Quadratic Equations: Fundamentals and Nature of Roots – Free JEE Math Tutorial

Quadratic Equations: Fundamentals and Nature of Roots

Key Definitions

1. Quadratic Expression: A polynomial of the form f(x) = ax² + bx + c, where a, b, c ∈ ℂ and a ≠ 0. If a, b, c ∈ ℝ, it is a real quadratic expression.
2. Quadratic Equation: The statement ax² + bx + c = 0 (a ≠ 0) is called a quadratic equation. The values of x that satisfy this equality are called the roots of the equation.
3. Identities: An equation of the form ax² + bx + c = 0 is an identity in x if it is satisfied by more than two distinct values of x. This occurs if and only if a = b = c = 0.
4. Discriminant (D): For the equation ax² + bx + c = 0, the quantity D = b² – 4ac is the discriminant, which characterizes the nature of the roots.

Theory and Concepts

1. Roots of the Quadratic Equation

The roots of the equation ax² + bx + c = 0 are given by the quadratic formula:

Let the roots be α and β.

2. Nature of Roots (For a, b, c ∈ ℝ)

• If D > 0: Roots are real and distinct.
• If D = 0: Roots are real and equal (coincident). The value of each root is -b/2a.
• If D < 0: Roots are imaginary and occur in conjugate pairs (i.e., p + iq and p - iq).

3. Special Cases for Rational Roots

If a, b, c are rational numbers:

• If D is a perfect square of a rational number, the roots are rational.
• If D is not a perfect square, the roots are irrational and occur in conjugate pairs (i.e., p + √q and p – √q).

4. Condition for Common Roots

Consider two equations a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0.

• Both roots common: a₁/a₂ = b₁/b₂ = c₁/c₂
• One root (α) common: By eliminating x² and the constant term, we get the condition:
  (a₁c₂ – a₂c₁)² = (a₁b₂ – a₂b₁)(b₁c₂ – b₂c₁)

Solved Examples

Solved Example 1.1

Find the value of k for which the equation (k-1)x² – (k+1)x + (k+1) = 0 has real and equal roots.

Solution

For real and equal roots, D = 0.

Solving 3k² – 2k – 5 = 0:

However, if k=1, it is not quadratic. If k=-1, the coefficients suggest a trivial case. Thus, for a quadratic with equal roots, k = 5/3.

Solved Example 1.2

If α, β are the roots of x² – px + q = 0, find the value of α⁴ + β⁴.

Solution

We know α + β = p and αβ = q.
α² + β² = (α + β)² – 2αβ = p² – 2q.
α⁴ + β⁴ = (α² + β²)² – 2α²β²

Substituting values:

Solved Example 1.3

Prove that the roots of (b-c)x² + (c-a)x + (a-b) = 0 are always rational for rational a, b, c.

Solution

Observe that the sum of coefficients (b-c) + (c-a) + (a-b) = 0.
This implies x=1 is always a root.
Since the product of roots αβ = (a-b)/(b-c), and α=1, the other root is (a-b)/(b-c).
Since a, b, c are rational, both roots {1, (a-b)/(b-c)} are rational.

Solved Example 1.4

If x² + ax + b = 0 and x² + bx + a = 0 (a ≠ b) have a common root, find the value of a+b.

Solution

Let α be the common root.
α² + aα + b = 0 and α² + bα + a = 0.
Subtracting the two equations: (a-b)α + (b-a) = 0 ⇒ (a-b)α = (a-b).
Since a ≠ b, α = 1.
Substitute α=1 in the first equation: 1 + a + b = 0 ⇒ a + b = -1.

Solved Example 1.5

Determine the nature of the roots of the equation 3x² – 10x + 3 = 0.

Solution

Since D = 8² (a positive perfect square) and coefficients are rational, roots are real, distinct, and rational.

Solved Example 1.6

For what values of m will x² – 2x(1+3m) + 7(3+2m) = 0 have equal roots?

Solution

Set D = 0:

Solved Example 1.7

If a+b+c=0, show that the roots of ax²+bx+c=0 are real.

Solution

D = b² – 4ac. Since b = -(a+c),
D = (a+c)² – 4ac = (a-c)².
Since (a-c)² ≥ 0, roots are real.

Solved Example 1.8

Solve for x: x² – 5|x| + 6 = 0.

Solution

For x ≥ 0, x² – 5x + 6 = 0 ⇒ x=2, 3.
For x < 0, x² + 5x + 6 = 0 ⇒ x=-2, -3.
Solutions: x ∈ {-3, -2, 2, 3}.

Solved Example 1.9

Find the condition that the roots of ax² + bx + c = 0 are in the ratio m:n.

Solution

Let roots be mα, nα. Sum = (m+n)α = -b/a; Product = mnα² = c/a.
Eliminating α:

Solved Example 1.10

If α is a root of 4x² + 2x – 1 = 0, prove the other root is 4α³ – 3α.

Solution

Sum α + β = -1/2. Given 4α² = 1 – 2α.
4α³ – 3α = α(4α² – 3) = α(1-2α-3) = -2α² – 2α.
Since -2α² = α – 1/2, the expression becomes (α – 1/2) – 2α = -1/2 – α = β.