Introduction to Quadratic Equations for JEE Mains and Advance – Free JEE Math Tutorial

Introduction to Quadratic Equations

Key Definitions

  1. Polynomial Expression: An expression of the form f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, where n is a non-negative integer and aᵢ are constants.
  2. Degree of an Equation: The highest power of the variable x in a polynomial equation f(x) = 0. For a quadratic equation, the degree is exactly 2.
  3. Roots (Zeros): A value α is a root of f(x) = 0 if f(α) = 0. Geometrically, roots are the x-intercepts of the graph y = f(x).
  4. Identity: An equation satisfied by every value of the variable. A quadratic ax² + bx + c = 0 is an identity if a = b = c = 0.

Theory and Concepts

1. The Concept of Equation vs. Expression
An expression is a mathematical phrase (e.g., x² + 5x + 6), whereas an equation is a statement of equality (e.g., x² + 5x + 6 = 0).

2. Quadratic Equation Standard Form
The standard form is ax² + bx + c = 0, where a, b, c are coefficients and a ≠ 0.

3. Fundamental Theorem of Algebra
Every polynomial equation of degree n has exactly n roots. Thus, a quadratic equation always has exactly two roots (real or imaginary).

4. Geometrical Interpretation
The graph of y = ax² + bx + c is a parabola. The roots are the points where this parabola crosses or touches the x-axis.

Figure 1.1: Parabola f(x) intersecting x-axis at roots α, β

Solved Examples

Solved Example 1.1

If the equation (k²-1)x² + (k-1)x + (k²-4k+3) = 0 is an identity in x, find the value of k.

Solution: For a quadratic to be an identity, all coefficients must be zero:

  • k² – 1 = 0 ⇒ k = ±1
  • k – 1 = 0 ⇒ k = 1
  • k² – 4k + 3 = 0 ⇒ (k-1)(k-3) = 0 ⇒ k = 1, 3

The common value satisfying all three conditions is k = 1.

Solved Example 1.2

Find the degree of the equation: √(x+1) – √(x-1) = 1.

Solution: Rearranging: √(x+1) = 1 + √(x-1).
Squaring both sides: x + 1 = 1 + (x – 1) + 2√(x-1).
Simplifying: 1 = 2√(x-1) ⇒ 1/4 = x – 1 ⇒ x = 5/4.
The equation leads to a linear form. The degree is 1.

Solved Example 1.3

Determine the roots of x² – 5x + 6 = 0 using the factorization method.

Solution: Split the middle term: x² – 3x – 2x + 6 = 0.
x(x – 3) – 2(x – 3) = 0 ⇒ (x – 2)(x – 3) = 0.
Roots are x = 2 and x = 3.

Solved Example 1.4

Show that x = 1 is a root of (b-c)x² + (c-a)x + (a-b) = 0.

Solution: Substitute x = 1:
f(1) = (b – c)(1)² + (c – a)(1) + (a – b) = b – c + c – a + a – b = 0.
Since f(1) = 0, x = 1 is a root.

Solved Example 1.5

If f(x) is a quadratic polynomial such that f(1) = f(-1), what can we conclude about the coefficient ‘b’?

Solution: Let f(x) = ax² + bx + c.
f(1) = a + b + c and f(-1) = a – b + c.
Given a + b + c = a – b + c ⇒ 2b = 0 ⇒ b = 0.

Solved Example 1.6

How many real roots does the equation x² + |x| – 6 = 0 have?

Solution: Let |x| = t (where t ≥ 0). Then t² + t – 6 = 0.
(t + 3)(t – 2) = 0 ⇒ t = -3 (rejected) or t = 2.
|x| = 2 ⇒ x = ±2. There are 2 real roots.

Solved Example 1.7

Find the value of m if x² – mx + 1 = 0 has two equal roots.

Solution: For equal roots, D = 0.
D = (-m)² – 4(1)(1) = 0 ⇒ m² = 4 ⇒ m = ±2.

Solved Example 1.8

Represent the roots of x² – 4 = 0 on the complex plane.

Solution: x = ±2. These are real numbers. On the Argand plane, they lie on the Real axis at (2, 0) and (-2, 0).

Solved Example 1.9

Identify the quadratic equation whose roots are reciprocal to the roots of ax² + bx + c = 0.

Solution: Replace x with 1/x:
a(1/x)² + b(1/x) + c = 0 ⇒ a/x² + b/x + c = 0.
Multiply by x²: cx² + bx + a = 0.

Solved Example 1.10

Solve for x: x⁴ – 5x² + 4 = 0.

Solution: Let y = x². Then y² – 5y + 4 = 0 ⇒ (y-1)(y-4) = 0.
y = 1 ⇒ x = ±1.
y = 4 ⇒ x = ±2.
Solution set: { -2, -1, 1, 2 }.

Concept Application Exercise 1.1

  1. Find the value of a if (a²-3a+2)x² + (a²-5a+6)x + (a²-4) = 0 is an identity in x.
  2. If x² + px + q = 0 has roots α and β, find the equation whose roots are -α and -β.
  3. Solve for x ∈ ℝ: |x² – 3x + 2| = 0.
  4. Find the degree of the polynomial P(x) = (x-1)(x-2)(x-3) – x³.
  5. If f(x) is a quadratic and f(0)=3, f(1)=2, f(2)=3, find the expression for f(x).
  6. Determine the number of real roots of x2/3 + x1/3 – 2 = 0.
  7. Discuss the degree and number of roots for (x+1)² = x² + 2x + 1.
  8. Find the sum of all real roots of the equation |x-2|² + |x-2| – 2 = 0.
  9. If α is a root of ax² + bx + c = 0, prove that kα is a root of ax² + bkx + ck² = 0.
  10. Solve for x: (x-1)/(x-2) – (x-2)/(x-3) = (x-5)/(x-6) – (x-6)/(x-7).