Higher degree quadratic equation jee questions Free JEE Math Tutorial

Higher Degree Equations

Key Definitions

  1. Polynomial Equation: An equation of the form P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀ = 0, where n ∈ ℕ, aₙ ≠ 0, and coefficients are constants.
  2. Degree of Polynomial: The highest power of x (here n) in the polynomial.
  3. Multiplicity of a Root: If (x-α)ᵏ is a factor of P(x) but (x-α)ᵏ⁺¹ is not, then α is a root of multiplicity k.
  4. Monical Polynomial: A polynomial where the leading coefficient aₙ is equal to 1.

Theory and Concepts

1. Remainder and Factor Theorems

  • Remainder Theorem: If a polynomial P(x) is divided by (x-h), the remainder is P(h).
  • Factor Theorem: A polynomial P(x) has a factor (x-h) if and only if P(h) = 0.

2. Relations between Roots and Coefficients (Vieta’s Formulas)

For an nᵗʰ degree equation aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₀ = 0 with roots α₁, α₂, …, αₙ:

  • Sum of roots (one at a time): Σαᵢ = -aₙ₋₁ / aₙ
  • Sum of roots (two at a time): Σαᵢαⱼ = aₙ₋₂ / aₙ
  • Product of roots: α₁α₂…αₙ = (-1)ⁿ (a₀ / aₙ)

3. Formation of Higher Degree Equations

If roots α₁, α₂, …, αₙ are known, the equation is:

(x – α₁)(x – α₂)…(x – αₙ) = 0
xⁿ – (Σαᵢ)xⁿ⁻¹ + (Σαᵢαⱼ)xⁿ⁻² – … + (-1)ⁿ(Product) = 0

4. Geometrical Representation of a Cubic

A cubic polynomial y = ax³ + bx² + cx + d can have one or three real roots depending on its intersections with the x-axis.

Figure 8.1: Cubic function with 3 distinct real roots

Solved Examples

Solved Example 1.1

Problem: Find the remainder when x¹⁰⁰ – 2x⁹⁹ + 3 is divided by (x-2).

Solution

Let P(x) = x¹⁰⁰ – 2x⁹⁹ + 3. By Remainder Theorem, R = P(2).
R = 2¹⁰⁰ – 2(2⁹⁹) + 3 = 2¹⁰⁰ – 2¹⁰⁰ + 3 = 3.

Solved Example 1.2

Problem: If α, β, γ are the roots of x³ – px² + qx – r = 0, find the value of Σα².

Solution

From Vieta’s: α+β+γ = p and αβ+βγ+γα = q.
Using (α+β+γ)² = α²+β²+γ² + 2(αβ+βγ+γα):
p² = Σα² + 2q ⇒ Σα² = p² – 2q.

Solved Example 1.3

Problem: Form a cubic equation whose roots are 1, 2, and 3.

Solution

Equation: (x-1)(x-2)(x-3) = 0
Expanding: (x²-3x+2)(x-3) = x³ – 3x² – 3x² + 9x + 2x – 6 = 0
Result: x³ – 6x² + 11x – 6 = 0.

Solved Example 1.4

Problem: If the roots of x³ – 12x² + 39x – 28 = 0 are in A.P., find the roots.

Solution

Let roots be a-d, a, a+d.
Sum: 3a = 12 ⇒ a = 4.
Product: (a-d)(a)(a+d) = 28 ⇒ 4(16 – d²) = 28
16 – d² = 7 ⇒ d² = 9 ⇒ d = ±3.
Roots are {1, 4, 7}.

Solved Example 1.5

Problem: If x-1 is a factor of P(x) = aₙxⁿ + … + a₀, show the sum of coefficients is zero.

Solution

By factor theorem, P(1) = 0.
P(1) = aₙ(1)ⁿ + aₙ₋₁(1)ⁿ⁻¹ + … + a₀ = aₙ + aₙ₋₁ + … + a₀.
Thus, the sum of all coefficients must be 0.

Solved Example 1.6

Problem: Find the sum of the cubes of the roots of x³ – px² + qx – r = 0.

Solution

Using Σα³ – pΣα² + qΣα – 3r = 0:
Σα³ = p(p² – 2q) – q(p) + 3r = p³ – 3pq + 3r.

Solved Example 1.7

Problem: Two roots of x⁴ – 2x³ – 4x² + 10x – 5 = 0 are √5 and -√5. Find others.

Solution

Product of known factors: (x – √5)(x + √5) = (x² – 5).
Dividing polynomial by (x² – 5) gives (x² – 2x + 1).
(x-1)² = 0 ⇒ x = 1, 1.
Other roots are 1, 1.

Solved Example 1.8

Problem: Solve x³ – 7x² + 14x – 8 = 0 given that roots are in G.P.

Solution

Let roots be a/r, a, ar. Product: a³ = 8 ⇒ a = 2.
Sum: 2/r + 2 + 2r = 7 ⇒ 2/r + 2r = 5
2r² – 5r + 2 = 0 ⇒ (2r-1)(r-2) = 0 ⇒ r = 2 or 1/2.
Roots are 1, 2, 4.

Solved Example 1.9

Problem: If roots of x³ + qx + r = 0 are α, β, γ, find roots (β+γ), (γ+α), (α+β).

Solution

Since Σα = 0, then β+γ = -α, γ+α = -β, α+β = -γ.
Transform x → -x in the original equation:
(-x)³ + q(-x) + r = 0 ⇒ -x³ – qx + r = 0 ⇒ x³ + qx – r = 0.

Solved Example 1.10

Problem: Find a, b if x⁴ + ax³ + bx² – 8x + 1 is a perfect square (x² + kx + 1)².

Solution

(x² + kx + 1)² = x⁴ + 2kx³ + (k²+2)x² + 2kx + 1.
Comparing x coefficients: 2k = -8 ⇒ k = -4.
a = 2k = -8; b = k²+2 = (-4)² + 2 = 18.

Concept Application Exercise 1.1

  1. Find the remainder when x²⁰²⁶ + x²⁰²⁵ + 1 is divided by x+1.

  2. If α, β, γ are roots of x³ – 6x² + 11x – 6 = 0, find Σ(1/α).

  3. If a+b+c=0, prove that a³+b³+c³ = 3abc using factor theorem.

  4. Find the condition that the roots of x³ + px² + qx + r = 0 are in A.P.

  5. If one root of x³ – 7x + 6 = 0 is 1, find the other roots.

  6. Find the sum of roots of the equation x⁴ – 5x³ + 5x² + 5x – 6 = 0.

  7. Solve x³ – 3x² + 4 = 0 if it has a repeated root.

  8. Find the number of real roots of x⁴ + 2x² + 1 = 0.

  9. If α, β, γ are roots of x³ – px² + qx – r = 0, find Σα²β.

  10. Form a fourth degree equation with rational coefficients having one root 1+√2+i.