Math Olympiad Secrets: Excelling in Rational Equations

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Math Olympiad Secrets: Excelling in Rational Equations
Welcome to "Math Olympiad Secrets: Excelling in Rational Equations!" In this video, we delve into the essential strategies and techniques to master rational equations, a critical topic for Math Olympiad competitions. Whether you're a beginner or looking to polish your skills, this comprehensive guide will help you solve even the trickiest rational equations with confidence.
In this tutorial, you'll learn:
1- Fundamental concepts and definitions of rational equations
2- Step-by-step methods to solve various types of rational equations
3- Common pitfalls and how to avoid them
4- Expert tips and tricks for solving problems quickly and accurately
5- Practice problems with detailed solutions
Timestamps:
0:00 Introduction
0:50 Substitution
2:12 Pascal's triangle
3:01 Binomial expansion
9:18 Cubic equation
10:56 Quadratic formula
12:06 Solution
12:15 Verification
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Join us as we unlock the secrets to excelling in rational equations and take your Math Olympiad prep to the next level. Don't forget to like, subscribe, and hit the bell icon for more Math Olympiad prep videos. Let's conquer those equations together!
Thanks for Watching!!
‪@infyGyan‬

[@mohammedsaysrashid3587]

Thanks for sharing....x=3

[@StaR-uw3dc]

Nice solution.
Some remark: numerator and denominator should be multiplied by 2 as we add a^6 to a^6 and subsequent similar terms. Of course this 2 will be cancelled but we should include it.

[@mathsneat]

Thanks for this video. I couldn't have solved this without you. I factored 63a^6 + 900a^4 + 720a^2 by by greatest common factor, 9u^2, before the substitution for a^2. Using the mean to start and the verification were helpful.

[@tunneloflight]

Sub x = y + 3. Converts then to ((y+2)^6 + (y-2)^6) / ((y+1)^6 + (y-1)^6) = 2^6. At y = 0, (2 * (2^6)) / (2 * 1^6) = 2^6. check. Then X = 0 + 3 = 3. As the shape of the relationship is maximum at y = 0, no other real roots exist.

[@user-kp2rd5qv8g]

Let x-3=t. The equation becomes (2^6-1)t^6 +(60)(15)t^4 +(48)(15)t^2 = 0. So, either t=0 > x=3 or 7t^4 + 100t^2 +80 = 0. This does not have positive solutions for t^2 (x cannot be real). So, x=3.

[@tejpalsingh366]

X= 3

[@user-nd7th3hy4l]

X=3