Quadratic Equations
Overview
Quadratic equations form a cornerstone topic in upper-primary and secondary mathematics, appearing consistently in Bihar TET Paper II. A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2. This topic bridges basic algebra with more advanced mathematical reasoning and has direct applications in real-world problems involving area, projectile motion, and optimization.
For Bihar TET, you must master three core skills: identifying quadratic equations, finding their roots using multiple methods, and applying these concepts to word problems. The syllabus specifically emphasizes "roots of quadratic equations and applications," so expect questions testing both computational ability and conceptual understanding of the nature of roots.
This topic connects closely with algebraic expressions and linear equations from your syllabus. Strong command here also supports geometry problems (area calculations) and science applications (motion under gravity).
Key Concepts
- **Standard Form**: A quadratic equation in variable x is written as ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0. The condition a ≠ 0 is essential—if a = 0, the equation becomes linear.
- **Roots/Solutions**: The values of x that satisfy the equation are called roots or zeros. A quadratic equation has exactly two roots (which may be equal, distinct, or complex).
- **Discriminant (D)**: The expression D = b² − 4ac determines the nature of roots. This is the most important diagnostic tool for quadratic equations.
- **Nature of Roots Based on D**:
- D > 0: Two distinct real roots
- D = 0: Two equal real roots (one repeated root)
- D < 0: No real roots (roots are complex/imaginary)
- **Sum and Product of Roots**: If α and β are roots of ax² + bx + c = 0, then:
- Sum of roots (α + β) = −b/a
- Product of roots (αβ) = c/a
- **Methods to Solve**: Factorization, completing the square, and quadratic formula are the three standard methods. For TET, factorization and the quadratic formula are most frequently tested.
- **Forming Equations**: Given roots α and β, the quadratic equation is x² − (α + β)x + αβ = 0.
Formulas / Key Facts
| Formula | Context | |---------|---------| | ax² + bx + c = 0 | Standard form of quadratic equation | | x = (−b ± √(b² − 4ac)) / 2a | Quadratic formula (Shridharacharya's rule) | | D = b² − 4ac | Discriminant formula | | α + β = −b/a | Sum of roots | | αβ = c/a | Product of roots | | x² − (sum)x + (product) = 0 | Forming equation from roots |
**Key Facts to Remember**:
- Every quadratic equation has exactly 2 roots (counting multiplicity)
- If one root is irrational (like 2 + √3), the other root is its conjugate (2 − √3)
- If coefficients are rational and one root is p + √q, the other is p − √q
- A quadratic equation with rational coefficients cannot have exactly one irrational root
Worked Examples
**Example 1: Solving by Factorization**
Solve: x² − 7x + 12 = 0
*Step 1*: Find two numbers whose product is 12 and sum is −7 Numbers: −3 and −4 (since −3 × −4 = 12 and −3 + −4 = −7)
*Step 2*: Factorize x² − 3x − 4x + 12 = 0 x(x − 3) − 4(x − 3) = 0 (x − 3)(x − 4) = 0
*Step 3*: Apply zero product property x = 3 or x = 4
**Answer**: Roots are 3 and 4
---
**Example 2: Using Quadratic Formula**
Solve: 2x² + 5x − 3 = 0
*Step 1*: Identify a = 2, b = 5, c = −3
*Step 2*: Calculate discriminant D = b² − 4ac = 25 − 4(2)(−3) = 25 + 24 = 49
*Step 3*: Apply formula x = (−5 ± √49) / (2 × 2) = (−5 ± 7) / 4
x = (−5 + 7)/4 = 2/4 = 1/2 x = (−5 − 7)/4 = −12/4 = −3
**Answer**: Roots are 1/2 and −3
---
**Example 3: Application Problem**
The product of two consecutive positive integers is 306. Find the integers.
*Step 1*: Let integers be x and x + 1 x(x + 1) = 306
*Step 2*: Form equation x² + x − 306 = 0
*Step 3*: Factorize (find factors of 306 differing by 1) 306 = 17 × 18 x² + 18x − 17x − 306 = 0 (x + 18)(x − 17) = 0
*Step 4*: x = 17 or x = −18 Since integers are positive, x = 17
**Answer**: The integers are 17 and 18
---
**Example 4: Finding Nature of Roots**
For what value of k does 2x² − 6x + k = 0 have equal roots?
*Solution*: For equal roots, D = 0 b² − 4ac = 0 (−6)² − 4(2)(k) = 0 36 − 8k = 0 k = 36/8 = 9/2
**Answer**: k = 9/2 or 4.5
Common Mistakes
- **Forgetting a ≠ 0**: Students sometimes accept a = 0 in standard form → Remember, if a = 0, it's not quadratic but linear.
- **Sign errors in quadratic formula**: Writing −b as just b, or miscalculating under the square root → Always write the formula first, then substitute carefully with parentheses.
- **Confusing sum and product formulas**: Mixing up which has the negative sign → Remember: Sum has the negative (−b/a), Product is straightforward (c/a).
- **Ignoring context in word problems**: Accepting negative roots when the problem asks for length, age, or count → Always verify if roots make sense in the given context.
- **Incomplete factorization**: Finding only one root and stopping → A quadratic always has two roots; check both factors.
- **Discriminant calculation errors**: Forgetting to square b or miscalculating 4ac → Write D = b² − 4ac separately before substituting into the main formula.
Quick Reference
- Standard form: ax² + bx + c = 0 (a ≠ 0)
- Quadratic formula: x = (−b ± √D) / 2a where D = b² − 4ac
- D > 0 → real distinct; D = 0 → real equal; D < 0 → no real roots
- Sum of roots = −b/a; Product of roots = c/a
- For equal roots, set discriminant = 0 and solve for unknown
- Always verify roots by substituting back into original equation