Cracking the Challenge of a Reciprocal Equation

Cracking the Challenge of a Reciprocal Equation

🔍 Ready to crack the code? Join us in "Cracking the Challenge of a Reciprocal Equation" and unlock the secrets of algebraic mastery! 🧠🔢 In this immersive tutorial, we dive deep into the complexities of reciprocal equations, providing step-by-step solutions and insights. Whether you're a math enthusiast or gearing up for an algebraic challenge, this video is your key to conquering the uncrackable. Let's break through together! 🚀🎓 #ReciprocalEquation #AlgebraChallenge #ProblemSolving #CrackTheCode #MathMastery #EquationSolutions #LearnMath #AlgebraicThinking #Mathematics #EducationalVideo #SolveLikeAPro #MathEnthusiast #ChallengeAccepted #AlgebraicProwess #MathJourney #CrackTheChallenge

Topics covered:
Algebra Challenge
Reciprocal equation
How to simplify Expressions?
Math Olympiad
Algebra
Math Tricks
Algebraic identities
Algebraic manipulations
Substitutions
Real Solutions
Exponents
Quadratic equations
Algebraic Challenging Equations and Expressions
Expression
Math Olympiad Preparation

Time-stamps:
0:00 Introduction
0:52 Algebraic identities
1:58 Substitution
4:00 Algebraic manipulations
6:02 Quadratic equation
7:26 Solutions
8:33 Evaluating Expression

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Thanks for Watching !!
@infyGyan ​

[Skyler]

I have a question, if anyone faced a problem like this in the real world, why not just solve for x directly? I tried, and the equation produces ugly square roots and high exponents, but a computer can solve them numerically. The only benefit is we can prove that this particular expression is sqrt(10) but if we got the same answer via some complicated polynomial root, what difference does it make? Especially if we had no reason to suspect the expression has a simpler form?

[sunil.shegaonkar]

at 4:07, can you suggest any method to break 488 into 512-24?

[walterwen]

Cracking the Challenge of a Reciprocal Equation: x⁶ + 1/x⁶ = 488; x + 1/x = ?
x⁶ + 1/x⁶ = 488; x ≠ 0
Let: y = x + 1/x ≠ 0, y³ = (x + 1/x)³ = x³ + 1/x³ + 3(x + 1/x) = x³ + 1/x³ + 3y
x³ + 1/x³ = y³ – 3y, (y³ – 3y)² = (x³ + 1/x³)² = x⁶ + 1/x⁶ + 2 = 488 + 2 = 490
y³ – 3y = ± √490 = ± 7√10, y³ – 3y –/+ 7√10 = 0
y³ –/+ 10√10 – 3y +/– 3√10 = y³ –/+ (√10)³ – 3(y –/+ √10) = 0
(y –/+ √10)(y² ± y√10 + 10 – 3) = (y –/+ √10)(y² ± y√10 + 7) = 0; y² ± y√10 + 7 ≠ 0
y –/+ √10 = 0; y = x + 1/x = ± √10
Final answer:
x + 1/x = √10 or x + 1/x = – √10