An Amazing Algebra Challenge | Can You Crack This?

An Amazing Algebra Challenge | Can You Crack This?

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In this algebraic video a simple and effective technique for solving rational equation is explained in an easy-to-understand way, making this video an excellent resource for beginners looking to build a strong foundation in algebra. This video breaks down complex concepts into manageable steps, ensuring viewers can confidently tackle rational equations with newfound ease.

Watch as we break down the steps to solve this equation, providing clear explanations and tips along the way. Whether you're a math enthusiast or a student aiming to excel in Olympiad competitions, this video will help you understand and master rational equations.

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In this tutorial, you'll learn:

1- Fundamental concepts and definitions of rational equations
2- Step-by-step methods to solve rational equation
3- Common pitfalls and how to avoid them
4- Expert tips and tricks for solving problems quickly and accurately
5- Practice problem with detailed solution

Additional Resources:

#matholympiad #rationalequations #mathtutorial #substitution #math #mathskills #learnmath #education #algebra

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!

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[gregevgeni]

Happy New Year to everyone 🎉
The given equation is equivalent to
(1/x -2)² + (1/x -3)³ + (1/x -4)² = 2 (1).
Let t = 1/x -2 (*) . Then the (1)
written as t⁴ +(t-1)³ +(t-2)² = 2 <=>
t⁴+t³-2t²-t+1=0 <=>
t⁴+t³-t²-t²-t+1=0 <=>
t²(t²+t-1)-(t²+t-1)=0 <=>
(t²+t-1)(t²-1)=0 <=>
t=±1, t=(-1±√5)/2 (2)
Due to (*) and (2) =>
x=1, x=1/3, x=2/(3±√5)=(3±√5)/2

[wes]

Substitute x=1/(y+2) or 1/x=y+2 into the given equation:
By inspection y=±1. Divide by y^2–1: y^2+y–1, which has roots y=(–1±√5)/2
From x=1/(y+2) we have x=1, x=1/3, or x=2/(3±√5)=(3∓√5)/2

[RashmiRay-cy]

Let 1/x-3=a. The given equation then becomes (a+1)^4+a^3+(a-1)^2=2 or a[a^3+5a^2+7a+2]=0. If, a=0 or x=1/3. If a^3+5a^2+7a+2 = 0, a=-2, i.e., x=1 or a= 1/2[-3 +/- √5], which means x=[(√5 +/-1)/2]^2. So, x=1/3, 1, (√5 +/-1)/2]^2.

[허공]

Using RRT and SDM,
t^4 + t^3 - 3t^2 - t + 1 = (t + 1)(t - 1)(t^2 + t - 1) = 0
t = { -1, 1, (-1±√5)/2 } => x = { 1, 1/3, (3±√5)/2 }

[Fjfurufjdfjd]

Η εξισωση γραφεται;

Θετω (1/χ)-3=ψ οποτε η εξισωση γραφεται :
(ψ+1)^4+ψ^3+(ψ-1)^2=2 ...
ψ^4+5ψ^3+7ψ^2+2ψ=0
ψ(ψ^3+5ψ^2+7ψ+2)=0
ψ(ψ+2)(ψ^2+3ψ+1)=0
ψ=0 ή ψ=-2 ή ψ=[-3+ -(5)^(1/2)]/2 οποτε
χ=1/3 ή χ=1 ή χ=[3-(5)^(1/2)]/2 ή χ=[3+(5)^(1/2)]/2.

[RealQinnMalloryu]

(1 ➖ 16x^4/x^4)+(1 ➖ 27x^3/x^3)+(1 ➖ 16x^2/x^2) 7^8x^1 7^2^3x^1 3^4^1^1^1x^1 3^2^2x^1 3^1^2x^1 3^2x (x ➖ 3x+2).

[潘博宇-kl]

X=(1, 1/3, [3+(5)^(1/2)]/2, [3-(5)^(1/2)]/2.)