A nice Math Olympiad Algebra Equation. Can you dare solve this ? Find the Value of x ?

Hello,
If you love❤️ this video, subscribe, give it a thumbs-up 👍, and turn on the notification bell 🔔 to get new updates when I upload fresh video.
In this Math Olympiad Algebra Problem, you'll learn tips and tricks of solving International Math Olympiad exams quickly. #IMO #matholympiad #algebra #radicalequations #simplify #exponential
A nice Math Olympiad Algebra Equation. Can you dare solve this ? Find the Value of x ?

[@shannonmcdonald7584]

Your videos help me to think differently about equations. Im learning a lot. Thank you so much.

[@superacademy247]

You are welcome.🤩🤩🤩

[@guyhoghton399]

Good solution, but in this case it seems easier just to subtract the terms, cross multiply, expand and simplify.
We have _(⁽ˣ⁺¹⁾/ₓ)² - (⁽ˣ⁺¹⁾/₍ₓ₋₁₎)² = 1_
⇒ _(x + 1)²[⅟ₓ² - ⅟₍ₓ₋₁₎²] = 1_
⇒ _(x + 1)²[(x - 1)² - x²]/[x²(x - 1)²] = 1_
⇒ _(x + 1)²(1 - 2x) = x²(x - 1)²_
⇒ _-2x³ - 3x² + 1 = x⁴ - 2x³ + x²_
⇒ _(x²)² + 4x² - 1 = 0_
⇒ _x² = -2 ± √5_
⇒ *_x = ±√(-2 ±√5)_*

[@superacademy247]

Awesome 👍💯

[@user-u3e3kx5m6pg]

Use ((x+1)/x)² =
=((x+1)/(x-1))²+1
=2(x²+1)/(x-1)²
Simplify it and get
(x² -1)² =2x²(x²+1) =>
X⁴+4x²-1=0 => x²=-2±5½. So x=±(-2+5½)½, ±(2+5½)½ i.

[@user-st4vk5lj7o]

Very good solution.many many thanks sir.❤

[@superacademy247]

Most welcome

[@anestismoutafidis4575]

If I put for x=-0,5, in the equation, then the solution of the equation is 1,11'. For this I try with the number -0,485 and put it for x. Then the solution of the equation is:
[(-0,485+1)^2/-0,485^2] - [(-0,485+1)^2/(-0,485-1)^2 =1,006 x=-0,485

[@NadiehFan]

The equation to solve is
((x + 1)/x)² − ((x + 1)/(x − 1))² = 1
Start by multiplying both sides by x²(x − 1)² to get rid of the fractions and we have
(x + 1)²(x − 1)² − (x + 1)²x² = x²(x − 1)²
(x + 1)²(x − 1)² = (x + 1)²x² + x²(x − 1)²
(x² − 1)² = x²((x + 1)² + (x − 1)²)
(x² − 1)² = x²(2x² + 2)
x⁴ − 2x² + 1 = 2x⁴ + 2x²
x⁴ + 4x² − 1 = 0
x² = −2 + √5 ⋁ x² = −2 − √5
x = √(−2 + √5) ⋁ x = −√(−2 + √5) ⋁ x = i√(2 + √5) ⋁ x = −i√(2 + √5)

[@superacademy247]

It's stunning how you break down this gigantic equation to simple quadratic of fourth degree(quartic equation). Thanks 👍💯 for your input

[@ArwindSah]

Easy way to solve this equation
Equation has solution for x not equal to 0 or 1
(X+1)^2/((1/X^2-1/(X-1)^2) =1
Or,
( (X+1)^2)((X-1)^2-x^2))/((X^2)(X-1)^2) =1
Or,
((X+1)^2)(X-1+X)(X-1-X) =(X^2)(X-1)^2
Or,
(X^2+2X+1)(2X-1)(-1) =(X^2)(X^2-2X+1)
Or,
(X^2+2X+1)(1-2X) =X^4-2X^3+X^2
Or,
X^2+2X+1-2X^3-4X^2-2X = X^4-2X^3+X^2
Simplifying we get
X^4+4X^2-1 =0
Put X^2=t
Then X^4=t^2
Equation t^2+4t-1=0
t=-2+√5 or t = -2-√5
Case 1
X^2=-2+√5
X1= √(-2+√5) or
X2 = -√(-2+√5)
Case 2
X^2= -2-√5 =(-1)(2+√5)
X3=i√(2+√5) or
X4= -i√(2+√5)

[@superacademy247]

Thanks 👍💯😊 for your support

[@اقرء]

Good

[@superacademy247]

Thanks

[@abdrrahmanaaboun4168]

Very good work

[@superacademy247]

Thank you so much 😀

[@walterwen2975]

A nice Math Olympiad Algebra Equation:
[(x + 1)/x]² - [(x + 1)/(x - 1)² = 1; x = ?
x ≠ 0, x ≠ 1; (x + 1)²[(x - 1)² - x²]/[(x²)(x - 1)²] = 1
[(x + 1)²(1 - 2x)]/[(x²)(x - 1)²] = (1 - 3x² - 2x³)/(x⁴ - 2x³ + x²) = 1
1 - 3x² - 2x³ = x⁴ - 2x³ + x², x⁴ + 4x² - 1 = 0, (x² + 2)² = 5 = (√5)²
x² = - 2 + √5 or x² = - 2 - √5 = - (2 + √5); Imaginary value roots
x² = - 2 + √5 = 0.236; x = ± √(- 2 + √5) = ± √0.236 = ± 0.486
x² = - (2 + √5) = - 4.236; x = ± i√(2 + √5) = ± i√4.236 = ± 2.058i
The calculation was achieved on a smartphone with a standard calculator app
Answer check:
[(x + 1)/x]² - [(x + 1)/(x - 1)]² = (1 - 3x² - 2x³)/(x⁴ - 2x³ + x²)
x² = - 2 ± √5: x⁴ + 4x² - 1 = 0, x⁴ = 1 - 4x²
(1 - 3x² - 2x³)/(x⁴ - 2x³ + x²) = (1 - 3x² - 2x³)/(1 - 4x² - 2x³ + x²)
= (1 - 3x² - 2x³)/(1 - 3x² - 2x³) = 1; Confirmed
Final answer:
x = √(- 2 + √5) = 0.486, x = - √(- 2 + √5) = - 0.486
x = i√(2 + √5) = 2.058i or x = - i√(2 + √5) = - 2.058i

[@superacademy247]

Nice 👍 calculations. Thanks 😊