A Very Nice Math Olympiad Problem | Can You Solve for x? | Algebra
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- เผยแพร่เมื่อ 9 ต.ค. 2024
- In this video, I'll be showing you step by step on how to solve this Olympiad Maths Algebra problem using a simple trick.
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It’s super 👌🏿
Excellent thank you
You're welcome 😊
Thank you for explaining. I will show my method.
Letting, f(x)=(x^2-2)^2, g(x)=x+2, f(2)=g(2)=2 and f(-1)=g(-1)=1. Therefore, f(x)-g(x) is divided by (x-2)(x+1) [=x^2-x-2].
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∴ x^4 - 4x^2 - x + 2 = (x^2-x-2)(x^2+ax+b) ∴ x^2+ax+b = x^2 + x - 1 From x^2+x-1=0, x=(-1±√5)/2 ∴ x = 2, -1, (-1±√5)/2
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I guess the test-maker respected the test-takers can calculate or find x=-1,2 at the beginning when he/she made this problem.
Judging from the left side of the given equation is (・・・)^2, if x has integer solutions, the value of (x^2-2)^2 [=x+2] is 0, 1, 4, 9, ・・・.
The case of =0 should be rejected because x^2-2 cannot be 0 if x is an integer. And, we can find x=-1 and x=2 easily by trial and error.
Thus, we can get 2 integer solutions at the beginning without complicated calculation.
・・・ Hence, I guess the test-maker made the problem because he/she tried to know the test-takers' can notice the calculation that I typed above.
Simpler method: Once you get the equation at 2:42 you can use the Rational Roots Theorem and see immediately that the only possible rational roots are 1,-1,2 or -2. Testing these options and finding x_1 and x_2 only takes a few seconds. Then you can divide by (x+1)(x-2) and get the second degree equation that gives you x_3 and x_4.
Nice problem
other solution are
Method 1
(x^-2)^2 = x+2
Let y= x^ 2 - 2
then y^2 = x+2
adding these equations
x^2+x =y^2+y
(x-y)(x+y+1) = 0
(x-x^2+2)(x+x^2-2+1) = 0
(x^2-x-2)(x^2+x-1) = 0
(x+1) (x-2) (x^2+x-1)=0
Method 2
subtracting x^2 both sides and factorising LHS
(x^2-2+x ) ( x^2-2-x) = x+2-x^2
transposing RHS to LHS and taking common
(x^2-x-2)(x^2+x-1) = 0
Method 3
x4-4x^2-x+2 = 0
suppose that factorisation of expression is
(x^2+px+a )(x^2-px+b) then
a+b-p^2= - 4
P (b-a ) = - 1
ab = 2
Now (a + b)^2 - (a - b)^2 = 4ab hence
(P^2-4)^2 -1/p^2 = 8 let t = p^2
t^2-8t+16-1/t = 8
t^3-8t ^2+16t-9 = 0
t = 1 because sum of coefficients is zero
hence p = 1
a+b = -3 , a - b = 1
hence a = -1 , b = -2
(x^2+x-1)(x^2-x-2) = 0
Method 4
let a = 2 then
x^4+a^2-2ax^2-x-a = 0
a^2 - (2x^2+1)a + (x^4-x)= 0
Discriminant is
4x^4+4x^2+1-4x^4+4x = (2x+1)^2
a =(2x^2+1+2x+1)/2 , (2x^2+1-2x-1)/2
hence 2 = x^2+x+1 and x^2-x
x^2+x-1 =0 , x^2-x-2 = 0
Superb! 👌
(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
Second case: (x² - x - 2) = 0
x² - x - 2 = 0
Δ = (- 1)² - (4 * - 2) = 9
x = (1 ± 3)/2
Solution = { (- 1 - √5)/2 ; - 1 ; (- 1 + √5)/2 ; 2 }
Excellent job well done 👏