A Nice Rational Equation Solved with Substitution | Math Olympiad 

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

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.

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.

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

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

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

{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).

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? 😅

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

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