In-Depth: Quadratic Equations
CLAT Application & Relevance
Importance: Low. While "Basic Algebra" is listed, direct, complex quadratic equation solving is very rare in CLAT. However, understanding how to form and solve simple quadratic equations (especially by factorization) can be necessary if a word problem in a caselet naturally leads to a quadratic form. Focus on recognizing the structure and knowing the fundamental methods rather than advanced problem-solving.
How it's tested: Indirectly, as an intermediate step in a larger word problem derived from a passage. You might need to set up an equation that turns out to be quadratic.
Section 1: Core Concepts & Solving Methods
A quadratic equation is a polynomial equation of the second degree. The general form of a quadratic equation is ax² + bx + c = 0, where 'a', 'b', and 'c' are real numbers, and 'a' is not equal to zero.
Key Definitions
- Roots (or Solutions): The values of the variable (x) that satisfy the equation. A quadratic equation typically has two roots.
- Discriminant (Δ): The expression
b² - 4ac. It determines the nature of the roots.
Methods for Solving Quadratic Equations
Method 1: Factorization Method (Splitting the Middle Term)
- Rewrite the quadratic expression
ax² + bx + cby splitting the middle term 'bx' into two terms whose sum is 'bx' and whose product is 'ac'. - Factor out common terms from the pairs to get two linear factors.
- Set each linear factor to zero and solve for 'x'.
Method 2: Quadratic Formula
The roots of the quadratic equation ax² + bx + c = 0 are given by:
x = (-b ± √(b² - 4ac)) / 2a
Nature of Roots (based on Discriminant Δ = b² - 4ac):
- If
Δ > 0: Two distinct real roots. - If
Δ = 0: Two equal real roots (or one repeated root). - If
Δ < 0: No real roots (roots are complex/imaginary). CLAT typically deals with real roots.
Section 2: Solved CLAT-Style Examples
Example 1: Forming and Solving a Quadratic by Factorization (Caselet)
Passage Context: "A public park, maintained by the city's legal department, is rectangular in shape. The length of the park is 5 meters more than its width. The area of the park is 300 square meters."
Question: "What is the width of the park?"
Detailed Solution:
1. Define Variables:
Let the width of the park = w meters.
Then the length of the park = w + 5 meters.
2. Formulate the Equation (Area = Length × Width):
(w + 5) * w = 300
w² + 5w = 300
w² + 5w - 300 = 0. This is a quadratic equation.
3. Solve by Factorization: Find two numbers that multiply to -300 and add up to +5. These numbers are +20 and -15.
w² + 20w - 15w - 300 = 0
w(w + 20) - 15(w + 20) = 0
(w - 15)(w + 20) = 0
4. Find Possible Values for w:
w - 15 = 0 => w = 15
w + 20 = 0 => w = -20
5. Choose the Valid Solution: Since width cannot be negative, w = 15 meters.
Answer: The width of the park is 15 meters.
Example 2: Using the Quadratic Formula (Illustrative)
Passage Context: "A hypothetical scenario in a legal simulation involves a variable 'x' representing a certain quantity. A calculation yields the equation 2x² - 7x + 3 = 0, where 'x' must be a positive integer related to the number of documents."
Question: "Find the value(s) of 'x'."
Detailed Solution:
1. Identify a, b, c: For 2x² - 7x + 3 = 0,
a = 2, b = -7, c = 3.
2. Apply the Quadratic Formula: x = (-b ± √(b² - 4ac)) / 2a
x = ( -(-7) ± √((-7)² - 4 * 2 * 3) ) / (2 * 2)
x = ( 7 ± √(49 - 24) ) / 4
x = ( 7 ± 5 ) / 4
3. Calculate the two roots:
x1 = (7 + 5) / 4 = 12 / 4 = 3
x2 = (7 - 5) / 4 = 2 / 4 = 0.5
4. Consider problem constraints: The problem states 'x' must be a positive integer.
Answer: The valid value for 'x' is 3.