An Amazing Rational Equation | Give It a Try! | Algebra

Подписаться 6 тыс.

Просмотров 1,4 тыс.

50% 1

An Amazing Rational Equation | Give It a Try! | Algebra
Welcome to another algebraic equation challenge! In this video, we tackle a nice rational equation using the powerful method of substitution. This problem is perfect for sharpening your algebra skills and preparing for competitive math exams.
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.
Don't forget to like, comment, and subscribe for more math challenges and solutions. Happy solving!
In this tutorial, you'll learn:
1- Fundamental concepts and definitions of rational equations
2- Step-by-step method of substitutions 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 problems with detailed solutions
Additional Resources:
• Math Olympiad Secrets:...
• Ready for a Math Chall...
• Conquer a Difficult Ra...
• Overcoming Rational Eq...
#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!
Thanks for Watching!!

Опубликовано:

10 окт 2024

Ссылка:

Скачать:

Готовим ссылку...

Добавить в:

Мой плейлист

Посмотреть позже

Комментарии : 7

@RashmiRay-c1y 4 дня назад

Writing x+1/x + 1 =a, we get (a-4)^2 + (a+4)^2 = 36 > a^2 =2 > a=+/-√2 > x^2-(√2-1)x+1=0 or x^2+(√2+1)x+1=0. The first does not have real roots. So, the second equation yields x= 1/2[-(√2+1) +/- √(2√2-1)].

@gregevgeni1864 4 дня назад

Set x^2 + x + 1 = t (*). The original equation is equivalent (( t - 4 x)/2 x)^2 + ((t + 4 x)/2 x)^2 = 9 ... (2 t^2 + 32 x^2)/(4 x^2) = 9 (t^2)/(2 x^2) + 8 = 9 t^2 = 2 x^2 (x^2 + x + 1)^2 - (√2 x)^2 = 0 ( due to (*) ) [x^2 + (1+√2)x +1]•[x^2 + (1-√2)x +1] = 0 etc etc . .

@9허공 4 дня назад

(x^2 - 3x + 1)^2 + (x^2 + 5x + 1)^2 - 36x^2 = 0 let t = x^2 + x + 1 then (t - 4x)^2 + (t + 4x)^2 - 36x^2 = 2t^2 + 4x^2 - 36x^2 = 2(t^2 - 2x^2) = 2(t + √2x)(t - √2x) = 0 x^2 + x + 1 = √2x or x^2 + x + 1 = -√2x ...

@Quest3669 4 дня назад

Le t 1st term in bracket is t then eqn becomes t^2+(t+4)^2= 9 gives t= 1/2(-4+-√2) X= 1/2[( √2+1) + - √-1+2√2] real solns.

@RealQinnMalloryu4 4 дня назад

({x^4 ➖ 9x^2}+2/4x^2)+({x^4+10x^2}+2/4x^2)=({9x^2+2}/4x^2)+(10x^6+2/4x^2)={11x^2/4x^2+12x^6/4x^2}=23x^8/8x^4=2.7x^2 1.7^1x^2 1.1^1x^2 1x^2(x ➖ 2x+1).

@RajeshKumar-wu7ox 4 дня назад

No real roots

@key_board_x 4 дня назад

[(x² - 3x + 1)/2x]² + [(x² + 5x + 1)/2x]² = 3² [(x² + 1 - 3x)/2x]² + [(x² + 1 + 5x)/2x]² = 9 → let: a = x² + 1 [(a - 3x)/2x]² + [(a + 5x)/2x]² = 9 [(a/2x) - (3x/2x)]² + [(a/2x) + (5x/2x)]² = 9 [(a/2x) - (3/2)]² + [(a/2x) + (5/2)]² = 9 (1/4).[(a/x) - 3]² + (1/4).[(a/x) + 5]² = 9 [(a/x) - 3]² + [(a/x) + 5]² = 36 → let: b = a/x [b - 3]² + [b + 5]² = 36 b² - 6b + 9 + b² + 10b + 25 = 36 2b² + 4b - 2 = 0 b² + 2b - 1 = 0 b² + 2b = 1 b² + 2b + 1 = 2 (b + 1)² = 2 b + 1 = ± √2 b = - 1 ± √2 → recall: b = a/x a/x = - 1 ± √2 → recall: a = x² + 1 (x² + 1)/x = - 1 ± √2 x² + 1= (- 1 ± √2).x x² - (- 1 ± √2).x + 1 = 0 First case: x² - (- 1 + √2).x + 1 = 0 x² + (1 - √2).x + 1 = 0 Δ = (1 - √2)² - (4 * 1) = 1 - 2√2 + 2 - 4 = - 1 - 2√2 = (1 + 2√2).i² x = [- (1 - √2) ± i.√(1 + 2√2)]/2 → x = [(√2 - 1) ± i.√(2√2 + 1)]/2 Second case: x² - (- 1 - √2).x + 1 = 0 x² + (1 + √2).x + 1 = 0 Δ = (1 + √2)² - (4 * 1) = 1 + 2√2 + 2 - 4 = - 1 + 2√2 x = [- (1 + √2) ± √(- 1 + 2√2)]/2 → x = [- (1 + √2) ± √(2√2 - 1)]/2