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A Nice Radical Equation | Math Olympiad Prep!
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Ай бұрынA Nice Radical Equation | Math Olympiad Prep!
Welcome to our Math Olympiad Prep series! In this video, we tackle a fascinating radical equation that's perfect for honing your problem-solving skills. Whether you're preparing for a Math Olympiad or just love challenging algebra problems, this video is for you. We'll walk you through each step of the solution, providing clear explanations and tips to help you master radical equations.
🔢 What You'll Learn:
Techniques for solving radical equations
Tips for simplifying complex expressions
Step-by-step solutions and explanations
#matholympiad #radicalequations #mathchallenge #algebra #mathcompetition #learnmaths #mathtutorial #problemsolving #advancedmaths #mathskills #matholympiadpreparation
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Let's see if you can solve this amazing radical equation. Good luck!
Additional Resources:
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• An Amazing Radical Cha...
• Nice Algebra Challenge...
• A Fascinating Radical ...
• Math Olympiad Strategi...
Thanks for Watching !
@user-kp2rd5qv8g Ай бұрын
Let y=x^2+1. The given equation then reads y-x = sqrt(4/3xy) > x^2+y^2-10/3xy=0 > y = x/3, 3x. The first does not yield any real x as solutions. If y=3x, x^2-3x+1=0 > x = 1/2[3+/-sqrt(5)], both of which are valid solutions.
@richardleveson6467 Ай бұрын
Quartic nicely avoided! Clever move - thanks!
@RajeshKumar-wu7ox Ай бұрын
(3+√5)/2,(3-√5)/2, (1+√35 i)/2,(1-√35 i)/2
@kassuskassus6263 Ай бұрын
Two real solutions x=(3+or-sqrt5)/2 and two complex ones x=(1+or-isqrt35)/6. The other real solution (double) x=-1 is rejected.
@tejpalsingh366 Ай бұрын
X= (3+ -√5)/2 only real solns. Rest is complex.
@ZhilinChen-my7tp Ай бұрын
I have got the correct answer, happy.
@RealQinnMalloryu4 Ай бұрын
3x^9/3x^9 1x^1 (x ➖ 1x+1) 4x^3+4x^3 ➖4x+4x ➖ = {8x^6+8x^2}=18x^8 18x^8/3=6x2^6.2 3^2x2^3^2^2.2 1^1x1^3^1^.2 x^3^.2 (x ➖ 3x+2)
@user-ny6jf9is3t Ай бұрын
χ=(3+ -ριζα5)/2 στο R
@user-ee7nw2rx9s Ай бұрын
1.Почему без проверки может ли х=0, делится на х. А если х=0, является корнем исходного уравнения? 2. Почему в одних примерах находятся комплексные корни, а в других нет. Какой критерий? Разумеется если это не определяется условиями примеров.
@moeberry8226 Ай бұрын
There is no criteria for him to ask for real solutions or non real solutions, that is at his discretion. The reason you can divide by x because he’s saying we know obviously x=0 is not a solution because you can put it into the original equation and you will not get a true statement. However even if you didn’t know if x=0 was a solution you can still divide by x stating x cannot be 0 and then handle that case separately at the end.
@SidneiMV Ай бұрын
3(x² - x + 1)² = 4(x³ + x) = 4x(x² + 1) x² + 1 = u 3(u - x)² = 4ux 3(u + x)² = 16ux ux = (3/16)(u + x)² u + x = b ux = (3/16)b² t² - bt + (3/16)b² = 0 t = (b ± b/2)/2 t = 3b/4 or t = b/4 => u = 3x or u = x/3 x² + 1 = u x² + 1 = 3x x² - 3x + 1 = 0 *x = (3 ± √5)/2* x² + 1 = u x² + 1 = x/3 x² - x/3 + 1 = 0 x = [1/3 ± √(1/9 - 4)]/2
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