2.1 Advanced Algebra, Progressions & Word Problems

2.1 Advanced Algebra, Progressions & Word Problems

Advanced algebra and progression series constitute the computational foundation of the Applied Mathematics cluster in the Philippine Civil Engineering Licensure Examination (CELE). Board exam problems frequently combine polynomial algebra, progression theory, and rate-based word problems into multi-step engineering scenarios. Mastery of both algebraic derivations and high-speed calculator-supported solution strategies is essential for completing the examination within the strict allotted time.

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Quadratic Equations & Polynomial Theory

Standard Form and Discriminant Analysis

A quadratic equation in standard form is expressed as: $ax^2 + bx + c = 0 \quad (a \ne 0)$

Its roots are obtained analytically via the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The algebraic nature of the roots is governed entirely by the discriminant $\Delta = b^2 - 4ac$:

Vieta's Relations for Polynomial Roots

Vieta's formulas establish direct relationships between the coefficients of a polynomial and symmetric sums of its roots without requiring explicit factorization:

• Quadratic ($ax^2 + bx + c = 0$): $\text{Sum of roots: } x_1 + x_2 = -\frac{b}{a}$ $\text{Product of roots: } x_1 x_2 = \frac{c}{a}$
• Cubic ($ax^3 + bx^2 + cx + d = 0$): $x_1 + x_2 + x_3 = -\frac{b}{a}$ $x_1 x_2 + x_2 x_3 + x_3 x_1 = \frac{c}{a}$ $x_1 x_2 x_3 = -\frac{d}{a}$

Remainder & Factor Theorems

• Remainder Theorem: If a polynomial $P(x)$ is divided by a linear divisor $(x - r)$, the scalar remainder $R$ is identical to the functional evaluation $R = P(r)$.
• Factor Theorem: A linear binomial $(x - r)$ is a factor of $P(x)$ if and only if $P(r) = 0$.
• Synthetic Division: A rapid tabular shortcut used on board exam scratch papers to evaluate $P(r)$ and determine the depressed polynomial quotient.
• Descartes' Rule of Signs: The number of positive real roots of a real polynomial $P(x)$ equals the number of sign variations between consecutive non-zero coefficients or is less by an even positive integer. The number of negative real roots equals the sign variations in $P(-x)$ or is less by an even positive integer.

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The Binomial Theorem & Term Expansion

For any positive integer $n$, the expansion of the binomial $(a + b)^n$ is given by: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \dots + \binom{n}{n}b^n$

Where the binomial coefficient is evaluated as: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Properties of the Expansion

1. Total Number of Terms: An expansion of degree $n$ contains exactly $n + 1$ terms.
2. Sum of Coefficients: Evaluated by substituting all variable terms with unity: $\text{Sum} = (a + b)^n \big|_{a=1, b=1} = 2^n$.
3. General $(r+1)$-th Term Formula: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ (Note: The $r$-value is always one less than the ordinal term position; for example, the 5th term uses $r = 4$.)
4. Middle Terms:
   • If $n$ is even, there is a single middle term at ordinal position $\frac{n}{2} + 1$, where $r = \frac{n}{2}$.
   • If $n$ is odd, there are two symmetric middle terms at positions $\frac{n+1}{2}$ and $\frac{n+3}{2}$, where $r = \frac{n-1}{2}$ and $r = \frac{n+1}{2}$.
5. Constant Term (Term Independent of $x$): Formulated by equating the combined exponent of the variable $x$ in $T_{r+1}$ to zero and solving for integer $r$.

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Progression Series: Arithmetic, Geometric & Harmonic

Progressions model recurring patterns, depreciation schedules, drainage runoff steps, and structural load distributions.

Infinite Geometric Series

When the common ratio satisfies the strict convergence criterion $|r| < 1$, the higher-order terms $r^n \to 0$ as $n \to \infty$. The infinite sum converges to: $S_\infty = \frac{a_1}{1 - r} \quad (|r| < 1)$ If $|r| \ge 1$, the series diverges and has no finite sum.

Pythagorean Means Inequality

For any set of positive real numbers $a$ and $b$: $A \ge G \ge H$ $\frac{a + b}{2} \ge \sqrt{ab} \ge \frac{2ab}{a + b}$ The equality holds if and only if $a = b$. Furthermore, the geometric mean is the exact geometric mean of the arithmetic and harmonic means: $G^2 = A \cdot H$

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Standard CELE Board Exam Word Problem Archetypes

CELE word problems test the ability to translate physical and engineering descriptions into algebraic models. Five classical problem archetypes appear regularly on the examination:

1. Work and Rate Problems

Work problems rely on reciprocal unit rates. If an agent completes a task in $t$ time units, their rate of production is $R = \frac{1}{t}$ tasks per unit time.

• Combined Simultaneous Work: $\sum R_i = \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} = \frac{1}{T_{\text{total}}}$
• Hydraulic Pipes (Inlets & Outlets): Filling pipes contribute positive rates ($+1/t$), whereas drainage outlets contribute negative rates ($-1/t$): $R_{\text{net}} = \sum \frac{1}{t_{\text{in}}} - \sum \frac{1}{t_{\text{out}}}$
• Interrupted / Sequential Work: If crew A works for time $t_A$ and crew B works for time $t_B$ to finish the complete project: $R_A t_A + R_B t_B = 1.0$

2. Mixture and Dilution Problems

Governed by the conservation of solute mass. In any combination of volumes: $C_1 V_1 + C_2 V_2 = C_{\text{final}}(V_1 + V_2)$ Where $C$ represents solute concentration (percentage or decimal fraction) and $V$ is volume.

• Successive Dilution (Repeated Replacement): If a container of volume $V$ initially full of pure liquid is repeatedly diluted by removing volume $y$ and replacing it with pure solvent, the concentration remaining after $n$ iterations is: $x_n = x_0 \left(1 - \frac{y}{V}\right)^n$

3. Motion and Relative Velocity Problems

Governed by $d = v \cdot t$.

• Upstream and Downstream Motion: $v_{\text{downstream}} = v_{\text{still}} + v_{\text{current}}$ $v_{\text{upstream}} = v_{\text{still}} - v_{\text{current}}$
• Average Speed Over Equal Distances: When traveling distance $D$ at speed $v_1$ and returning the same distance $D$ at speed $v_2$, total time is $t = D/v_1 + D/v_2$. The overall average speed is the harmonic mean: $v_{\text{avg}} = \frac{2D}{\frac{D}{v_1} + \frac{D}{v_2}} = \frac{2 v_1 v_2}{v_1 + v_2}$ (Exam Alert: Never compute simple arithmetic average for round-trip speeds!)

4. Age Relationship Problems

Structured by setting up linear equations across distinct temporal frames: past ($x - N$), present ($x$), and future ($x + M$). The age difference between two individuals remains invariant across all time frames.

5. Clock Hands Problems

A standard circular clock dial is divided into 60 minute-spaces ($360^\circ$).

• Minute Hand Speed: $360^\circ / 60\text{ min} = 6^\circ/\text{min} = 1.0\text{ minute-space/min}$.
• Hour Hand Speed: $30^\circ / 60\text{ min} = 0.5^\circ/\text{min} = \frac{1}{12}\text{ minute-space/min}$.
• Relative Speed of Separation: $\Delta v = 6^\circ - 0.5^\circ = 5.5^\circ/\text{min} = \frac{11}{12}\text{ minute-spaces/min}$
• Time Formula: To gain an angular separation of $S$ minute-spaces starting from hour $H$ (where the hour hand is initially at $5H$ minute-spaces): $m = \frac{12}{11} \times (\text{net minute-spaces to be gained})$

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Step-by-Step Worked Problem Examples

Worked Example 1: Multi-Pipe Water Reservoir Filling

Problem: A water storage reservoir for a municipal distribution system has a capacity of $2,400\text{ m}^3$. Inlet Pipe A can fill the reservoir alone in $8\text{ hours}$. Inlet Pipe B can fill it alone in $12\text{ hours}$. An emergency drainage Pipe C can empty the full reservoir alone in $16\text{ hours}$. At 6:00 AM, Pipe A and Pipe C are opened simultaneously. At 9:00 AM, Pipe B is also opened. At what exact clock time will the reservoir be completely filled?

Solution:

1. Establish individual hourly rates: $R_A = +\frac{1}{8}\text{ res/hr}, \quad R_B = +\frac{1}{12}\text{ res/hr}, \quad R_C = -\frac{1}{16}\text{ res/hr}$
2. Compute volume filled between 6:00 AM and 9:00 AM ($\Delta t_1 = 3\text{ hours}$): $R_{\text{initial}} = R_A + R_C = \frac{1}{8} - \frac{1}{16} = \frac{2 - 1}{16} = \frac{1}{16}\text{ res/hr}$ $\text{Fraction filled} = \frac{1}{16} \times 3 = \frac{3}{16}$
3. Determine remaining fraction to fill: $\text{Remaining fraction} = 1 - \frac{3}{16} = \frac{13}{16}$
4. Compute combined net rate with all three pipes active from 9:00 AM onwards: $R_{\text{net}} = R_A + R_B + R_C = \frac{1}{8} + \frac{1}{12} - \frac{1}{16} = \frac{6 + 4 - 3}{48} = \frac{7}{48}\text{ res/hr}$
5. Solve for remaining time $t_2$: $t_2 = \frac{\text{Remaining fraction}}{R_{\text{net}}} = \frac{13/16}{7/48} = \frac{13}{16} \times \frac{48}{7} = \frac{39}{7}\text{ hours} = 5\frac{4}{7}\text{ hours}$ $5\text{ hours} + \left(\frac{4}{7} \times 60\right)\text{ minutes} = 5\text{ hours } 34.29\text{ minutes} \approx 5\text{ hr } 34\text{ min } 17\text{ sec}$
6. Add time elapsed to 9:00 AM: $\text{Clock time} = \text{9:00 AM} + 5\text{ hr } 34\text{ min } 17\text{ sec} = \text{2:34:17 PM}$

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Worked Example 2: Clock Hands Perpendicularity

Problem: In how many minutes after 2:00 PM will the hands of a clock be perpendicular ($90^\circ$ apart) for the first time?

Solution:

1. At 2:00 PM, the minute hand is at 12 (0 minute-spaces) and the hour hand is at 2 ($2 \times 5 = 10\text{ minute-spaces}$).
2. For the hands to be perpendicular ($90^\circ$), they must be separated by: $\text{Separation} = \frac{90^\circ}{6^\circ/\text{minute-space}} = 15\text{ minute-spaces}$
3. For the first time after 2:00 PM, the minute hand must overtake the hour hand and pull $15\text{ minute-spaces}$ ahead of it (since initially it is only 10 spaces behind, it cannot be 15 spaces behind while remaining after 2:00 PM).
4. The total distance the minute hand must gain relative to the hour hand is: $\text{Distance to catch up} + \text{Lead required} = 10 + 15 = 25\text{ minute-spaces}$
5. Using the clock relative speed multiplier $\frac{12}{11}$: $m = \frac{12}{11} \times 25 = \frac{300}{11} = 27\frac{3}{11}\text{ minutes} \approx 27.27\text{ minutes}$
6. The hands will be perpendicular for the first time at 2:27:16 PM.

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CELE Board Exam Traps & Strategic Checklists