Recognize that 2 is an obvious solution, as is -1. Expanding the equation and moving it all to the LHS, then doing synthetic division with 2 and -1 results in a quadratic. Solving that reveals the final two roots to be -phi and 1/phi. All four check out as correct. x = {-phi, -1, 1/phi, 2}.
10:33 -1 est solution évidente on factorise le polynôme cubique par x+1 on obtient un polynôme de second degré qui est facilement résoluble … et on peut aller dormir tranquillement
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]. > ∴ 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 ・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・ 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.
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
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.
Before watching the video (or reading other comments) ... (x² - 2)² = x + 2 Expand with only 0 on the rhs x⁴ - 4x² - x + 2 = 0 by inspection x = 2 is a solution (sub in to lhs and rhs) so (x - 2) is a factor. do the long division ( by (x - 2) ) x⁴ - 4x² - x + 2 = (x - 2)(x³ + 2x² - 1) by inspection x = - 1 is a solution so (x + 1) is a factor x⁴ - 4x² - x + 2 = (x - 2)(x + 1)(x² + x - 1) Solutions are the above two x = 2, x = - 1 and (using the quadratic formula on the last bracket) x = -(√5+1)/2, x = (√5-1)/2 NB I see others have done the same.
I think you can try some values and notice x=2 is root of original equation. Then you are left with x^3+2x^2-1=0 which has root -1 which you can factor out, considering polynomial p(x)=x^3+2x^2-1 has integer roots which are divisors of free term 1
When we know the golden ratio φ=(1+√5)/2≈1.618, by plotting the graph of the function f(x)=x⁴−4x²−x+2, in addition to the solutions x=-1 and x=2, we easily recognize the 2 other solutions x≈(-1.618)≈(-φ) and x≈(0.618)≈(φ−1) or x=-(1+√5)/2 and x=(1+√5)/2−1
Recognize that 2 is an obvious solution, as is -1. Expanding the equation and moving it all to the LHS, then doing synthetic division with 2 and -1 results in a quadratic. Solving that reveals the final two roots to be -phi and 1/phi. All four check out as correct. x = {-phi, -1, 1/phi, 2}.
exactly!!
Agreed. Those factoring choices to find 2 and -1 are just contrived otherwise.
10:33 -1 est solution évidente
on factorise le polynôme cubique par x+1 on obtient un polynôme de second degré qui est facilement résoluble …
et on peut aller dormir tranquillement
@@victordunord7261
Et x=2 est aussi d’OL évidente, ça simplifie encore le pb!
Отлично! С удовольствием разобрался вместе с Вами. Привет из Москвы! 😊❤❤😊
Thanks. I'm glad you enjoyed it 😊
(x^2 - 2)^2 = x + 2
x^4 - 4x^2 - x + 2 = 0
(x + 1)(x^3 - x^2 - 3x + 2) = 0
(x + 1)(x - 2)(x^2 + x - 1) = 0
x = -1, 2, (-1 +/- ✓5)/2
Excellent
@@SpencersAcademy 🙂🙏
once you have x^3 + 2x^2 - 1, rather sum of cubes rigamarole, do
x^3 + 2x^2 - 1
= x^3 + x^2 + x^2 - 1
= (x+1)(x^2) + (x+1)(x-1)
= (x+1)(x^2 + x - 1)
i was about to write the same )
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].
>
∴ 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
・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・・
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.
Excellent job 👏
Linda ecuación de 4to grado, resuelta solo con álgebra elemental y un poco de creatividad 😊
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! 👌
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 one. I like that! 👍
Simple, use 4th degree equation formula. Our professor made US derive and learn It, in case It would be helpful in future.
x^2-2=a => a^2 = x+2 => 2*2=2+2 => a=2 and x=2
It takes a long time you solve a problem
(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 x⁴ 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 👏
Appreciate the detail but incredibly long!
@@clmkc5393
it's precisely because it takes so long that I'm passing on this solution.
Before watching the video (or reading other comments) ...
(x² - 2)² = x + 2
Expand with only 0 on the rhs
x⁴ - 4x² - x + 2 = 0
by inspection x = 2 is a solution (sub in to lhs and rhs) so (x - 2) is a factor.
do the long division ( by (x - 2) )
x⁴ - 4x² - x + 2 = (x - 2)(x³ + 2x² - 1)
by inspection x = - 1 is a solution so (x + 1) is a factor
x⁴ - 4x² - x + 2 = (x - 2)(x + 1)(x² + x - 1)
Solutions are the above two
x = 2, x = - 1
and (using the quadratic formula on the last bracket)
x = -(√5+1)/2, x = (√5-1)/2
NB I see others have done the same.
Beautiful! 👍
It’s super 👌🏿
Excellent thank you
You're welcome 😊
X=2 or (-1)
This is quintic with two trivial roots. Kindergarden stuff
You write a lot, choose the font with the largest size. thank you
Sir. U show a+b formula n use a-b formula
You can also use that.
(x^2-2)^2 -4 = x+2 -4
x^2 (x^2- 4) =x-2
(x-2)(x^3+2x^2- 1)=0
(x-2)(x+1)(x^2 +x-1)=0
x=2,-1,(-1+-sqrt5)/2
Great job! 👏
X,2x+5=8")
Y=X^2-2， Y^2=X+2， Y+Y^2=X^2+X， （X-Y）（X+Y+1）=0，
(x^4 ➖ 4)={x^0+x0 ➖ }=x^1(x ➖ 1x+1).{x+x ➖ }+{2+2 ➖ }={x^2+4}=4x^2 2^2x^2 1^1x^2 1x^2 (x ➖ 2x+1).
I think you can try some values and notice x=2 is root of original equation.
Then you are left with x^3+2x^2-1=0 which has root -1 which you can factor out, considering polynomial p(x)=x^3+2x^2-1 has integer roots which are divisors of free term 1
Nice one! 👍
Write with biggest fonts
X = 3....demore menos de 5 segundos
Nah
@@ServoDoFJ 😀😀😀😀😃
@@yanssala2214 Melhor do que fazer toneladas de equações
Once again, the Golden Ratio shows up.
When we know the golden ratio φ=(1+√5)/2≈1.618, by plotting the graph of the function f(x)=x⁴−4x²−x+2, in addition to the solutions x=-1 and x=2, we easily recognize the 2 other solutions x≈(-1.618)≈(-φ) and x≈(0.618)≈(φ−1) or x=-(1+√5)/2 and x=(1+√5)/2−1
Nice one 👍