-(½)(1+√5), case 2, could also be considered a solution if we treat the ½-power as a multivalued function, since it can have positive & negative values. If you only want the positive values of ½-power, you should use the "√" (radical) symbol. So, @22:00 for the 1st term on the left side we can consider the negative root & for the 2nd term, the positive root, so that we get -1.618+2.618=1.
x = 1/golden ratio = 1/phi original equation then simplifies to 1/phi + 1/phi^2 = 1 implying that phi + 1 = phi^2, the defining equation of the golden ratio.
Thanks for your insteresting problem that I solved as here below. Of course, I didn't look the way you solved it. I hope you like my solution. Greetings ! (E) sqrt(x-x^3)+sqrt(x^2-x^3)=1 First, we have to define the existence domain of this equation D(E). x € D(E) if (x-x^3)>0 and (x^2-x^3)>0 then x(1-x)(1+x)>0 and x^2(1-x)>0 then -1
@@superacademy247 Thanks for your nice comment ! This message to say I have to correct a mistake. The validity domain for the solutions is not well determined at first. Let me correct: A) x(1-x)(1+x)>0 and x^2(1-x)>0 leads to the following set {-inf
Uma linda questão difícil, muito difícil!
-(½)(1+√5), case 2, could also be considered a solution if we treat the ½-power as a multivalued function, since it can have positive & negative values. If you only want the positive values of ½-power, you should use the "√" (radical) symbol. So, @22:00 for the 1st term on the left side we can consider the negative root & for the 2nd term, the positive root, so that we get -1.618+2.618=1.
x = 1/golden ratio = 1/phi
original equation then simplifies to 1/phi + 1/phi^2 = 1
implying that phi + 1 = phi^2, the defining equation of the golden ratio.
Hey man!! (X^2)^(1/2)=X if X >0 otherwise = abs(X).
check the expressions u found not the approximations.
(x-x^3)≥ 0 && (x^2-x^3)≥ 0 ⇒ -∞ 1
Let x=Sin(t ) then x-x^3=Sin(t)•(Cos(t))^2, x^2-x^3=(1-Sin(t))•(Sin(t))^2, 0 (Sin(t))^2 + Sin(t) -1=0 => Sin(t)=(√5 - 1)/2
Finally x=(√5 - 1)/2
First condition:
x - x³ ≥ 0
x.(1 - x²) ≥ 0
x.(1 + x).(1 - x) ≥ 0
x ∈ ] - ∞ ; - 1[ U [0 ; 1]
Second condition
x² - x³ ≥ 0
x².(1 - x) ≥ 0 → you know that: x² ≥ 0
1 - x ≥ 0
- x ≥ - 1
x ≤ 1
√(x - x³) + √(x² - x³) = 1
√(x - x³) + √[x².(1 - x)] = 1
√(x - x³) + x√(1 - x) = 1 → you square both sides
[√(x - x³) + x√(1 - x)]² = 1²
(x - x³) + 2x√[(x - x³).(1 - x)] + x².(1 - x) = 1
2x√[(x - x³).(1 - x)] = 1 - (x - x³) - x².(1 - x)
2x√[(x - x³).(1 - x)] = 1 - x + x³ - x² + x³
2x√[(x - x³).(1 - x)] = 2x³ - x² - x + 1 → you square both sides
{ 2x√[(x - x³).(1 - x)] }² = (2x³ - x² - x + 1)²
4x².(x - x³).(1 - x) = 4x⁶ - 2x⁵ - 2x⁴ + 2x³ - 2x⁵ + x⁴ + x³ - x² - 2x⁴ + x³ + x² - x + 2x³ - x² - x + 1
4x³ - 4x⁴ - 4x⁵ + 4x⁶ = 4x⁶ - 4x⁵ - 3x⁴ + 6x³ - x² - 2x + 1
4x³ - 4x⁴ - 4x⁵ + 4x⁶ = 4x⁶ - 4x⁵ - 3x⁴ + 6x³ - x² - 2x + 1
x⁴ + 2x³ - x² - 2x + 1 = 0 → the aim, if we are to continue effectively, is to eliminate terms to the 3rd power
Let: x = z - (b/4a) → where:
b is the coefficient for x³, in our case: 2
a is the coefficient for x⁴, in our case: 1
x⁴ + 2x³ - x² - 2x + 1 = 0 → let: x = z - (2/4) → x = z - (1/2)
[z - (1/2)]⁴ + 2.[z - (1/2)]³ - [z - (1/2)]² - 2.[z - (1/2)] + 1 = 0
[z - (1/2)]².[z - (1/2)]² + 2.[z - (1/2)]².[z - (1/2)] - [z² - z + (1/4)] - 2z + 1 + 1 = 0
[z² - z + (1/4)].[z² - z + (1/4)] + 2.[z² - z + (1/4)].[z - (1/2)] - z² + z - (1/4) - 2z + 1 + 1 = 0
[z⁴ - z³ + (1/4).z² - z³ + z² - (1/4).z + (1/4).z² - (1/4).z + (1/16)] + 2z³ - z² - 2z² + z + (1/2).z - (1/4) - z² + z - (1/4) - 2z + 1 + 1 = 0
[z⁴ - 2z³ + (3/2).z² - (1/2).z + (1/16)] + 2z³ - 4z² + (1/2).z + (3/2) = 0
z⁴ - (5/2).z² + (25/16) = 0
[z² - (5/4)]² = 0
z² - (5/4) = 0
z² = 5/4
z = ± (√5)/2 → recall: x = z - (1/2)
x = ± (√5)/2) - (1/2)
x = (± √5 - 1)/2
x = (- 1 ± √5)/2 → recall the conditions about x
x = (- 1 + √5)/2
(1x ➖ 1x^1^.1)+(1x^1 ➖ 1x^1^.1)={0+0 ➖} x^1.1+{0+0 ➖} x^1.{0+0 ➖ }={1x^1^.1+1^1.1}= 2x^2.2 1x^1.2.1 x^1.2 (x ➖ 2x+1).
Stanford University Entrance Aptitude Test: √(x - x³) + √(x² - x³) = 1; x =?
1 > √(x² - x³) > √(x - x³) > 0; 1 > x ≠ 0
Let: u = √(x - x³), v = √(x² - x³); u + v = 1, u² - v² = (u + v)(u - v) = u - v = x - x²
u + v = 1, u - v = x - x²; (u + v) + (u - v) = 2u = 1 + x - x²
[2√(x - x³)]² = 4x - 4x³ = (1 + x - x²)² = 1 + x² + x⁴ + 2x - 2x³ - 2x²
x⁴ + 2x³ - x² - 2x + 1 = 0, x⁴ + (2x³ - 2x²) + (x² - 2x + 1) = x⁴ + 2x²(x - 1) + (x - 1)² = 0
(x² + x - 1)² = 0, x² + x - 1 = 0; x = (- 1 ± √5)/2; Double roots
Answer check:
x = (- 1 ± √5)/2, x² + x - 1 = 0; x = 1 - x², x² = 1 - x, x + x² = 1
√(x - x³) + √(x² - x³) = √[x(1 - x²)] + √[x²(1 - x)] = √(x²) + √(x⁴) = x + x² = 1; Confirmed
Final answer:
x = (- 1 + √5)/2 = 0.618, x = (- 1 - √5)/2 = - 1.618; Double roots
The calculation was achieved on a smartphone with a standard calculator app
this is better! i solved in this method.
we get , x^4+2x^3-x^2-2x+1=0 , /:x^2 , x^2+1/x^2+2x-2/x-1=0 , / (x+1/x)^2-2=(x-1/x)^2+2 / , (x-1/x)^2+2(x-1/x)+1=0 ,
let u=(x-1/x) , u^2+2u+1=0 , u= -1 , x-1/x=-1 , --> , x^2+x-1=0 , x= (-1+V5)/2 , / (-1-V5)/2 , not a solu /, solu , x=(-1+V5)/2 ,
Thanks for your insteresting problem that I solved as here below.
Of course, I didn't look the way you solved it.
I hope you like my solution.
Greetings !
(E) sqrt(x-x^3)+sqrt(x^2-x^3)=1
First, we have to define the existence domain of this equation D(E).
x € D(E) if (x-x^3)>0 and (x^2-x^3)>0
then x(1-x)(1+x)>0 and x^2(1-x)>0
then -1
Thanks for details and resourceful explanation 🙏🤩💕
@@superacademy247 Thanks for your nice comment !
This message to say I have to correct a mistake.
The validity domain for the solutions is not well determined at first.
Let me correct:
A)
x(1-x)(1+x)>0 and x^2(1-x)>0
leads to the following set {-inf