Algebra — Study Notes for UP Police Constable

Overview

Algebra forms a crucial component of the Numerical & Mental Ability section in UP Police Constable exam, typically contributing 3-5 questions. Unlike the heavy computational arithmetic topics, algebra tests your ability to work with variables, solve equations, and apply mathematical identities. The exam focuses on practical problem-solving rather than abstract theory.

For UP Police Constable preparation, you must master three core areas: linear equations (single and two variables), quadratic equations (factorization and formula method), and basic algebraic identities used for simplification. Questions are usually straightforward — finding the value of x, solving word problems that translate into equations, or simplifying expressions using identities. The key is recognizing problem patterns quickly and applying the correct method without calculation errors. Speed and accuracy matter more than theoretical depth.

Most algebra questions can be solved within 45-60 seconds if you know the formulas and have practiced standard problem types. This topic integrates well with other areas — percentage, profit-loss, and time-work problems often require forming and solving equations.

Key Concepts

• **Linear equations** involve variables to the first power (x, not x²) and can have one or two variables; solving means finding the value that makes the equation true.

• **Substitution method** for two-variable linear equations: solve one equation for a variable, substitute into the other, then back-substitute to find both values.

• **Elimination method** adds or subtracts equations after multiplying by suitable constants to eliminate one variable, solving for the other.

• **Quadratic equations** have the form ax² + bx + c = 0 where a ≠ 0; they produce two solutions (roots) which may be real, equal, or involve negative discriminants (rarely tested).

• **Factorization method** works when the quadratic can be expressed as a product of two binomials: (x + p)(x + q) = 0, giving roots x = -p and x = -q.

• **Quadratic formula** x = [-b ± √(b² - 4ac)] / 2a solves any quadratic equation; memorize this formula and the discriminant (b² - 4ac) which determines the nature of roots.

• **Algebraic identities** are proven formulas for expanding or factorizing expressions — they save time in simplification problems and are frequently tested.

• **Transposing** means moving terms across the equals sign (changing their sign) to isolate the variable — the foundation of equation solving.

Formulas / Key Facts

**Linear Equations:**

• Standard form: ax + b = 0, solution x = -b/a
• Two-variable form: ax + by = c and dx + ey = f

**Quadratic Formula:**

• ax² + bx + c = 0 → x = [-b ± √(b² - 4ac)] / 2a
• Discriminant D = b² - 4ac (D > 0: two distinct roots; D = 0: equal roots; D < 0: no real roots)

**Sum and Product of Roots:**

• For ax² + bx + c = 0, sum of roots = -b/a, product of roots = c/a

**Essential Algebraic Identities:** 1. (a + b)² = a² + 2ab + b² 2. (a - b)² = a² - 2ab + b² 3. a² - b² = (a + b)(a - b) 4. (a + b)³ = a³ + b³ + 3ab(a + b) 5. (a - b)³ = a³ - b³ - 3ab(a - b) 6. a³ + b³ = (a + b)(a² - ab + b²) 7. a³ - b³ = (a - b)(a² + ab + b²) 8. (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Worked Examples

**Example 1: Linear equation (single variable)**

Solve: 5x - 7 = 3x + 9

*Solution:*

• Step 1: Bring variable terms to one side: 5x - 3x = 9 + 7
• Step 2: Simplify: 2x = 16
• Step 3: Divide: x = 8

**Example 2: Linear equations (two variables)**

Solve: 2x + 3y = 13 and 3x - y = 3

*Solution (Elimination method):*

• Multiply second equation by 3: 9x - 3y = 9
• Add to first equation: (2x + 3y) + (9x - 3y) = 13 + 9
• Simplify: 11x = 22, so x = 2
• Substitute x = 2 in second equation: 3(2) - y = 3
• Solve: 6 - y = 3, so y = 3

**Example 3: Quadratic equation (factorization)**

Solve: x² - 7x + 12 = 0

*Solution:*

• Step 1: Find two numbers that multiply to 12 and add to -7: these are -3 and -4
• Step 2: Factor: (x - 3)(x - 4) = 0
• Step 3: Set each factor to zero: x = 3 or x = 4

**Example 4: Using algebraic identity**

If x + 1/x = 5, find x² + 1/x²

*Solution:*

• Use identity: (x + 1/x)² = x² + 2(x)(1/x) + 1/x²
• Substitute: (5)² = x² + 2 + 1/x²
• Simplify: 25 = x² + 1/x² + 2
• Therefore: x² + 1/x² = 23

Common Mistakes

**Wrong: Forgetting to change signs when transposing** → Correct: When moving a term across the equals sign, always flip its sign. If 2x + 5 = 13, then 2x = 13 - 5, not 13 + 5.

**Wrong: Solving quadratic by taking square root of both sides incorrectly** → Correct: √(x²) = |x|, not just x. Better to factor or use the quadratic formula than risk sign errors.

**Wrong: Multiplying only one term when solving two-variable equations** → Correct: When using elimination, multiply the ENTIRE equation by the constant, not just one term. If multiplying 2x + y = 5 by 3, you get 6x + 3y = 15.

**Wrong: Misapplying (a - b)² = a² - b²** → Correct: The middle term matters! (a - b)² = a² - 2ab + b², and a² - b² = (a + b)(a - b). These are different identities.

**Wrong: Calculation errors in quadratic formula with signs** → Correct: Carefully handle -b in the numerator. If b is already negative, -b becomes positive. For x² - 5x + 6 = 0, use -(-5) = +5 in the formula.

Quick Reference

• Linear equation solution in one step: isolate x by transposing all other terms and dividing by coefficient of x
• Two-variable linear equations need two equations; use elimination (faster) or substitution (safer) method
• Quadratic roots from factorization: split middle term so it becomes (x + p)(x + q) = 0
• Quadratic formula works always: x = [-b ± √(b² - 4ac)] / 2a — memorize this completely
• Identity shortcuts: (a + b)² ≠ a² + b²; always include 2ab term
• Check solutions by substituting back into original equation — catches 90% of calculation mistakes