Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Quick Summary
Basic Algebra extends Class 9–10 algebra into polynomials, inequalities, partial fractions, and absolute value. It underpins calculus, sequences, and geometry. Quadratic equations and inequalities are directly tested in board exams and entrance tests.
Polynomials
A polynomial p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, where aₙ ≠ 0.
Remainder Theorem: When p(x) is divided by (x − a), the remainder = p(a).
Factor Theorem: (x − a) is a factor of p(x) if and only if p(a) = 0.
Key Identities
| Identity | Form |
|---|---|
| (a + b)² | a² + 2ab + b² |
| (a − b)² | a² − 2ab + b² |
| (a + b)³ | a³ + 3a²b + 3ab² + b³ |
| a³ − b³ | (a − b)(a² + ab + b²) |
| a³ + b³ | (a + b)(a² − ab + b²) |
Quadratic Equations
For ax² + bx + c = 0 (a ≠ 0):
- Discriminant: Δ = b² − 4ac
- Δ > 0 → two distinct real roots
- Δ = 0 → one repeated real root
- Δ < 0 → two complex conjugate roots
- Quadratic formula: x = (−b ± √Δ) / 2a
- Sum of roots: α + β = −b/a
- Product of roots: αβ = c/a
Forming equation from roots: x² − (α + β)x + αβ = 0
Inequalities
Linear
Solve like equations but flip the sign when multiplying or dividing by a negative number.
Quadratic Inequalities
1. Bring all terms to one side. 2. Find roots r₁ < r₂. 3. For a > 0: ax² + bx + c > 0 when x < r₁ or x > r₂; < 0 when r₁ < x < r₂.
Modulus (Absolute Value)
| Expression | Meaning | ||
|---|---|---|---|
| \ | x\ | = a (a > 0) | x = a or x = −a |
| \ | x\ | < a | −a < x < a |
| \ | x\ | > a | x < −a or x > a |
| \ | x − a\ | < b | a − b < x < a + b |
| Triangle Inequality: \ | a + b\ | ≤ \ | a\ | + \ | b\ |
|---|
Partial Fractions
Used when degree of numerator < degree of denominator.
| Denominator Factor | Form |
|---|---|
| (ax + b) | A/(ax + b) |
| (ax + b)² | A/(ax+b) + B/(ax+b)² |
| ax² + bx + c (irreducible) | (Ax + B)/(ax² + bx + c) |
Method: Multiply both sides by the denominator, substitute roots to find constants.
Worked Examples
Example 1: Solve x² − 5x + 6 < 0. Roots: x = 2, x = 3. Parabola opens up. Negative between roots: 2 < x < 3.
Example 2: Decompose (3x + 5)/((x−1)(x+2)). 3x + 5 = A(x+2) + B(x−1). x = 1: A = 8/3. x = −2: B = 1/3. Result: 8/(3(x−1)) + 1/(3(x+2)).
−5 ≤ 2x − 3 ≤ 5 → −2 ≤ 2x ≤ 8 → −1 ≤ x ≤ 4.
| Example 3: Solve | 2x − 3 | ≤ 5. |
|---|
Important Questions
One Mark
- State the Factor Theorem.
- If roots of x² − 5x + k = 0 are equal, find k.
- Solve |x − 2| = 3.
- Write the partial fraction form for 1/((x+1)(x−2)).
Short Answer (3 marks)
- Find k if one root of 2x² − 5x + k = 0 is double the other.
- Solve the inequality x² − x − 6 > 0.
- Decompose (2x + 3)/((x+1)(x−1)) into partial fractions.
Long Answer (5 marks)
- If α, β are roots of x² − 3x + 2 = 0, find α² + β², α³ + β³, and the equation with roots 1/α and 1/β.
- Solve: x² − 4x + 3 ≤ 0 and x² − 6x + 8 > 0. Find the common solution set.
Common Mistakes
- Not flipping the inequality sign when multiplying/dividing by a negative number.
- Applying the quadratic formula without writing the equation in standard form first.
- Partial fractions: using A/(ax+b) form for irreducible quadratic factors instead of (Ax+B)/(ax²+bx+c).
- Misreading |x| > a as −a < x < a (correct form: x < −a or x > a).
Board Exam Tips
- Sum and product of roots appear almost every year — know the formulas cold.
- Quadratic inequality solutions require a number line sketch — show it in your answer.
- Partial fractions also appear as a prerequisite step inside integration problems in Chapter 11.
- JEE: Modulus inequalities and polynomial factorisation are standard short-answer questions.