IIT JEE Main Maths -Unit 2- Relations between roots and coefficients- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 2- Relations between roots and coefficients – Study Notes – New syllabus
IIT JEE Main Maths -Unit 2- Relations between roots and coefficients – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Relations Between Roots and Coefficients & Formation of Quadratic Equation from Given Roots
Relations Between Roots and Coefficients & Formation of Quadratic Equation from Given Roots
1. Relations Between Roots and Coefficients
Let the quadratic equation be
\( ax^2 + bx + c = 0 \), where \( a \ne 0 \).
If \( \alpha \) and \( \beta \) are the roots of the equation, then by factorization:
\( a(x – \alpha)(x – \beta) = 0 \)
Expanding,
\( ax^2 – a(\alpha + \beta)x + a\alpha\beta = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get:
- Sum of roots: \( \alpha + \beta = -\dfrac{b}{a} \)
- Product of roots: \( \alpha\beta = \dfrac{c}{a} \)
These are called the relations between roots and coefficients.
2. Formation of Quadratic Equation from Given Roots
If \( \alpha \) and \( \beta \) are given roots, then the quadratic equation with these roots is:
\( x^2 – (\alpha + \beta)x + \alpha\beta = 0 \)
3. Special Cases
- (i) If the roots are real and distinct, the equation remains the same form.
- (ii) If the roots are equal (\( \alpha = \beta \)), the equation is \( (x – \alpha)^2 = 0 \).
- (iii) If the roots are complex conjugates \( p + iq \) and \( p – iq \), the equation is:
\( (x – (p + iq))(x – (p – iq)) = x^2 – 2px + (p^2 + q^2) = 0 \)
Example
If the roots of a quadratic equation are \( 2 \) and \( 3 \), find the equation.
▶️ Answer / Explanation
Step 1: Sum of roots = \( 2 + 3 = 5 \)
Step 2: Product of roots = \( 2 \times 3 = 6 \)
Step 3: Required equation: \( x^2 – (\alpha + \beta)x + \alpha\beta = 0 \)
\( \Rightarrow x^2 – 5x + 6 = 0 \)
Answer: \( x^2 – 5x + 6 = 0 \)
Example
The sum and product of the roots of a quadratic equation are \( \dfrac{1}{2} \) and \( \dfrac{1}{8} \) respectively. Find the equation.
▶️ Answer / Explanation
Step 1: \( \alpha + \beta = \dfrac{1}{2}, \; \alpha\beta = \dfrac{1}{8} \)
Step 2: Equation: \( x^2 – (\alpha + \beta)x + \alpha\beta = 0 \)
\( \Rightarrow x^2 – \dfrac{1}{2}x + \dfrac{1}{8} = 0 \)
Step 3: Multiply through by 8 to remove fractions: \( 8x^2 – 4x + 1 = 0 \)
Answer: \( 8x^2 – 4x + 1 = 0 \)
Example
Form the quadratic equation whose roots are \( 2 + 3i \) and \( 2 – 3i \).
▶️ Answer / Explanation
Step 1: Since the roots are complex conjugates:
\( \alpha = 2 + 3i, \quad \beta = 2 – 3i \)
Step 2: Sum = \( \alpha + \beta = 4 \)
Product = \( \alpha\beta = (2 + 3i)(2 – 3i) = 4 + 9 = 13 \)
Step 3: Equation: \( x^2 – (\alpha + \beta)x + \alpha\beta = 0 \)
\( \Rightarrow x^2 – 4x + 13 = 0 \)
Answer: \( x^2 – 4x + 13 = 0 \)