Procurement management professionals frequently encounter mathematical problems involving quadratic and simultaneous equations when analyzing cost functions, optimizing resource allocation, or forecasting demand and supply relationships. Mastery of these equations allows for accurate decision-making in contract pricing, budget planning, and supplier negotiations. This chapter develops skills in solving quadratic equations by formula and factorization methods, as well as solving simultaneous equations using substitution, elimination, and matrix methods. Each technique is illustrated with practical examples relevant to procurement scenarios in Kenya.

2.2 Solving quadratic equations

Quadratic equations often arise in procurement when modeling costs that depend on quantities squared, such as bulk order discounts or penalty calculations. Being able to solve these equations accurately is essential for determining optimal order quantities or evaluating contract terms.

2.2.1 Formula method

A quadratic equation has the form \(ax^{2} + bx + c = 0\), where \(a eq 0\). The formula method, also called the quadratic formula, finds the roots (solutions) of the equation using:

$$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$

This formula calculates the values of \(x\) where the quadratic expression equals zero.

Worked Examples

Example 1: A procurement officer models the cost savings \(C\) (in Ksh) on bulk buying as \(C = 2x^{2} - 8x + 6\), where \(x\) is the number of bulk units ordered. Find the order size \(x\) that results in zero savings.

Given: \(a=2\), \(b=-8\), \(c=6\)

$$x = \frac{-(-8) \pm \sqrt{(-8)^{2} - 4 \times 2 \times 6}}{2 \times 2}$$

$$x = \frac{8 \pm \sqrt{64 - 48}}{4}$$

$$x = \frac{8 \pm \sqrt{16}}{4}$$

$$x = \frac{8 \pm 4}{4}$$

For \(x_1\):

$$x_1 = \frac{8 + 4}{4} = \frac{12}{4} = 3$$

For \(x_2\):

$$x_2 = \frac{8 - 4}{4} = \frac{4}{4} = 1$$

Answer: The order sizes that yield zero savings are 1 and 3 units.

Example 2: A supplier’s penalty cost \(P\) (Ksh) for late delivery is modeled as \(P = x^{2} + 6x + 5\), where \(x\) is the days late. Find the days late when the penalty cost is zero.

Given: \(a=1\), \(b=6\), \(c=5\)

$$x = \frac{-6 \pm \sqrt{6^{2} - 4 \times 1 \times 5}}{2 \times 1}$$

$$x = \frac{-6 \pm \sqrt{36 - 20}}{2}$$

$$x = \frac{-6 \pm \sqrt{16}}{2}$$

$$x = \frac{-6 \pm 4}{2}$$

For \(x_1\):

$$x_1 = \frac{-6 + 4}{2} = \frac{-2}{2} = -1$$

For \(x_2\):

$$x_2 = \frac{-6 - 4}{2} = \frac{-10}{2} = -5$$

Answer: The penalty is zero at -1 and -5 days late, which are not practical. Hence, no days late yield zero penalty cost.

Example 3: The profit function \(P\) (Ksh) for a procurement batch is given by \(P = -3x^{2} + 12x - 9\), where \(x\) is the batch size in hundreds of units. Find the batch sizes where profit is zero.

Given: \(a=-3\), \(b=12\), \(c=-9\)

$$x = \frac{-12 \pm \sqrt{12^{2} - 4 \times (-3) \times (-9)}}{2 \times (-3)}$$

$$x = \frac{-12 \pm \sqrt{144 - 108}}{-6}$$

$$x = \frac{-12 \pm \sqrt{36}}{-6}$$

$$x = \frac{-12 \pm 6}{-6}$$

For \(x_1\):

$$x_1 = \frac{-12 + 6}{-6} = \frac{-6}{-6} = 1$$

For \(x_2\):

$$x_2 = \frac{-12 - 6}{-6} = \frac{-18}{-6} = 3$$

Answer: The profit is zero at batch sizes of 1 and 3 hundred units.

Example 4: The cost function in procurement is \(C = 4x^{2} - 20x + 25\). Determine the values of \(x\) where the cost is zero.

Given: \(a=4\), \(b=-20\), \(c=25\)

$$x = \frac{-(-20) \pm \sqrt{(-20)^{2} - 4 \times 4 \times 25}}{2 \times 4}$$

$$x = \frac{20 \pm \sqrt{400 - 400}}{8}$$

$$x = \frac{20 \pm 0}{8}$$

$$x = \frac{20}{8} = 2.5$$

Answer: The cost is zero at \(x = 2.5\) units (a single root).

2.2.2 Factorization

Factorization solves quadratic equations by expressing the quadratic as a product of two binomials equal to zero. This method is efficient when factors are integers or simple fractions.

The general form is:

$$ax^{2} + bx + c = (mx + n)(px + q) = 0$$

Solve by setting each factor equal to zero:

$$mx + n = 0 \quad \text{or} \quad px + q = 0$$

Worked Examples

Example 1: Solve \(x^{2} - 5x + 6 = 0\) by factorization.

Given: \(a=1\), \(b=-5\), \(c=6\)

Find two numbers multiplying to 6 and adding to -5: -2 and -3.

$$x^{2} - 5x + 6 = (x - 2)(x - 3) = 0$$

Set each factor to zero:

$$x - 2 = 0 \Rightarrow x = 2$$

$$x - 3 = 0 \Rightarrow x = 3$$

Answer: The solutions are \(x = 2\) and \(x = 3\).

Example 2: Solve \(2x^{2} + 7x + 3 = 0\) by factorization.

Given: \(a=2\), \(b=7\), \(c=3\)

Multiply \(a \times c = 6\). Find two numbers that multiply to 6 and add to 7: 6 and 1.

Rewrite middle term:

$$2x^{2} + 6x + x + 3 = 0$$

Group:

$$(2x^{2} + 6x) + (x + 3) = 0$$

Factor each group:

$$2x(x + 3) + 1(x + 3) = 0$$

Factor out common binomial:

$$(2x + 1)(x + 3) = 0$$

Set each factor to zero:

$$2x + 1 = 0 \Rightarrow x = -\frac{1}{2}$$

$$x + 3 = 0 \Rightarrow x = -3$$

Answer: The solutions are \(x = -\frac{1}{2}\) and \(x = -3\).

Example 3: Solve \(3x^{2} - 2x - 8 = 0\) by factorization.

Given: \(a=3\), \(b=-2\), \(c=-8\)

Calculate \(a \times c = -24\). Find two numbers that multiply to -24 and add to -2: 4 and -6.

Rewrite middle term:

$$3x^{2} + 4x - 6x - 8 = 0$$

Group:

$$(3x^{2} + 4x) - (6x + 8) = 0$$

Factor each group:

$$x(3x + 4) - 2(3x + 4) = 0$$

Factor out common binomial:

$$(x - 2)(3x + 4) = 0$$

Set each factor to zero:

$$x - 2 = 0 \Rightarrow x = 2$$

$$3x + 4 = 0 \Rightarrow x = -\frac{4}{3}$$

Answer: The solutions are \(x = 2\) and \(x = -\frac{4}{3}\).

Example 4: Solve \(x^{2} + 3x - 10 = 0\) by factorization.

Given: \(a = 1\), \(b = 3\), \(c = -10\)

Find two numbers multiplying to -10 and adding to 3: 5 and -2.

$$x^{2} + 3x - 10 = (x + 5)(x - 2) = 0$$

Set each factor to zero:

$$x + 5 = 0 \Rightarrow x = -5$$

$$x - 2 = 0 \Rightarrow x = 2$$

Answer: The solutions are \(x = -5\) and \(x = 2\).

Chapter Summary

This chapter covered essential mathematical techniques for solving quadratic equations using the formula method and factorization, enabling the determination of roots in various civil engineering problems. It explored methods for solving simultaneous equations, including substitution, elimination, and matrix approaches, which are vital for handling multiple variable relationships in practical scenarios. The concept of breakeven analysis was introduced to identify the point where costs and revenues balance, supporting effective financial decision-making. Calculus fundamentals were addressed through differentiation and integration, providing tools for analyzing rates of change and accumulated quantities. Finally, the chapter demonstrated the formulation of total revenue, total cost, and profit equations, linking mathematical principles to business and economic evaluations in engineering projects. Together, these topics equip students with a robust foundation in applying statistical and mathematical equations to solve real-world technical and financial challenges in the field.