A Nice Rational Equation Solved with Substitution

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A Nice Rational Equation Solved with Substitution | Math Olympiad

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Күн бұрын

A Nice Rational Equation Solved with Substitution | Math Olympiad
Welcome to another Math Olympiad 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 substitution 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
Time-stamps:
00:00 Introduction
00:33 Substitution
06:50 Quadratic equation
07:18 Quadratic formula
10:42 Solutions
11:57 Verification
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!
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Пікірлер: 12

@woobjun2582 Ай бұрын

By letting (2x² +3x -5)/(x +1) =y the given becomes y² + (y +2)² = 4, and then y² + y² +4y +4 = 4; 2y² +4y +4 =4; 2y² +4y =0; y² +2y =0; y(y+2) =0, that is, y =0 or y =-2 Thus, recalling 'y' (2x² +3x -5)/(x +1) =0 (e1) or (2x² +3x -5)/(x +1) =-2 (e2) Solving (e1) 2x² +3x -5 =0; (x -1)(2x +5) =0; x = 1, -5/2 Solving (e2) 2x² +3x -5 =-2(x +1); 2x² +5x -3 = 0; (x +3)(2x -1) =0; x = -3, 1/2 Over all, x = -3, -5/2, 1/2, 1

@vacuumcarexpo Ай бұрын

I solved this as follows: Let t=2(x+1)-6/(x+1), then the given equation is (t-1)^2+(t+1)^2=4⇔t=±1. 2(x+1)-6/(x+1)=±1 ⇔2x^2+3x-5=0 or 2x^2+5x-3=0 It is intriguing that x's such that either of the numerators is zero are solutions.

@user-kp2rd5qv8g Ай бұрын

Note that [2x^2+5x-3]/(x+1) = [2x^2+3x-5]/(x+1 +2. So, let t= [2x^2+3x-5]/(x+1) +1 = [2x^2+4x-4]/(x+1). Then the given equation becomes (t+1)^2 + (t-1)^2 = 4 > t^2=1 > t = +/-1. If t=1, 2x*2+3x-5=0 > x= -5/2, 1. If t=-1, 2x^2+5x-3=0 > x=-3, 1/2. Thus, x=-3, -5/2, 1/2, 1.

@kassuskassus6263 Ай бұрын

x=-3, x=-5/2, x=1/2 and x=1

@user-kt1dm9jz5t Ай бұрын

X=1, -3, 1/2; -5/2.

@tejpalsingh366 Ай бұрын

X=1; -5/2; 1/2; -3 Ironically; both upper terms of l. H. S are the solns.

@MrGeorge1896 Ай бұрын

Multiply both sides by 4 (x + 1)²: (4x² + 6x - 10)² + (4x² + 10x - 6)² = 16 (x + 1)² (2x + 5)² (2x - 2)² + (2x - 6)² (2x + 1)² = 16 (x + 1)² I got stuck here so I just solved the four brackets: 2x + 5 = 0 -> x = -5/2 2x - 2 = 0 -> x = 1 2x - 6 = 0 -> x = 3 2x + 1 = 0 -> x = -1/2 and all four solutions work. But I am not quiet sure why? 😅

@RealQinnMalloryu4 Ай бұрын

{8x^2+9x^2➖} (5)^2)=| 17x^4 ➖ 25}》= 8x^4/2x^2 =4x 2^2x^2 1^1x^2 1x^2 (x ➖ 2x+1) {8x^2+10x^2} (3)^2 ={18x^4 ➖ 9} =9x^4 /2x^2 =4 1x^2 2^2.1^1x^2^1.1^1x^2^1 x^2^1 (x ➖ 2x+1).

@user-ny6jf9is3t Ай бұрын

Θετωy=2(x)^2+4x-4. και εχω y/(x+1)=+ -1 αρα x=-5/2, 1, -3, 1/2

@walterwen2975 28 күн бұрын

A Nice Rational Equation Solved with Substitution, Math Olympiad: [(2x² + 3x - 5)/(x + 1)]² + [(2x² + 5x - 3)/(x + 1)]² = 4, x ϵR, x ≠ - 1; x = ? Let: y = (2x² + 4x - 4)/(x + 1) (2x² + 3x - 5)/(x + 1) = y - 1, (2x² + 5x - 3)/(x + 1) = y + 1 [(2x² + 3x - 5)/(x + 1)]² + [(2x² + 5x - 3)/(x + 1)]² = (y - 1)² + (y + 1)² = 4 2(y² + 1) = 4, y² + 1 = 2, y² =1; y = ± 1 = (2x² + 4x - 4)/(x + 1) (2x² + 4x - 4)/(x + 1) = 1 or (2x² + 4x - 4)/(x + 1) = - 1 2x² + 4x - 4 = x + 1, 2x² + 3x - 5 = 0, (x - 1)(2x + 5) = 0 x - 1 = 0; x = 1 or 2x + 5 = 0; x = - 5/2 2x² + 4x - 4 = - (x + 1), 2x² + 5x - 3 = 0, (x + 3)(2x - 1) = 0 x + 3 = 0; x = - 3 or 2x - 1 = 0; x = 1/2 x =1, x = 1/2, x = - 5/2, x = - 3 Answer check: [(2x² + 3x - 5)/(x + 1)]² + [(2x² + 5x - 3)/(x + 1)]² = 4 x = 1: [(2 + 3 - 5)/(1 + 1)]² + [(2 + 5 - 3)/(1 + 1)]² = 0 + 2² = 4; Confirmed x = 1/2: [(1/2 + 3/2 - 5)/(1/2 + 1)]² + [(1/2 + 5/2 - 3)/(1/2 + 1)]² = (- 2)² + 0 = 4; Confirmed x = - 5/2: [(25/2 - 15/2 - 5)/(- 5/2 + 1)]² + [(25/2 - 25/2 - 3)/(- 5/2 + 1)]² = (- 2)² + 0 = 4; Confirmed x = - 3: [(18 - 9 - 5)/(- 3 + 1)]² + [(18 - 15 - 3)/(- 3 + 1)]² = 4; Confirmed Final answer: x = 1, x = 1/2, x = - 5/2 or x = - 3

@52soccerstar Ай бұрын

My method shows I have no talent

@paulortega5317 Ай бұрын

[((2x^2+4x-4) - (x+1)) / (x+1)]^2 + [((2x^2+4x-4) + (x+1)) / (x+1)]^2 = 4. Let u = (2x^2+4x-4)/(x+1)]^2. (u/(x+1) - 1)^2 + (u/(x+1)+1)^2=4. Let v = u/(x+1). (v-1)^2 + (v+1)^2 = 4. 2*v^2+2=4. v= +/- 1. Etc.