Conquer a Difficult Rational Equation | Olympiad Prep

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Conquer a Difficult Rational Equation | Olympiad Prep
Ready to take on a tough math challenge? In this video, we tackle a difficult rational equation, perfect for those preparing for Math Olympiads or anyone looking to sharpen their algebra skills. Follow along as we break down the problem step-by-step, offering tips and strategies to help you conquer similar equations on your own.
Don't forget to like, share, and subscribe for more challenging math problems and Olympiad prep content. Share your solutions and questions in the comments below!
📚 Key Highlights:
Step-by-step breakdown of the solution process
Insights into the unique aspects of the rational equation
Tips and tricks for mastering equation through manipulations
Application of algebraic identities and quadratic formula
#mathchallenge #rationalequations #matholympiad #olympiadpreparation #algebra #mathproblems #brainteasers #mathskills #advancedmaths #mathematics #education
Time-stamps:
0:00 Introduction
2:32 Solving Quintic equation
4:00 Synthetic division
5:20 Quartic equation
8:44 Cubic equation
11:14 Quadratic equation
12:08 Solutions
12:45 Verification
🎓 Suitable for students, math enthusiasts, and anyone looking to explore the wonders of rational equations. Let's unravel the mystery together! 💡✨
We'd love to hear from you! Did you manage to solve the equation? What other math problems would you like us to cover? Let us know in the comments below!
🎓 Happy learning, and see you in the next video! 🎉
Thanks for Watching !

[@mohammedsaysrashid3587]

Thanks for sharing...it was a wonderful explanation....x=1/2:-1:+i:-i

[@golddddus]

No need to check 1 and -1 all the time. When they are rejected for the first time, it is impossible for them to appear as solutions. Plus, synthetic division is great for calculations, too. If the last term is not 0, that part of the solution is deleted, and the leading coefficients are kept.😎

[@shmuelzehavi4940]

After reducing the equation to: 4x^4 - 4x^3 + 5x^2 - 4x +1 = 0 , we may write:
4x^4 - 4x^3 + 5x^2 - 4x +1 = 4x^4 - 4x^3 + x^2 + 4x^2 - 4x +1 = (x^2 + 1)(2x - 1)^2 = 0.
The rest is trivial, giving us the aditional roots (includind the complex ones: ±i).

[@tejpalsingh366]

x= 1/2; -1; + - i

[@woobjun2582]

From the given, x ≠ 0, -1/2.
Expanding into
4x^5 +2x^3 +x^2 -3x +1 =0.
By RRT and SDMs,
(x+1)•(x+1/2)•(x-1/2)
•(4x^2 +4) = 0,
that is,
x = -1,
x = -1/2 (= rejected)
x = 1/2
x = ±i (cmplx, rejected)

[@shmuelzehavi4940]

-1/2 is not a root, and the original equation may be expanded into 4x^5 + x^3 + x^2 - 3x +1 = 0, and not 4x^5 + 2x^3 + x^2 - 3x + 1= 0.

[@offrampc]

x = -1, 1/2

[@user-nd7th3hy4l]

X=-1