Brazil | A Nice Algebra Problem | Math Olympiad

math olympiad
olympiad math
math olympiad question
math olympiad questions
math olympiad problem
math olympiad problems
olympiad math problems
algebra olympiad problems
math
maths
algebra
algebra problem
nice algebra problem
a nice algebra problem
equation
exponential equation
a nice exponential equation
radical equation
a nice radical equation
math tricks
imo
Find the value of x?
How to solve (x^2-2)^2=x+2
In this video, we'll show you How to Solve Math Olympiad Question A Nice Algebra Problem (x^2-2)^2=x+2 in a clear , fast and easy way. Whether you are a student learning basics or a professtional looking to improve your skills, this video is for you. By the end of this video, you'll have a solid understanding of how to solve math olympiad exponential equations and be able to apply these skills to a variety of problems.
#matholympiad #maths #math #algebra

[@LITHICKROSHANMS-gw2lx]

Good solution sir 👍

[@SALogics]

Thanks for liking! ❤

[@АндрейПергаев-з4н]

Попробуйте сделать замену 2=у
Получаем простое квадратное уравнение относительно у, которое решается легко

[@1234larry1]

or simply use synthetic division to test all possible values on the 4th degree polynomial, break it down to two binomials and one quadratic and solve for x in each.

[@SALogics]

Yes, it will work fine! ❤

[@key_board_x]

(x² - 2)² = x + 2
x⁴ - 4x² + 4 = x + 2
x⁴ - 4x² + 4 - x - 2 = 0
x⁴ - 4x² - x + 2 = 0 ← it would be interesting to have 2 squares on the left side (because power 4 and power 2)
Let's rewrite the equation by introducing a variable "λ" which, if carefully chosen, will produce 2 squares on the left side
Let's tinker a bit with 25x⁴ as the beginning of a square: x⁴ = (x² + λ)² - 2λx² - λ²
x⁴ - 4x² - x + 2 = 0 → where: x⁴ = (x² + λ)² - 2λx² - λ²
(x² + λ)² - 2λx² - λ² - 4x² - x + 2 = 0
(x² + λ)² - [2λx² + λ² + 4x² + x - 2] = 0 → let"s try to get a second member as a square
(x² + λ)² - [x².(2λ + 4) + x + (λ² - 2)] = 0 → a square into […] means that Δ = 0 → let"s calculate Δ
Δ = (1)² - 4.[(2λ + 4).(λ² - 2)] → then, Δ = 0
(1)² - 4.[(2λ + 4).(λ² - 2)] = 0
4.[(2λ + 4).(λ² - 2)] = 1
8.[(λ + 2).(λ² - 2)] = 1
(λ + 2).(λ² - 2) = 1/8
λ³ - 2λ + 2λ² - 4 - (1/8) = 0
λ³ + 2λ² - 2λ - (33/8) = 0
λ = - 3/2
Restart
(x² + λ)² - [x².(2λ + 4) + x + (λ² - 2)] = 0 → when λ = - 3/2, a square will appear
[x² - (3/2)]² - [x².(2.{- 3/2} + 4) + x + ({- 3/2}² - 2)] = 0
[x² - (3/2)]² - [x².(- 3 + 4) + x + ({9/4} - 2)] = 0
[x² - (3/2)]² - [x² + x + (1/4)] = 0 ← we can see a square
[x² - (3/2)]² - [x + (1/2)]² = 0 → recall: a² - b² = (a + b).(a - b)
{ [x² - (3/2)] + [x + (1/2)] }.{ [x² - (3/2)] - [x + (1/2)] } = 0
[x² - (3/2) + x + (1/2)].[x² - (3/2) - x - (1/2)] = 0
[x² + x - (2/2)].[x² - x - (4/2)] = 0
(x² + x - 1).(x² - x - 2) = 0
First case: (x² + x - 1) = 0
x² + x - 1 = 0
Δ = (1)² - (4 * - 1) = 5
x = (- 1 ± √5)/2
→ x = (- 1 + √5)/2
→ x = (- 1 - √5)/2
Second case: (x² - x - 2) = 0
x² - x - 2 = 0
Δ = (- 1)² - (4 * - 2) = 9
x = (1 ± 3)/2
→ x = 2
→ x = - 1

[@SALogics]

Very nice trick! I really appreciate that ❤

[@prollysine]

by faktoring , (x+1)(x^3-x^2-3x+2)=0 , x= -1 , x^3-x^2-3x+2=0 , (x-2)(x^2+x-1)=0 , x=2 , x^2+x-1=0 , x= (-1+V5)/2 , (-1-V5)/2 ,

[@SALogics]

Very nice! ❤

[@walterwen2975]

Math Olympiad: (x² - 2)² = x + 2; x =?
x⁴ - 4x² + 4 = x + 2, x⁴ - 4x² - x + 2 = x²(x² - 4) - x + 2 = x²(x + 2)(x - 2) - (x - 2) = 0
(x - 2)(x³ + 2x² - 1) = 0, x - 2 = 0 or x³ + 2x² - 1 = 0, (x³ + 1) + 2(x² - 1) = 0
x = 2 or (x + 1)(x² - x + 1) + 2(x + 1)(x - 1) = (x + 1)(x² - x + 1 + 2x - 2) = 0
x + 1 = 0, x = - 1 or x² - x + 1 + 2x - 2 = x² + x - 1 = 0, x = (- 1 ± √5)/2
Answer check:
x = 2: (x² - 2)² = (4 - 2)² = 4 = 2 + 2 = x + 2; Confirmed
x = - 1: (1 - 2)² = 1 = (- 1) + 2 = x + 2; Confirmed
x = (- 1 ± √5)/2, x² + x - 1 = 0, x² = 1 - x
(1 - x - 2)² = (x + 1)² = x² + 2x + 1 = (1 - x) + 2x + 1 = x + 2; Confirmed
Final answer:
x = 2, x = - 1, x = (- 1 + √5)/2 or x = (- 1 - √5)/2

[@SALogics]

Very nice trick! I really appreciate that ❤