Apply statistical equations in procurement management involves solving equations that arise in cost analysis, demand forecasting, and contract optimization. Accurate mathematical solutions help procurement professionals in Kenya make data-driven decisions, optimize resource allocation, and manage risks effectively. This chapter focuses on solving quadratic equations and simultaneous equations using methods applicable in procurement contexts such as budgeting, supplier evaluation, and order quantity determination.

2.2 Solving quadratic equations

Quadratic equations frequently appear in procurement when calculating cost functions, order quantities, or profit maximization problems. Mastery of different solution methods enables procurement managers to analyze situations where the relationship between variables is nonlinear. This section covers the formula method and factorization method, both essential in solving quadratic equations encountered in procurement scenarios.

2.2.1 Formula method

The formula method, also known as the quadratic formula, solves any quadratic equation of the form \( ax^{2} + bx + c = 0 \) by applying a direct formula. This is particularly useful when factorization is difficult or impossible due to the coefficients’ values.

The quadratic formula is:

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

Worked Examples

Example 1: A procurement officer needs to determine the optimal order quantity \(x\) for a product where the cost function is given by \(2x^{2} - 12x + 10 = 0\). Find \(x\).

Given: \(a=2, b=-12, c=10\)

$$ x = \frac{-(-12) \pm \sqrt{(-12)^{2} - 4 \times 2 \times 10}}{2 \times 2} $$

$$ x = \frac{12 \pm \sqrt{144 - 80}}{4} $$

$$ x = \frac{12 \pm \sqrt{64}}{4} $$

$$ x = \frac{12 \pm 8}{4} $$

Two solutions:

$$ x_1 = \frac{12 + 8}{4} = \frac{20}{4} = 5 $$

$$ x_2 = \frac{12 - 8}{4} = \frac{4}{4} = 1 $$

Answer: \(x = 5\) or \(x = 1\) units

Example 2: In supplier selection, the performance score \(x\) satisfies \(3x^{2} + 4x - 7 = 0\). Find the possible values of \(x\).

Given: \(a=3, b=4, c=-7\)

$$ x = \frac{-4 \pm \sqrt{4^{2} - 4 \times 3 \times (-7)}}{2 \times 3} $$

$$ x = \frac{-4 \pm \sqrt{16 + 84}}{6} $$

$$ x = \frac{-4 \pm \sqrt{100}}{6} $$

$$ x = \frac{-4 \pm 10}{6} $$

Two solutions:

$$ x_1 = \frac{-4 + 10}{6} = \frac{6}{6} = 1 $$

$$ x_2 = \frac{-4 - 10}{6} = \frac{-14}{6} = -\frac{7}{3} \approx -2.33 $$

Answer: \(x = 1\) or \(x \approx -2.33\) (Negative value may be discarded in procurement context)

Example 3: A procurement budget constraint is represented by \(5x^{2} - 20x + 15 = 0\). Find \(x\).

Given: \(a=5, b=-20, c=15\)

$$ x = \frac{-(-20) \pm \sqrt{(-20)^{2} - 4 \times 5 \times 15}}{2 \times 5} $$

$$ x = \frac{20 \pm \sqrt{400 - 300}}{10} $$

$$ x = \frac{20 \pm \sqrt{100}}{10} $$

$$ x = \frac{20 \pm 10}{10} $$

Two solutions:

$$ x_1 = \frac{20 + 10}{10} = \frac{30}{10} = 3 $$

$$ x_2 = \frac{20 - 10}{10} = \frac{10}{10} = 1 $$

Answer: \(x = 3\) or \(x = 1\) units

Example 4: The profit function for a procurement project is \(x^{2} - 6x + 8 = 0\). Find the break-even points \(x\).

Given: \(a=1, b=-6, c=8\)

$$ x = \frac{-(-6) \pm \sqrt{(-6)^{2} - 4 \times 1 \times 8}}{2 \times 1} $$

$$ x = \frac{6 \pm \sqrt{36 - 32}}{2} $$

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

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

Two solutions:

$$ x_1 = \frac{6 + 2}{2} = \frac{8}{2} = 4 $$

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

Answer: \(x = 4\) or \(x = 2\) units

Example 5: A procurement cost equation is \(4x^{2} + 4x + 1 = 0\). Find \(x\).

Given: \(a=4, b=4, c=1\)

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

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

$$ x = \frac{-4 \pm \sqrt{0}}{8} $$

$$ x = \frac{-4 \pm 0}{8} $$

Single solution:

$$ x = \frac{-4}{8} = -0.5 $$

Answer: \(x = -0.5\) units

2.2.2 Factorization

Factorization solves quadratic equations by expressing the quadratic expression as a product of two binomials. This method is effective when factors are integers, enabling quick solutions in procurement problems involving simple cost or quantity relationships.

The general approach is to write \( ax^{2} + bx + c = 0 \) as \( (mx + n)(px + q) = 0 \), then solve for \(x\).

Worked Examples

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

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

Find two numbers that multiply to 6 and add to -5: -2 and -3

Express:

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

Set each factor to zero:

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

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

Answer: \(x = 2\) or \(x = 3\) units

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

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

Find two numbers that multiply to \(2 \times 3 = 6\) and add to 7: 6 and 1

Rewrite middle term:

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

Group terms:

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

Factor each:

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

Factor out common binomial:

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

Set each factor to zero:

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

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

Answer: \(x = -3\) or \(x = -\frac{1}{2}\) units

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

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

Find two numbers that multiply to -10 and add to 3: 5 and -2

Express:

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

Set each factor to zero:

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

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

Answer: \(x = -5\) or \(x = 2\) units

Example 4: Solve \( 3x^{2} - 8x + 4 = 0 \) by factorization.

Given: \(a=3, b=-8, c=4\)

Find two numbers that multiply to \(3 \times 4 = 12\) and add to -8: -6 and -2

Rewrite middle term:

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

Group terms:

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

Factor each:

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

Factor out common binomial:

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

Set each factor to zero:

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

$$ 3x - 2 = 0 \Rightarrow x = \frac{2}{3} $$

Answer: \(x = 2\) or \(x = \frac{2}{3}\) units

Example 5: Solve \( 4x^{2} + 4x - 3 = 0 \) by factorization.

Given: \(a=4, b=4, c=-3\)

Find two numbers that multiply to \(4 \times (-3) = -12\) and add to 4: 6 and -2

Rewrite middle term:

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

Group terms:

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

Factor each:

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

Factor out common binomial:

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

Set each factor to zero:

$$ 2x + 3 = 0 \Rightarrow x = -\frac{3}{2} $$

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

Answer: \(x = -\frac{3}{2}\) or \(x = \frac{1}{2}\) units

Chapter Summary

This chapter focused on applying various mathematical techniques essential for solving practical problems in engineering and business contexts. It began with methods for solving quadratic equations using the formula method and factorization, providing tools to handle second-degree polynomial equations effectively. The study then progressed to solving simultaneous equations through substitution, elimination, and matrix methods, enabling the determination of multiple unknowns in linear systems. Breakeven analysis was introduced to assess the point where total costs and total revenues are equal, a vital concept in financial decision-making. Calculus was explored through differentiation and integration, equipping students with techniques to analyze rates of change and accumulated quantities. Finally, the chapter covered the formulation of total revenue, total cost, and profit equations, linking mathematical expressions directly to business performance evaluation. Together, these topics form a comprehensive foundation for applying statistical equations in various technical and economic scenarios.