A Nice Rational Equation Solved with Substitution | Math Olympiad
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...
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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
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.
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.
X=1, -3, 1/2; -5/2.
X=1; -5/2; 1/2; -3 Ironically; both upper terms of l. H. S are the solns.
x=-3, x=-5/2, x=1/2 and x=1
Θετωy=2(x)^2+4x-4. και εχω y/(x+1)=+ -1 αρα x=-5/2, 1, -3, 1/2
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? 😅
{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).
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
[((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.
My method shows I have no talent