CELE Engineering Mathematics — Algebra and FundamentalsStudy Notes

Exponents and radicals are compact notation systems for repeated multiplication and fractional powers. Understanding these laws is essential because engineering formulas often involve powers (e.g., stress concentration factors, deflection coefficients, and flow rate calculations). **Fundamental Laws of Exponents:** 1. **Product Rule:** $a^m \cdot a^n = a^{m+n}$ - When multiplying powers with the same base, add the exponents. - Example: $x^3 \cdot x^5 = x^{3+5} = x^8$ 2. **Quotient Rule:** $\frac{a^m}{a^n} = a^{m-n}$ (where $a \neq 0$) - When dividing powers with the same base, subtract the exponents. - Example: $\frac{y^7}{y^2} = y^{7-2} = y^5$ 3. **Power of a Power:** $(a^m)^n = a^{mn}$ - When raising a power to another power, multiply the exponents. - Example: $(2^3)^4 = 2^{3 \times 4} = 2^{12} = 4096$ 4. **Power of a Product:** $(ab)^n = a^n b^n$ - Distribute the exponent to each factor. - Example: $(3x)^2 = 3^2 x^2 = 9x^2$ 5. **Power of a Quotient:** $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$ (where $b \neq 0$) - Distribute the exponent to both numerator and denominator. - Example: $\left(\frac{2}{5}\right)^3 = \frac{2^3}{5^3} = \frac{8}{125}$ 6. **Zero Exponent:** $a^0 = 1$ (where $a \neq 0$) - Any non-zero number raised to the power of 0 equals 1. - Example: $7^0 = 1$; $(x^2)^0 = 1$ 7. **Negative Exponent:** $a^{-n} = \frac{1}{a^n}$ (where $a \neq 0$) - A negative exponent means take the reciprocal. - Example: $3^{-2} = \frac{1}{3^2} = \frac{1}{9}$; $x^{-3} = \frac{1}{x^3}$ **Radicals and Fractional Exponents:** Radicals are written as $\sqrt[n]{a}$ (the $n$-th root of $a$) and are equivalent to fractional exponents: $\sqrt[n]{a} = a^{1/n}$. - **Square root:** $\sqrt{a} = a^{1/2}$ - **Cube root:** $\sqrt[3]{a} = a^{1/3}$ - **General fractional exponent:** $a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$ **Radical Laws:** 1. **Product Rule:** $\sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b}$ - Example: $\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}$ 2. **Quotient Rule:** $\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}$ (where $b \neq 0$) - Example: $\sqrt{\frac{25}{4}} = \frac{\sqrt{25}}{\sqrt{4}} = \frac{5}{2}$ 3. **Power of a Radical:** $(\sqrt[n]{a})^m = \sqrt[n]{a^m} = a^{m/n}$ - Example: $(\sqrt{5})^4 = (5^{1/2})^4 = 5^2 = 25$ 4. **Rationalization:** Remove radicals from denominators by multiplying by a convenient form of 1. - Example: $\frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ **Worked Example 1: Simplify $\frac{(2x^3)^2 \cdot (3x^{-1})^3}{(x^2)^4}$** **Step 1:** Apply the power of a product rule to each term: - $(2x^3)^2 = 2^2 (x^3)^2 = 4x^6$ - $(3x^{-1})^3 = 3^3 (x^{-1})^3 = 27x^{-3}$ - $(x^2)^4 = x^8$ **Step 2:** Substitute into the original expression: $$\frac{4x^6 \cdot 27x^{-3}}{x^8} = \frac{108x^{6-3}}{x^8} = \frac{108x^3}{x^8}$$ **Step 3:** Apply the quotient rule: $$\frac{108x^3}{x^8} = 108x^{3-8} = 108x^{-5} = \frac{108}{x^5}$$ **Worked Example 2: Simplify $\sqrt[3]{54} - \sqrt[3]{2}$** **Step 1:** Factor the radicand in the first term: $$54 = 27 \times 2 = 3^3 \times 2$$ **Step 2:** Apply the product rule for radicals: $$\sqrt[3]{54} = \sqrt[3]{3^3 \times 2} = \sqrt[3]{3^3} \cdot \sqrt[3]{2} = 3\sqrt[3]{2}$$ **Step 3:** Combine like radicals: $$\sqrt[3]{54} - \sqrt[3]{2} = 3\sqrt[3]{2} - \sqrt[3]{2} = 2\sqrt[3]{2}$$

Logarithms are the inverse of exponential functions. They transform multiplicative operations into additive ones, making complex calculations manageable—a critical skill in engineering where compound growth, decay, and stress-strain relationships often follow exponential or logarithmic patterns. **Definition:** If $b^x = y$ (where $b > 0$, $b \neq 1$), then $\log_b y = x$. - $b$ is the **base** - $y$ is the **argument** - $x$ is the **logarithm** (the exponent) **Common Bases:** - **Base 10 (Common logarithm):** $\log_{10} y$ or $\log y$ (used in engineering decibel calculations, pH scales) - **Base $e$ (Natural logarithm):** $\log_e y$ or $\ln y$ (used in growth/decay, control systems) - **Base 2 (Binary logarithm):** $\log_2 y$ (used in computer science and information theory) **Fundamental Logarithmic Laws:** 1. **Product Rule:** $\log_b(MN) = \log_b M + \log_b N$ - The logarithm of a product equals the sum of logarithms. - Example: $\log_2(8 \times 4) = \log_2 8 + \log_2 4 = 3 + 2 = 5$ - Verification: $8 \times 4 = 32 = 2^5$, so $\log_2 32 = 5$ ✓ 2. **Quotient Rule:** $\log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N$ - The logarithm of a quotient equals the difference of logarithms. - Example: $\log_{10}\left(\frac{1000}{10}\right) = \log_{10} 1000 - \log_{10} 10 = 3 - 1 = 2$ - Verification: $\frac{1000}{10} = 100 = 10^2$, so $\log_{10} 100 = 2$ ✓ 3. **Power Rule:** $\log_b(M^p) = p \log_b M$ - The logarithm of a power equals the power times the logarithm of the base. - Example: $\log_3(9^2) = 2 \log_3 9 = 2 \times 2 = 4$ - Verification: $9^2 = 81 = 3^4$, so $\log_3 81 = 4$ ✓ 4. **Change of Base Formula:** $\log_b x = \frac{\log_c x}{\log_c b} = \frac{\ln x}{\ln b}$ - Allows conversion between different bases using natural log or common log. - Example: $\log_2 16 = \frac{\ln 16}{\ln 2} = \frac{2.773}{0.693} \approx 4$ - Verification: $2^4 = 16$ ✓ 5. **Special Values:** - $\log_b 1 = 0$ (because $b^0 = 1$) - $\log_b b = 1$ (because $b^1 = b$) - $\log_b b^x = x$ (because $b^x = b^x$) - $b^{\log_b x} = x$ (inverse relationship) **Exponential Functions:** Exponential growth and decay follow the form $A(t) = A_0 e^{kt}$, where: - $A_0$ is the initial amount - $k$ is the growth rate (positive for growth, negative for decay) - $t$ is time - $e \approx 2.71828$ is Euler's number Alternatively: $A(t) = A_0 b^t$ where $b = 1 + r$ ($r$ is the rate of change). **Worked Example 3: Expand $\log_3\left(\frac{27x^2}{y}\right)$ using logarithmic laws** **Step 1:** Apply the quotient rule to separate numerator and denominator: $$\log_3\left(\frac{27x^2}{y}\right) = \log_3(27x^2) - \log_3(y)$$ **Step 2:** Apply the product rule to the first term: $$\log_3(27x^2) - \log_3(y) = \log_3 27 + \log_3(x^2) - \log_3(y)$$ **Step 3:** Evaluate $\log_3 27$ and apply the power rule: - $27 = 3^3$, so $\log_3 27 = 3$ - $\log_3(x^2) = 2\log_3 x$ **Step 4:** Combine: $$3 + 2\log_3 x - \log_3 y$$ **Worked Example 4: Solve $2^{x+1} = 8$** **Step 1:** Express both sides as powers of the same base: - $2^{x+1} = 8$ - $8 = 2^3$ - Therefore: $2^{x+1} = 2^3$ **Step 2:** Equate exponents (since the bases are equal): $$x + 1 = 3$$ **Step 3:** Solve for $x$: $$x = 2$$ **Verification:** $2^{2+1} = 2^3 = 8$ ✓ **Alternative Method Using Logarithms:** **Step 1:** Take the natural log of both sides: $$\ln(2^{x+1}) = \ln(8)$$ **Step 2:** Apply the power rule: $$(x+1)\ln 2 = \ln 8$$ **Step 3:** Solve for $x$: $$x + 1 = \frac{\ln 8}{\ln 2} = \log_2 8 = 3$$ $$x = 2$$ **Worked Example 5: A concrete sample undergoes exponential decay of compressive strength due to sulfate attack. If initial strength is 40 MPa and strength decreases by 5% annually, find the strength after 4 years.** **Step 1:** Identify the exponential decay model: - Initial strength: $A_0 = 40$ MPa - Annual decay rate: 5% means $r = -0.05$ - Strength at time $t$ (years): $A(t) = A_0(1 + r)^t = 40(1 - 0.05)^t = 40(0.95)^t$ **Step 2:** Calculate strength after 4 years: $$A(4) = 40(0.95)^4$$ **Step 3:** Compute $(0.95)^4$: $$(0.95)^4 = 0.8145 \text{ (approximately)}$$ **Step 4:** Final answer: $$A(4) = 40 \times 0.8145 = 32.58 \text{ MPa}$$ **Verification using logarithms:** We could also express this as $A(t) = 40 e^{-0.0513t}$ where $\ln(0.95) \approx -0.0513$, giving $A(4) \approx 40 e^{-0.2052} \approx 32.6$ MPa.

Quadratic equations appear throughout engineering practice—from parabolic arch designs in NSCP 2015 to deflection curves and optimization problems. A quadratic equation has the standard form: $$ax^2 + bx + c = 0 \quad (a \neq 0)$$ where $a$, $b$, and $c$ are constants and $x$ is the variable. **The Quadratic Formula:** The roots (solutions) of any quadratic equation are given by: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ This formula gives up to two solutions corresponding to the $\pm$ sign. **The Discriminant:** The discriminant is $\Delta = b^2 - 4ac$. It determines the **nature and number of roots**: - **$\Delta > 0$ (positive):** Two **distinct real roots** - $x_1 = \frac{-b + \sqrt{\Delta}}{2a}$ and $x_2 = \frac{-b - \sqrt{\Delta}}{2a}$ - The parabola crosses the $x$-axis at two points. - Example: $x^2 - 5x + 6 = 0$ → $\Delta = 25 - 24 = 1 > 0$ → two distinct roots - **$\Delta = 0$ (zero):** One **repeated real root** (or a double root) - $x = \frac{-b}{2a}$ (a single value) - The parabola touches the $x$-axis at exactly one point (vertex). - Example: $x^2 - 4x + 4 = 0$ → $\Delta = 16 - 16 = 0$ → $x = 2$ (double root) - **$\Delta < 0$ (negative):** Two **complex conjugate roots** - $x = \frac{-b \pm i\sqrt{|\Delta|}}{2a}$ (involving the imaginary unit $i$) - The parabola does not cross the $x$-axis (remains entirely above or below it). - These roots are not used in most engineering contexts requiring real physical solutions. - Example: $x^2 + x + 1 = 0$ → $\Delta = 1 - 4 = -3 < 0$ → complex roots **Vieta's Formulas (Relationships Between Roots and Coefficients):** For a quadratic $ax^2 + bx + c = 0$ with roots $x_1$ and $x_2$: - **Sum of roots:** $x_1 + x_2 = -\frac{b}{a}$ - **Product of roots:** $x_1 \cdot x_2 = \frac{c}{a}$ These relationships are useful for constructing equations or verifying solutions. **Factoring Method:** If the quadratic can be factored, it may be solved more quickly: $$ax^2 + bx + c = a(x - x_1)(x - x_2) = 0$$ This gives roots $x_1$ and $x_2$ immediately. **Worked Example 6: Solve $x^2 - 5x + 6 = 0$ using both factoring and the quadratic formula** **Method 1: Factoring** **Step 1:** Look for two numbers that multiply to $c = 6$ and add to $b = -5$. - Candidates: $(−1, −6)$, $(−2, −3)$, $(−3, −2)$, $(−6, −1)$ - Check $(−2, −3)$: $(-2) + (-3) = -5$ ✓ and $(-2) \times (-3) = 6$ ✓ **Step 2:** Factor: $$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$$ **Step 3:** Solve: $$x = 2 \quad \text{or} \quad x = 3$$ **Method 2: Quadratic Formula** **Step 1:** Identify coefficients: $a = 1$, $b = -5$, $c = 6$ **Step 2:** Calculate the discriminant: $$\Delta = b^2 - 4ac = (-5)^2 - 4(1)(6) = 25 - 24 = 1$$ **Step 3:** Apply the quadratic formula: $$x = \frac{-(-5) \pm \sqrt{1}}{2(1)} = \frac{5 \pm 1}{2}$$ **Step 4:** Compute the two roots: $$x_1 = \frac{5 + 1}{2} = \frac{6}{2} = 3$$ $$x_2 = \frac{5 - 1}{2} = \frac{4}{2} = 2$$ **Verification using Vieta's formulas:** - Sum: $2 + 3 = 5 = -(-5)/1$ ✓ - Product: $2 \times 3 = 6 = 6/1$ ✓ **Worked Example 7: Solve $2x^2 + 3x - 5 = 0$** **Step 1:** Identify coefficients: $a = 2$, $b = 3$, $c = -5$ **Step 2:** Calculate the discriminant: $$\Delta = b^2 - 4ac = 3^2 - 4(2)(-5) = 9 + 40 = 49$$ **Step 3:** Since $\Delta = 49 > 0$, there are two distinct real roots. Apply the quadratic formula: $$x = \frac{-3 \pm \sqrt{49}}{2(2)} = \frac{-3 \pm 7}{4}$$ **Step 4:** Compute the roots: $$x_1 = \frac{-3 + 7}{4} = \frac{4}{4} = 1$$ $$x_2 = \frac{-3 - 7}{4} = \frac{-10}{4} = -\frac{5}{2}$$ **Verification:** - For $x = 1$: $2(1)^2 + 3(1) - 5 = 2 + 3 - 5 = 0$ ✓ - For $x = -5/2$: $2(-5/2)^2 + 3(-5/2) - 5 = 2(25/4) - 15/2 - 5 = 25/2 - 15/2 - 10/2 = 0$ ✓ **Worked Example 8: Find the condition on $m$ such that $x^2 - (m+1)x + m = 0$ has real roots** **Step 1:** For real roots, the discriminant must be non-negative: $\Delta \geq 0$ **Step 2:** Set up the discriminant inequality: $$\Delta = [-(m+1)]^2 - 4(1)(m) \geq 0$$ $$(m+1)^2 - 4m \geq 0$$ **Step 3:** Expand and simplify: $$m^2 + 2m + 1 - 4m \geq 0$$ $$m^2 - 2m + 1 \geq 0$$ $$(m - 1)^2 \geq 0$$ **Step 4:** Interpret the result: Since $(m - 1)^2$ is always non-negative (a perfect square), the inequality holds for **all real values of $m$**. Therefore, the equation always has real roots. **Special case:** When $m = 1$, $(m-1)^2 = 0$, giving a repeated root. **Polynomial Equations (Higher Degree):** For polynomials of degree $n$: $$P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0$$ Solving cubic and quartic equations can be complex, but several strategies help: 1. **Rational Root Theorem:** If $p/q$ is a rational root (in lowest terms), then $p$ divides $a_0$ and $q$ divides $a_n$. 2. **Factor by grouping:** Rearrange and group terms to factor out common expressions. 3. **Synthetic division or polynomial long division:** Once one root is found, divide to reduce the degree. 4. **Special forms:** Recognize patterns like difference of squares, sum/difference of cubes, etc. **Worked Example 9: Solve $x^3 - 6x^2 + 11x - 6 = 0$** **Step 1:** Apply the Rational Root Theorem. Possible rational roots are divisors of 6: $\pm 1, \pm 2, \pm 3, \pm 6$ **Step 2:** Test $x = 1$: $$1 - 6 + 11 - 6 = 0$$ ✓ So $x = 1$ is a root, and $(x - 1)$ is a factor. **Step 3:** Perform synthetic division or polynomial long division to find the quotient when dividing by $(x - 1)$: $$x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6)$$ **Step 4:** Factor the resulting quadratic: $$x^2 - 5x + 6 = (x - 2)(x - 3)$$ **Step 5:** Complete factorization: $$x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3) = 0$$ **Step 6:** Solutions: $$x = 1, \quad x = 2, \quad x = 3$$

Progressions (sequences with specific patterns) are common in engineering analysis. Arithmetic progressions model linear change (constant difference), while geometric progressions model exponential growth or decay (constant ratio). Both appear in loan calculations, stress distributions, and sampling designs. **Arithmetic Progression (AP):** An AP is a sequence where the difference between consecutive terms is constant. This constant is called the **common difference**, denoted $d$. General sequence: $a_1, a_2, a_3, \ldots, a_n$ where $a_n - a_{n-1} = d$ for all $n \geq 2$. **Formula for the $n$-th term:** $$a_n = a_1 + (n-1)d$$ where: - $a_n$ is the $n$-th term - $a_1$ is the first term - $d$ is the common difference - $n$ is the term number **Sum of the first $n$ terms of an AP:** $$S_n = \frac{n}{2}[2a_1 + (n-1)d] = \frac{n}{2}(a_1 + a_n)$$ Both forms are useful: use the first when you know $a_1$ and $d$; use the second when you know both endpoints. **Worked Example 10: AP with known first term and common difference** Find the sum of the first 15 terms of the AP: $3, 7, 11, 15, \ldots$ **Step 1:** Identify the parameters: - First term: $a_1 = 3$ - Common difference: $d = 7 - 3 = 4$ - Number of terms: $n = 15$ **Step 2:** Find the 15th term using $a_n = a_1 + (n-1)d$: $$a_{15} = 3 + (15 - 1)(4) = 3 + 56 = 59$$ **Step 3:** Calculate the sum: $$S_{15} = \frac{15}{2}(a_1 + a_{15}) = \frac{15}{2}(3 + 59) = \frac{15}{2}(62) = 15 \times 31 = 465$$ **Verification using the alternative formula:** $$S_{15} = \frac{15}{2}[2(3) + (15-1)(4)] = \frac{15}{2}[6 + 56] = \frac{15}{2}(62) = 465$$ ✓ **Worked Example 11: Finding a specific term in an AP** In an AP, the 5th term is 13 and the 12th term is 34. Find the first term and the common difference. **Step 1:** Set up equations using $a_n = a_1 + (n-1)d$: - For $n = 5$: $a_5 = a_1 + 4d = 13$ ... (1) - For $n = 12$: $a_{12} = a_1 + 11d = 34$ ... (2) **Step 2:** Subtract equation (1) from equation (2): $$(a_1 + 11d) - (a_1 + 4d) = 34 - 13$$ $$7d = 21$$ $$d = 3$$ **Step 3:** Substitute $d = 3$ into equation (1): $$a_1 + 4(3) = 13$$ $$a_1 = 1$$ **Verification:** - $a_5 = 1 + 4(3) = 13$ ✓ - $a_{12} = 1 + 11(3) = 34$ ✓ - The AP is: $1, 4, 7, 10, 13, \ldots$ **Geometric Progression (GP):** A GP is a sequence where the ratio between consecutive terms is constant. This constant is called the **common ratio**, denoted $r$. General sequence: $a_1, a_2, a_3, \ldots, a_n$ where $\frac{a_n}{a_{n-1}} = r$ for all $n \geq 2$. **Formula for the $n$-th term:** $$a_n = a_1 r^{n-1}$$ where: - $a_n$ is the $n$-th term - $a_1$ is the first term - $r$ is the common ratio - $n$ is the term number **Sum of the first $n$ terms of a GP:** $$S_n = \frac{a_1(r^n - 1)}{r - 1} \quad (r \neq 1)$$ Alternatively (for $r \neq 1$): $$S_n = \frac{a_1(1 - r^n)}{1 - r}$$ Both forms are equivalent; choose based on convenience. **Special case: $r = 1$** When $r = 1$, all terms are equal to $a_1$, so $S_n = n \cdot a_1$. **Sum of an Infinite GP (when $|r| < 1$):** If $|r| < 1$ (the common ratio is strictly between $-1$ and $1$), the infinite series converges: $$S_\infty = \frac{a_1}{1 - r}$$ This formula is critical in engineering for calculating present values, damping ratios, and convergent stresses. **If $|r| \geq 1$, the infinite series diverges** (does not converge to a finite sum). **Worked Example 12: Find the sum of the first 8 terms of the GP: $2, 6, 18, 54, \ldots$** **Step 1:** Identify parameters: - First term: $a_1 = 2$ - Common ratio: $r = 6/2 = 3$ - Number of terms: $n = 8$ **Step 2:** Since $r \neq 1$, use the GP sum formula: $$S_8 = \frac{a_1(r^n - 1)}{r - 1} = \frac{2(3^8 - 1)}{3 - 1} = \frac{2(6561 - 1)}{2} = 6560$$ **Verification (compute individual terms):** - $a_1 = 2$ - $a_2 = 2 \times 3 = 6$ - $a_3 = 2 \times 3^2 = 18$ - $a_4 = 2 \times 3^3 = 54$ - $a_5 = 2 \times 3^4 = 162$ - $a_6 = 2 \times 3^5 = 486$ - $a_7 = 2 \times 3^6 = 1458$ - $a_8 = 2 \times 3^7 = 4374$ - Sum: $2 + 6 + 18 + 54 + 162 + 486 + 1458 + 4374 = 6560$ ✓ **Worked Example 13: Find the sum of an infinite GP** Find the sum of the infinite series: $8 + 4 + 2 + 1 + 0.5 + \ldots$ **Step 1:** Identify parameters: - First term: $a_1 = 8$ - Common ratio: $r = 4/8 = 1/2 = 0.5$ - Check: $|r| = 0.5 < 1$ ✓ (series converges) **Step 2:** Apply the infinite sum formula: $$S_\infty = \frac{a_1}{1 - r} = \frac{8}{1 - 0.5} = \frac{8}{0.5} = 16$$ **Interpretation:** As we add more and more terms, their sum approaches (but never exceeds) 16. **Worked Example 14: Engineering application—Damping in structures** A vibrating beam experiences damping such that each oscillation amplitude is 80% of the previous amplitude. If the initial amplitude is 10 mm, what is the total displacement (sum of all oscillations) theoretically? **Step 1:** Identify the GP: - First amplitude: $a_1 = 10$ mm - Common ratio: $r = 0.8$ (each is 80% of previous) - Check: $|r| = 0.8 < 1$ ✓ (infinite sum converges) **Step 2:** Calculate the theoretical total displacement: $$S_\infty = \frac{a_1}{1 - r} = \frac{10}{1 - 0.8} = \frac{10}{0.2} = 50 \text{ mm}$$ **Interpretation:** Although the beam oscillates infinitely, the total displacement from all oscillations converges to 50 mm. This is crucial in predicting fatigue damage and stress concentration factors.

The binomial theorem provides a formula for expanding expressions of the form $(a + b)^n$ without multiplying them out term by term. This is invaluable for approximations in engineering (e.g., small deformation theory) and combinatorial counting. **The Binomial Theorem (General Form):** $$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$ where $\binom{n}{k}$ (read "$n$ choose $k$") is the **binomial coefficient**: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ Here, $n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$ is the factorial. **Expanded form:** $$(a + b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n$$ **Key Properties of Binomial Coefficients:** 1. $\binom{n}{0} = 1$ and $\binom{n}{n} = 1$ 2. $\binom{n}{k} = \binom{n}{n-k}$ (symmetry) 3. $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$ (Pascal's identity) **Pascal's Triangle:** Binomial coefficients can be arranged in Pascal's triangle: ``` n=0: 1 n=1: 1 1 n=2: 1 2 1 n=3: 1 3 3 1 n=4: 1 4 6 4 1 n=5: 1 5 10 10 5 1 n=6: 1 6 15 20 15 6 1 ``` Each entry is the sum of the two entries above it. This allows rapid lookup of binomial coefficients without calculating factorials. **General Term ($(r+1)$-th term):** The $(r+1)$-th term in the expansion of $(a+b)^n$ is: $$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$ where $r = 0, 1, 2, \ldots, n$. (Note: we count from $r=0$, so the "first" term has $r=0$.) **Worked Example 15: Expand $(2x + 3)^4$** **Step 1:** Identify $a = 2x$, $b = 3$, $n = 4$ **Step 2:** Write the general expansion: $$(2x + 3)^4 = \sum_{k=0}^{4} \binom{4}{k} (2x)^{4-k} (3)^k$$ **Step 3:** Calculate binomial coefficients using Pascal's triangle (or factorials): - $\binom{4}{0} = 1$ - $\binom{4}{1} = 4$ - $\binom{4}{2} = 6$ - $\binom{4}{3} = 4$ - $\binom{4}{4} = 1$ **Step 4:** Compute each term: - $k=0$: $\binom{4}{0}(2x)^4(3)^0 = 1 \times 16x^4 \times 1 = 16x^4$ - $k=1$: $\binom{4}{1}(2x)^3(3)^1 = 4 \times 8x^3 \times 3 = 96x^3$ - $k=2$: $\binom{4}{2}(2x)^2(3)^2 = 6 \times 4x^2 \times 9 = 216x^2$ - $k=3$: $\binom{4}{3}(2x)^1(3)^3 = 4 \times 2x \times 27 = 216x$ - $k=4$: $\binom{4}{4}(2x)^0(3)^4 = 1 \times 1 \times 81 = 81$ **Step 5:** Combine all terms: $$(2x + 3)^4 = 16x^4 + 96x^3 + 216x^2 + 216x + 81$$ **Worked Example 16: Find the 6th term in the expansion of $(3a - 2b)^9$** **Step 1:** The 6th term corresponds to $r + 1 = 6$, so $r = 5$ **Step 2:** Use the general term formula: $$T_6 = T_{5+1} = \binom{9}{5} (3a)^{9-5} (-2b)^5$$ **Step 3:** Calculate $\binom{9}{5}$: $$\binom{9}{5} = \frac{9!}{5! \times 4!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{3024}{24} = 126$$ **Step 4:** Substitute and simplify: $$T_6 = 126 \times (3a)^4 \times (-2b)^5$$ $$= 126 \times 81a^4 \times (-32b^5)$$ $$= 126 \times 81 \times (-32) \times a^4 b^5$$ $$= -326592 a^4 b^5$$ **Worked Example 17: Use the binomial theorem to find $(1.02)^5$ approximately** **Method 1: Direct expansion** Write $1.02 = 1 + 0.02$, so $(1.02)^5 = (1 + 0.02)^5$ Using $(a+b)^5$ with $a = 1$, $b = 0.02$: $$(1 + 0.02)^5 = \sum_{k=0}^{5} \binom{5}{k} (1)^{5-k} (0.02)^k$$ Expand: - $k=0$: $\binom{5}{0}(0.02)^0 = 1$ - $k=1$: $\binom{5}{1}(0.02)^1 = 5 \times 0.02 = 0.1$ - $k=2$: $\binom{5}{2}(0.02)^2 = 10 \times 0.0004 = 0.004$ - $k=3$: $\binom{5}{3}(0.02)^3 = 10 \times 0.000008 = 0.00008$ - Higher terms are negligible **Step 1:** Sum the significant terms: $$(1.02)^5 \approx 1 + 0.1 + 0.004 + 0.00008 \approx 1.10408$$ **Verification (calculator):** $(1.02)^5 = 1.10408080...$ ✓ **Method 2: First-order approximation (linear approximation)** For small $b$, $(1+b)^n \approx 1 + nb$ (keeping only the first two terms): $$(1.02)^5 \approx 1 + 5(0.02) = 1 + 0.1 = 1.1$$ This is a quick approximation accurate to two significant figures. **Worked Example 18: Find the term independent of $x$ in the expansion of $(x^2 + \frac{1}{x})^9$** **Step 1:** Use the general term formula: $$T_{r+1} = \binom{9}{r} (x^2)^{9-r} \left(\frac{1}{x}\right)^r = \binom{9}{r} x^{2(9-r)} \times x^{-r}$$ $$= \binom{9}{r} x^{18 - 2r - r} = \binom{9}{r} x^{18-3r}$$ **Step 2:** For the term independent of $x$, the exponent of $x$ must be zero: $$18 - 3r = 0$$ $$3r = 18$$ $$r = 6$$ **Step 3:** Calculate the term: $$T_7 = \binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84$$ So the term independent of $x$ is **84**.

Word problems test your ability to translate real-world scenarios into mathematical equations. Common types include work problems, mixture problems, age problems, and motion problems. These are frequent on the PRC examination and require careful setup. **Work Problems:** When multiple workers or machines complete a task, **add the rates** (not the times). Rate = 1 / time. **Strategy:** 1. Define the rate for each worker/machine (fraction of job per unit time). 2. Add the rates if working together. 3. If one is removed or added mid-task, adjust accordingly. **Worked Example 19: Workers and job completion** Worker A can complete a job in 6 days. Worker B can complete the same job in 4 days. If they work together, how long will it take? **Step 1:** Define rates: - Rate of A: $\frac{1}{6}$ job/day - Rate of B: $\frac{1}{4}$ job/day **Step 2:** Combined rate when working together: $$\text{Combined rate} = \frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12} \text{ job/day}$$ **Step 3:** Time to complete 1 job: $$\text{Time} = \frac{1 \text{ job}}{5/12 \text{ job/day}} = \frac{12}{5} = 2.4 \text{ days}$$ **Verification:** Work done by A in 2.4 days + Work done by B in 2.4 days = $\frac{2.4}{6} + \frac{2.4}{4} = 0.4 + 0.6 = 1$ job ✓ **Worked Example 20: Three workers with different roles** Pipe A fills a tank in 3 hours. Pipe B fills it in 5 hours. Drain C empties it in 4 hours. If all three are open, how long to fill the tank? **Step 1:** Define rates (positive for inflow, negative for outflow): - Rate of A: $+\frac{1}{3}$ tank/hour - Rate of B: $+\frac{1}{5}$ tank/hour - Rate of C: $-\frac{1}{4}$ tank/hour **Step 2:** Net rate with all three open: $$\text{Net rate} = \frac{1}{3} + \frac{1}{5} - \frac{1}{4}$$ Find common denominator (LCM of 3, 5, 4 is 60): $$= \frac{20}{60} + \frac{12}{60} - \frac{15}{60} = \frac{17}{60} \text{ tank/hour}$$ **Step 3:** Time to fill 1 tank: $$\text{Time} = \frac{1 \text{ tank}}{17/60 \text{ tank/hour}} = \frac{60}{17} \approx 3.53 \text{ hours}$$ **Mixture Problems:** When combining solutions of different concentrations, use: **Total amount of substance = Sum of amounts from each component**. **Strategy:** 1. Define variables for unknown amounts or concentrations. 2. Set up an equation: (concentration 1) × (amount 1) + (concentration 2) × (amount 2) = (final concentration) × (final amount). 3. Solve for the unknown. **Worked Example 21: Mixing solutions of different concentrations** A chemist has 5 liters of 30% acid solution. How many liters of 80% acid solution must be added to obtain 50% acid solution? **Step 1:** Define the variable: - Let $x$ = liters of 80% acid solution to be added **Step 2:** Set up the mixture equation: $$\text{Acid from 30% solution} + \text{Acid from 80% solution} = \text{Acid in final solution}$$ $$0.30 \times 5 + 0.80 \times x = 0.50 \times (5 + x)$$ **Step 3:** Solve: $$1.5 + 0.80x = 2.5 + 0.50x$$ $$0.80x - 0.50x = 2.5 - 1.5$$ $$0.30x = 1$$ $$x = \frac{1}{0.30} = \frac{10}{3} \approx 3.33 \text{ liters}$$ **Verification:** - Acid from 30% solution: $0.30 \times 5 = 1.5$ liters - Acid from 80% solution: $0.80 \times 3.33 = 2.67$ liters - Total acid: $1.5 + 2.67 = 4.17$ liters - Final volume: $5 + 3.33 = 8.33$ liters - Final concentration: $\frac{4.17}{8.33} \approx 0.50 = 50\%$ ✓ **Age Problems:** These typically involve relationships between people's ages at different times (past, present, future). **Worked Example 22: Age relationship** The present age of a father is three times the age of his son. Five years from now, the father will be 2.5 times as old as the son. Find their present ages. **Step 1:** Define variables: - Let $s$ = son's present age (years) - Then $f = 3s$ = father's present age (years) **Step 2:** Set up the equation using ages five years from now: - Son's age in 5 years: $s + 5$ - Father's age in 5 years: $3s + 5$ - Relationship: $3s + 5 = 2.5(s + 5)$ **Step 3:** Solve: $$3s + 5 = 2.5s + 12.5$$ $$3s - 2.5s = 12.5 - 5$$ $$0.5s = 7.5$$ $$s = 15 \text{ years}$$ **Step 4:** Find father's age: $$f = 3s = 3 \times 15 = 45 \text{ years}$$ **Verification:** - Present: Father is 45, son is 15; $45 = 3 \times 15$ ✓ - In 5 years: Father is 50, son is 20; $50 = 2.5 \times 20$ ✓ **Motion Problems (Distance, Rate, Time):** Use the fundamental relationship: **Distance = Rate × Time**, or $d = rt$. **Worked Example 23: Two objects moving toward each other** Two cities are 240 km apart. A car from City A travels toward City B at 60 km/h. A car from City B travels toward City A at 80 km/h. Starting at the same time, when will they meet? **Step 1:** Define the variable: - Let $t$ = time (hours) when they meet **Step 2:** Set up the equation: The sum of distances traveled by both cars equals the total distance: $$\text{Distance by car from A} + \text{Distance by car from B} = 240$$ $$60t + 80t = 240$$ **Step 3:** Solve: $$140t = 240$$ $$t = \frac{240}{140} = \frac{12}{7} \approx 1.71 \text{ hours} \approx 1 \text{ hour } 43 \text{ minutes}$$ **Verification:** - Distance by car from A: $60 \times \frac{12}{7} = \frac{720}{7} \approx 102.86$ km - Distance by car from B: $80 \times \frac{12}{7} = \frac{960}{7} \approx 137.14$ km - Total: $102.86 + 137.14 = 240$ km ✓ **Worked Example 24: Current affecting motion** A boat travels downstream (with current) for 30 km in 2 hours. It then travels upstream (against current) for 20 km in 2 hours. Find the boat's speed in still water and the speed of the current. **Step 1:** Define variables: - Let $b$ = boat's speed in still water (km/h) - Let $c$ = current's speed (km/h) **Step 2:** Set up equations using $d = rt$: - Downstream: $30 = (b + c) \times 2$ → $b + c = 15$ ... (1) - Upstream: $20 = (b - c) \times 2$ → $b - c = 10$ ... (2) **Step 3:** Solve the system: Add equations (1) and (2): $$(b + c) + (b - c) = 15 + 10$$ $$2b = 25$$ $$b = 12.5 \text{ km/h}$$ Substitute back into equation (1): $$12.5 + c = 15$$ $$c = 2.5 \text{ km/h}$$ **Verification:** - Downstream: speed $= 12.5 + 2.5 = 15$ km/h; distance in 2 h $= 15 \times 2 = 30$ km ✓ - Upstream: speed $= 12.5 - 2.5 = 10$ km/h; distance in 2 h $= 10 \times 2 = 20$ km ✓