IIT JEE Main Maths -Unit 2- Nature of roots and formation of equations- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 2- Nature of roots and formation of equations – Study Notes – New syllabus
IIT JEE Main Maths -Unit 2- Nature of roots and formation of equations – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Nature of Roots and Formation of Equations
- Nature of Roots in terms of Graph (Position of Parabola with respect to x-axis)
1. Nature of Roots (Based on Discriminant)
For a quadratic equation \( ax^2 + bx + c = 0 \), the discriminant is given by:
\( D = b^2 – 4ac \)
The nature of roots depends on the value of \( D \):
| Discriminant (D) | Nature of Roots | Example |
|---|---|---|
| \( D > 0 \) | Two distinct real roots | \( x^2 – 5x + 6 = 0 \) |
| \( D = 0 \) | Two equal real roots | \( x^2 – 4x + 4 = 0 \) |
| \( D < 0 \) | Two non-real complex conjugate roots | \( x^2 – 2x + 5 = 0 \) |
2. Formula for Roots
The roots of \( ax^2 + bx + c = 0 \) are given by:
\( x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
3. Formation of Equation from Given Roots
If the roots of a quadratic equation are \( \alpha \) and \( \beta \), the equation is:
\( x^2 – (\alpha + \beta)x + \alpha\beta = 0 \)
If one root is given in terms of the other, the same relation can be used to find both roots and form the equation.
Example
Find the nature of the roots of \( 2x^2 – 3x + 1 = 0 \).
▶️ Answer / Explanation
Step 1: Identify coefficients: \( a = 2, b = -3, c = 1 \)
Step 2: Discriminant \( D = b^2 – 4ac = (-3)^2 – 4(2)(1) = 9 – 8 = 1 \)
Step 3: Since \( D > 0 \), the equation has two distinct real roots.
Step 4: Roots are \( x = \dfrac{3 \pm 1}{4} \Rightarrow x = 1, \dfrac{1}{2} \)
Conclusion: Roots are real and unequal.
Example
Find the quadratic equation whose roots are \( 3 \) and \( 4 \).
▶️ Answer / Explanation
Step 1: Sum of roots \( = 3 + 4 = 7 \)
Product of roots \( = 3 \times 4 = 12 \)
Step 2: Required equation: \( x^2 – (\alpha + \beta)x + \alpha\beta = 0 \)
\( \Rightarrow x^2 – 7x + 12 = 0 \)
Conclusion: The required equation is \( x^2 – 7x + 12 = 0 \).
Example
Find the quadratic equation whose roots are \( 2 + 3i \) and \( 2 – 3i \).
▶️ Answer / Explanation
Step 1: For complex conjugate roots:
\( \alpha = 2 + 3i, \; \beta = 2 – 3i \)
Step 2: Sum \( = \alpha + \beta = 4 \)
Product \( = \alpha\beta = (2 + 3i)(2 – 3i) = 4 + 9 = 13 \)
Step 3: Equation: \( x^2 – (\alpha + \beta)x + \alpha\beta = 0 \)
\( \Rightarrow x^2 – 4x + 13 = 0 \)
Step 4: Discriminant \( D = (-4)^2 – 4(1)(13) = 16 – 52 = -36 \lt 0 \)
Conclusion: The equation \( x^2 – 4x + 13 = 0 \) has complex conjugate roots \( 2 \pm 3i \).
Nature of Roots in terms of Graph (Position of Parabola with respect to x-axis)
1. Standard Form of a Quadratic Equation
The general quadratic equation is given by:
\( y = ax^2 + bx + c \), where \( a \ne 0 \)
The graph of this equation is a parabola.
- If \( a > 0 \), the parabola opens upward.
- If \( a < 0 \), the parabola opens downward.
2. Discriminant and Graphical Meaning
The nature of the roots of the quadratic equation \( ax^2 + bx + c = 0 \) depends on the value of the discriminant \( D = b^2 – 4ac \):
| Value of \( D \) | Nature of Roots | Graphical Representation |
|---|---|---|
| \( D > 0 \) | Two distinct real roots | Parabola cuts the x-axis at two distinct points. |
| \( D = 0 \) | Two equal real roots | Parabola touches the x-axis (vertex on the x-axis). |
| \( D < 0 \) | No real roots (complex roots) | Parabola does not intersect the x-axis. |
3. Vertex of the Parabola
The vertex of the parabola \( y = ax^2 + bx + c \) is given by:
\( \left( \dfrac{-b}{2a}, \dfrac{-D}{4a} \right) \)
This shows how the discriminant \( D \) affects the vertical position of the vertex relative to the x-axis.
Example
Determine the nature of roots and sketch position of the parabola for \( y = x^2 – 5x + 6 \).
▶️ Answer / Explanation
Step 1: \( a = 1, b = -5, c = 6 \)
Step 2: \( D = b^2 – 4ac = (-5)^2 – 4(1)(6) = 25 – 24 = 1 \)
Step 3: Since \( D > 0 \), two distinct real roots exist.
Step 4: Roots are \( x = 2 \) and \( x = 3 \)
Step 5: Parabola opens upward (since \( a > 0 \)) and cuts the x-axis at two points \( x=2, 3 \).
Example
For the quadratic \( y = 2x^2 + 4x + 2 \), determine the nature of the roots and describe its graph.
▶️ Answer / Explanation
Step 1: \( a = 2, b = 4, c = 2 \)
Step 2: \( D = b^2 – 4ac = 16 – 16 = 0 \)
Step 3: Since \( D = 0 \), roots are real and equal.
Step 4: The vertex lies on the x-axis at \( x = -\dfrac{b}{2a} = -1 \)
Step 5: The parabola touches the x-axis at \( x = -1 \).
Example
Examine the position of the parabola for \( y = 3x^2 + 2x + 5 \) with respect to the x-axis.
▶️ Answer / Explanation
Step 1: \( a = 3, b = 2, c = 5 \)
Step 2: \( D = b^2 – 4ac = 4 – 60 = -56 \lt 0 \)
Step 3: Since \( D < 0 \), there are no real roots.
Step 4: The parabola opens upward (as \( a > 0 \)) and lies entirely above the x-axis.
Step 5: Its vertex is at \( \left( -\dfrac{b}{2a}, \dfrac{-D}{4a} \right) = \left( -\dfrac{1}{3}, \dfrac{56}{12} \right) \)
Conclusion: The parabola does not intersect the x-axis; roots are complex.