Apply statistical equations are fundamental in diverse Kenyan professional settings such as county hospitals, banks, universities, and agricultural cooperatives. Mastery of quadratic and simultaneous equations enables problem-solving in budgeting, resource allocation, data analysis, and operational planning. This chapter focuses on practical methods for solving quadratic equations and systems of simultaneous equations using substitution, elimination, and matrix techniques. These mathematical tools empower diploma-level students to apply quantitative reasoning effectively across multiple sectors.

2.2 Solving quadratic equations

Quadratic equations arise in various Kenyan industries, including calculating profit maximization in retail businesses and optimizing resource use in farms. These equations take the standard form \( ax^{2} + bx + c = 0 \) where \(a \neq 0\). Two primary methods to solve such equations are the formula method and factorization. Each method suits different problem types and complexity levels.

2.2.1 Formula method

The formula method uses the quadratic formula to find the roots of any quadratic equation. The quadratic formula is

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

where \(a\), \(b\), and \(c\) are coefficients from the quadratic equation \(ax^{2} + bx + c = 0\).

Worked Examples

Example 1: A county hospital plans to construct a rectangular garden where the area is described by the equation \(2x^{2} + 5x - 3 = 0\), with \(x\) representing length in meters. Find the possible lengths.

Given: \(a=2\), \(b=5\), \(c=-3\)

$$ x = \frac{-5 \pm \sqrt{5^{2} - 4 \times 2 \times (-3)}}{2 \times 2} $$

$$ x = \frac{-5 \pm \sqrt{25 + 24}}{4} $$

$$ x = \frac{-5 \pm \sqrt{49}}{4} $$

$$ x = \frac{-5 \pm 7}{4} $$

For \(x = \frac{-5 + 7}{4} = \frac{2}{4} = 0.5\)

For \(x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3\)

Answer: The possible lengths are 0.5 m and -3 m. Since length cannot be negative, the valid length is 0.5 m.

Example 2: A retail business models profit \(P\) (in Ksh) as \(P = -3x^{2} + 12x + 7\). Find the values of \(x\) when profit is zero.

Given: Set \(P=0\), so \(-3x^{2} + 12x + 7 = 0\), \(a = -3\), \(b=12\), \(c=7\)

$$ x = \frac{-12 \pm \sqrt{12^{2} - 4 \times (-3) \times 7}}{2 \times (-3)} $$

$$ x = \frac{-12 \pm \sqrt{144 + 84}}{-6} $$

$$ x = \frac{-12 \pm \sqrt{228}}{-6} $$

$$ \sqrt{228} \approx 15.1 $$

$$ x = \frac{-12 \pm 15.1}{-6} $$

For \(x = \frac{-12 + 15.1}{-6} = \frac{3.1}{-6} = -0.517\)

For \(x = \frac{-12 - 15.1}{-6} = \frac{-27.1}{-6} = 4.517\)

Answer: The profit is zero at \(x = -0.517\) (not valid) and \(x = 4.517\).

Example 3: A SACCO models its loan repayment schedule with the equation \(x^{2} - 4x - 21 = 0\). Find the repayment period \(x\) in months.

Given: \(a=1\), \(b=-4\), \(c=-21\)

$$ x = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4 \times 1 \times (-21)}}{2 \times 1} $$

$$ x = \frac{4 \pm \sqrt{16 + 84}}{2} $$

$$ x = \frac{4 \pm \sqrt{100}}{2} $$

$$ x = \frac{4 \pm 10}{2} $$

For \(x = \frac{4 + 10}{2} = \frac{14}{2} = 7\)

For \(x = \frac{4 - 10}{2} = \frac{-6}{2} = -3\)

Answer: The repayment period is 7 months (negative value discarded).

2.2.2 Factorization

Factorization involves expressing the quadratic equation as a product of two binomials set to zero. This method works well when the quadratic can be factored into integer roots.

Worked Examples

Example 1: A university calculates the number of students in two batches with the equation \(x^{2} + 5x + 6 = 0\). Find the values of \(x\).

Given: \(x^{2} + 5x + 6 = 0\)

Factorize the quadratic:

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

Set each factor to zero:

$$ x + 2 = 0 \implies x = -2 $$

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

Answer: The values of \(x\) are -2 and -3.

Example 2: A farm calculates the area of a plot with the equation \(x^{2} - x - 6 = 0\). Find the possible dimensions.

Given: \(x^{2} - x - 6 = 0\)

Factorize:

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

Set each factor to zero:

$$ x - 3 = 0 \implies x = 3 $$

$$ x + 2 = 0 \implies x = -2 $$

Answer: The possible dimensions are 3 m and -2 m (discard negative).

Example 3: A retail store analyses its sales with the equation \(2x^{2} - 8x = 0\). Find the values of \(x\).

Given: \(2x^{2} - 8x = 0\)

Factor out common terms:

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

Set each factor to zero:

$$ 2x = 0 \implies x = 0 $$

$$ x - 4 = 0 \implies x = 4 $$

Answer: The values of \(x\) are 0 and 4.

Chapter Summary

This chapter focused on applying various mathematical techniques essential for solving engineering and business problems. It began with methods for solving quadratic equations, exploring both the formula method and factorization to find roots efficiently. The chapter then covered simultaneous equations, demonstrating how substitution, elimination, and matrix methods can be used to solve systems of linear equations. Breakeven analysis was introduced to determine the point where total costs equal total revenue, a critical concept in financial decision-making. The calculus section addressed fundamental operations, including differentiation to find rates of change and integration to calculate areas under curves. Finally, the chapter concluded with formulating total revenue, total cost, and profit equations, providing a framework for analysing business performance mathematically. These topics collectively equip students with practical tools for quantitative analysis in technical and economic contexts.