + 49} \right) &= 7 \\
 2x + 3 + 2x + 1 + 4x + 10 + 4x^2 + 28x + 49 &= 7 \\
 4x^2 + 36x + 63 &= 7 \\
 4x^2 + 36x + 56 &= 0 \\
 4\left( {x^2 + 9x + 14} \right) &= 0 \\
 x^2 + 9x + 14 &= 0 \\
 \left( {x + 2} \right)\left( {x + 7} \right) &= 0 \\
 \end{align}\)

Because this is an equation, you can remove the \(4\) by dividing both sides by \(4\). Zero divided by a non-zero is still zero.


\(\begin{align}
 x + 2 &= 0 \\
 x &= -2
 \end{align}\)
\(\begin{align}
 x + 7 &= 0 \\
 x &= -7
 \end{align}\)

Check your answers.

Left side
 Right Side
\(\begin{array}{r}
 \left( {2x + 3} \right) + \left( {2x + 1} \right) + \left( {4x + 10} \right) + \left( {4x^2 + 28x + 49} \right) \\
 \left( {2\left( { - 2} \right) + 3} \right) + \left( {2\left( { - 2} \right) + 1} \right) + \left( {4\left( { - 2} \right) + 10} \right) + \left( {4\left( { - 2} \right)^2 + 28\left( { - 2} \right) + 49} \right) \\
 \left( { - 1} \right) + \left( { - 3} \right) + \left( 2 \right) + \left( {16 - 56 + 49} \right) \\
  - 2 + 9 \\
 7 \end{array}\)
 \(7\)
LS = RS                


Left side
 Right Side
\(\begin{array}{r}
 \left( {2x + 3} \right) + \left( {2x + 1} \right) + \left( {4x + 10} \right) + \left( {4x^2 + 28x + 49} \right) \\
 \left( {2\left( { - 7} \right) + 3} \right) + \left( {2\left( { - 7} \right) + 1} \right) + \left( {4\left( { - 7} \right) + 10} \right) + \left( {4\left( { - 7} \right)^2 + 28\left( { - 7} \right) + 49} \right) \\
 \left( { - 11} \right) + \left( { - 13} \right) + \left( { - 18} \right) + \left( {196 - 196 + 49} \right) \\
  - 42 + 49  \\
7 \end{array}\)
 \(7\)
LS = RS                 


There are two sets of solutions:

If \(x = -2\), then the first number is \(2x + 1 = 2(-2) + 1 = -3\).
The four consecutive odd numbers are therefore \(-3\), \(-1\), \(1\), and \(3\).

If \(x = -7\), then the first number is \(2x + 1 = 2(-7) + 1 = -13\).
The four consecutive odd numbers are therefore \(-13\), \(-11\), \(-9\), and \(-7\).