Advanced Algebraic Equations and Polynomials: Problem Solving Guide, Study Guides, Projects, Research of Mathematics

A series of advanced problems and solutions related to quadratic equations, polynomials, linear inequalities, algebraic identities, and simultaneous linear equations. It includes step-by-step solutions for each problem, making it a useful resource for students looking to enhance their problem-solving skills in algebra. The content covers topics such as solving quadratic equations using the quadratic formula, finding the sum of squares of roots, determining values for equal roots, applying the remainder theorem, factoring polynomials, solving linear inequalities, simplifying algebraic expressions, and solving simultaneous linear equations.

Typology: Study Guides, Projects, Research

2025/2026

Available from 12/30/2025

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TOPIC: Quadratic Equations – Advanced Problems
1. Solve the equation:
3x² – 7x – 6 = 0
Solution:
Using quadratic formula:
x = [7 ± √(49 + 72)] / 6
x = [7 ± √121] / 6
x = (7 + 11)/6 , (7 – 11)/6
x = 3 , x = –2/3
2. If α and β are roots of the equation:
2x² – 5x + 3 = 0
Find (α² + β²).
Solution:
α + β = 5/2
αβ = 3/2
α² + β² = (α + β)² – 2αβ
= (25/4) – 3
= 13/4
3. Find the value of k for which the equation:
x² + kx + 9 = 0
has equal roots.
Solution:
For equal roots:
b² – 4ac = 0
k² – 36 = 0
k = ±6
pf3
pf4
pf5

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TOPIC: Quadratic Equations – Advanced Problems

  1. Solve the equation: 3x² – 7x – 6 = 0 Solution: Using quadratic formula: x = [7 ± √(49 + 72)] / 6 x = [7 ± √121] / 6 x = (7 + 11)/6 , (7 – 11)/ x = 3 , x = –2/
  2. If α and β are roots of the equation: 2x² – 5x + 3 = 0 Find (α² + β²). Solution: α + β = 5/ αβ = 3/ α² + β² = (α + β)² – 2αβ = (25/4) – 3 = 13/
  3. Find the value of k for which the equation: x² + kx + 9 = 0 has equal roots. Solution: For equal roots: b² – 4ac = 0 k² – 36 = 0 k = ±

TOPIC: Polynomials – Advanced Applications

1. Find the remainder when:

f(x) = 2x³ – 5x² + 4x – 7

is divided by (x – 2).

Solution:

By Remainder Theorem:

f(2) = 2(8) – 5(4) + 8 – 7

2. Find the factor of the polynomial:

x³ – 4x² – x + 4

Solution:

Group terms:

(x³ – 4x²) – (x – 4)

x²(x – 4) –1(x – 4)

(x – 4)(x² – 1)

(x – 4)(x – 1)(x + 1)

3. Verify whether (x + 2) is a factor of:

x³ + 2x² – x – 2

Solution:

f(–2) = –8 + 8 + 2 – 2 = 0

Hence, (x + 2) is a factor.

TOPIC: Algebraic Identities – Advanced Level

1. Evaluate:

(x + y)³ – (x – y)³

Solution:

(x³ + 3x²y + 3xy² + y³)

– (x³ – 3x²y + 3xy² – y³)

= 6x²y + 2y³

= 2y(3x² + y²)

2. Simplify:

(a – b)² – (a + b)²

Solution:

(a² – 2ab + b²) – (a² + 2ab + b²)

= –4ab

3. Find value of:

x³ + y³ – 3xy(x + y)

if x + y = 5

Solution:

Identity:

x³ + y³ – 3xy(x + y) = (x + y)³

TOPIC: Simultaneous Linear Equations

1. Solve:

2x + 3y = 11

4x – y = 5

Solution:

Multiply second equation by 3:

12x – 3y = 15

Add with first:

14x = 26

x = 13/

Substitute in 4x – y = 5:

52/7 – y = 5

y = 17/

2. Solve:

x – y = 4

x² – y² = 32

Solution:

x² – y² = (x – y)(x + y)

4(x + y) = 32

x + y = 8

Solve:

x – y = 4

x + y = 8

x = 6 , y = 2