Olympiad Algebra Challenge: Two Methods, One Ultimate Solution!

Olympiad Algebra Challenge: Two Methods, One Ultimate Solution!

In this transformative video, we delve deep into the world of algebra, unveiling not one but two powerful methods that will elevate your algebraic prowess in Olympiads to new heights! Whether you're a student striving for excellence or someone looking to refresh their algebraic skills, this comprehensive tutorial has you covered.

🔑 Key Highlights:
1️⃣ Method 1: Calculus Wizardry - Learn the conventional approach to solving algebraic equations using first derivative test and graphs.
2️⃣ Method 2: Algebraic Alchemy - Discover a unique, intuitive method that simplifies complex algebraic expressions and equations, making them a breeze to solve.

By the end of this video, you'll have a firm grip on both methods, empowering you to choose the one that suits your learning style best. Algebra will cease to be a challenge, and you'll be equipped to tackle any problem that comes your way!

📚 Topics Covered:

Olympiad Challenge
Calculus
Graphs
Algebra
Math Olympiad
Quadratic equations
Algebraic Radical Equation
Quadratic equations
Surds
Math Tutorial
Math Olympiad Preparation
Complex solution
Real solution
Quadratic formula

🌟 Why Watch?

Gain a comprehensive understanding of algebra and calculus.
Choose the method that resonates with you.
Boost your confidence in tackling algebraic challenges.

🔥 Ready to become an algebraic genius? Hit that play button and embark on your journey to algebra mastery!

13 Key moments of this video:
0:00 Introduction
0:30 Calculus
0:45 Derivatives
1:18 Domain
3:13 Check and error method
4:10 Graphs
5:22 Algebraic approach
5:30 Substitution
6:20 Algebraic manipulations
6:30 Algebraic identities
8:02 Quadratic equation
8:15 Quadratic formula
9:10 Solutions

#AlgebraMastery #AlgebraicWizardry #AlgebraicAlchemy #Mathematics #MathSkills #AlgebraTips #Education #Learning #AlgebraicEquations #MathTutorial #ProblemSolving #StudySmart #MathematicalMethods #MathGenius #StudentSuccess #AlgebraSimplified #UltimateSolution

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Thanks for watching!

@infyGyan

[moeberry]

X>or equal to 1 so when x is 1, we have 1^(1/3)=1. Which is true. And there will be no other solution since this function is strictly increasing and it’s equal to a constant.

[abbasmehdi]

We can also solve graphically .
Draw curve of 1-(x)^(1/3) and 2(x-1)^(1/2) . We'll notice that for x>1, it has only one trivial intersection.

[walterwen]

Olympiad Algebra Challenge: 2√(x – 1) + ³√x = 1; x = ?
2√(x – 1) + ³√x = 1; x ≥ 1, real value or Complex roots, if acceptable
Let: y = ³√x; x = y^3, y ≥ 1, 2√(x – 1) + ³√x – 1 = 2√(y^3 – 1) + y – 1 = 0
√[4(y – 1)(y^2 + y + 1)] + [√(y – 1)]^2 = 0, [√(y – 1)][√(4y^2 + 4y + 4) + √(y – 1)] = 0
√(y – 1) = 0, y – 1 = 0; y = 1 or √(4y^2 + 4y + 4) + √(y – 1) = 0; Complex roots
³√x = y = 1; x = 1 or √(4y^2 + 4y + 4) = – √(y – 1), 4y^2 + 4y + 4 = y – 1
4y^2 + 3y + 5 = 0, y = {– 3 ± √[9 – 4(4)(5)]}/[2(4)] = (– 3 ± i√71)/8
y^2 = [(– 3 ± i√71)/8]^2 = [9 – 71 –/+ i(6√71)]/64 = [– 31 –/+ i(3√71)]/32
x = y^3 = y(y^2) = [(– 3 ± i√71)/8]{[– 31 –/+ i(3√71)]/32}
= [93 – 3(71)(i^2) ± i(22√71)]/[8(32)] = [153 ± i(11√71)]/128
Answer check:
x = 1, 2√(x – 1) + ³√x = 2√(1 – 1) + ³√1 = 0 + 1 = 1; Confirmed
x = [153 ± i(11√71)]/128; y = ³√x, 4y^2 + 3y + 5 = 0, 4y^3 = – 3y^2 – 5y
2√(x – 1) + ³√x = 2√(y^3 – 1) + y = √(4y^3 – 4) + y = √(– 3y^2 – 5y – 4) + y
= √[(1 – 2y + y^2) – (4y^2 + 3y + 5)] + y = √[(1 – y)^2] + y = 1 – y + y = 1; Confirmed
Final answer:
x= 1, x = [153 + i(11√71)]/128 or x = [153 – i(11√71)]/12