amazing problem keep it up without substitution we have to choose generally either Ferrari or Descartes' method for solving biquadratic equation.we put y^2 = 20 -x y = x(x+10) y^2+y = x^2+9 x+20 4 y^2+4 y = 4 x^2+36 x+80 4 y^2+4 y = (2x+9)^2-1 (2 y +1)^2 = (2x+9)^2 now we put y = x^2+10x 2 x^2+20 x +1 = 2 x +9 , 2 x^2 + 20 x +1 = - 20 x - 9 x^2+9 x-4= 0 , x^2+20 x+5 = 0 rest follows at once.
Let x = -a-5. Then √(a+25)=(a+5)(a-5) = a^2-25. Squaring both sides, a^4-50a^2-a-600=0. Since the cubic term is missing, we write (a^2+pa+q)(a^2-pa+r)=0 where qr=600, q+r-p^2=-50 and p(r-q)=-1. Then p=1, q=-24, r=-25 > (a^2+a-24)(a^2-a-25)=0 > a= 1/2[1 +/-√101], 1/2[-1+/-√97] Of these, only a=1/2[1+√101] and a = -1/2[√97+1] satisfy √(a+25)=a^2-25. So, x=1/2[√97 - 9], -1/2[11 +√101].
Nice solution. The equation √(25-y)=y²-25 can be solved also as follows: Let t=25: √(t-y)=y²-t After squaring we get the quadratic in t: t²-(2y²+1)t+y⁴+y=0 with delta = (2y²+1)²-4(y⁴+y)=(2y-1)² and two roots: t=y²+y or t=y²-y+1. After back substitution t=25 we get two quadratics in y: y²+y-25=0 and y²-y-24=0.
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amazing problem keep it up without substitution we have to choose generally either Ferrari or Descartes' method for solving biquadratic equation.we put y^2 = 20 -x
y = x(x+10)
y^2+y = x^2+9 x+20
4 y^2+4 y = 4 x^2+36 x+80
4 y^2+4 y = (2x+9)^2-1
(2 y +1)^2 = (2x+9)^2 now we put y = x^2+10x
2 x^2+20 x +1 = 2 x +9 ,
2 x^2 + 20 x +1 = - 20 x - 9
x^2+9 x-4= 0 ,
x^2+20 x+5 = 0
rest follows at once.
Let x = -a-5. Then √(a+25)=(a+5)(a-5) = a^2-25. Squaring both sides, a^4-50a^2-a-600=0. Since the cubic term is missing, we write (a^2+pa+q)(a^2-pa+r)=0 where qr=600, q+r-p^2=-50 and p(r-q)=-1. Then p=1, q=-24, r=-25 > (a^2+a-24)(a^2-a-25)=0 > a= 1/2[1 +/-√101], 1/2[-1+/-√97] Of these, only a=1/2[1+√101] and a = -1/2[√97+1] satisfy √(a+25)=a^2-25. So, x=1/2[√97 - 9], -1/2[11 +√101].
Let √20-x= t solving ewn gives
(t^2-t-25)(t^2+t+24)= 0
Gives x= (-11+-√101)/2; (-9+-√97)/2 but valid only sols
X=( -11-√101)/2; (√97-9)/2
[√(20 - x)]/(10 + x) = x → where: (20 - x) ≥ 0 → x ≤ 20 and where: (10 + x) ≠ 0 → x ≠ - 10
√(20 - x) = x.(10 + x)
[√(20 - x)]² = [x.(10 + x)]²
20 - x = x².(10 + x)²
20 - x = x².(100 + 20x + x²)
20 - x = 100x² + 20x³ + x⁴
x⁴ + 20x³ + 100x² + x - 20 = 0 → the aim is to eliminate terms to the 3rd power
Let: z = x - (b/4a) → where:
b is the coefficient for x², in our case: 20
a is the coefficient for x³, in our case: 1
x⁴ + 20x³ + 100x² + x - 20 = 0 → let: x = z - (20/4) → let: x = z - 5
(z - 5)⁴ + 20.(z - 5)³ + 100.(z - 5)² + (z - 5) - 20 = 0
(z - 5)².(z - 5) + 20.(z - 5)².(z - 5) + 100.(z² - 10z + 25) + z - 5 - 20 = 0
(z² - 10z + 25).(z² - 10z + 25) + 20.(z² - 10z + 25).(z - 5) + 100z² - 1000z + 2500 + z - 5 - 20 = 0
(z⁴ - 10z³ + 25z² - 10z³ + 100z² - 250z + 25z² - 250z + 625) + 20.(z³ - 5z² - 10z² + 50z + 25z - 125) + 100z² - 1000z + 2500 + z - 5 - 20 = 0
(z⁴ - 20z³ + 150z² - 500z + 625) + 20.(z³ - 15z² + 75z - 125) + 100z² - 1000z + 2500 + z - 5 - 20 = 0
z⁴ - 20z³ + 150z² - 500z + 625 + 20z³ - 300z² + 1500z - 2500 + 100z² - 1000z + 2500 + z - 5 - 20 = 0
z⁴ - 50z² + z + 600 = 0 ← no more term to the 3rd power
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 z⁴ as the beginning of a square: z⁴ = (z² + λ)² - 2λz² - λ²
z⁴ - 50z² + z + 600 = 0 → where: z⁴ = (z² + λ)² - 2λz² - λ²
(z² + λ)² - 2λz² - λ² - 50z² + z + 600 = 0
(z² + λ)² - [2λz² + λ² + 50z² - z - 600] = 0 → let's try to get a second member as a square
(z² + λ)² - [z².(2λ + 50) - z + (λ² - 600)] = 0 → a square into […] means that Δ = 0 → let's calculate Δ
Δ = (- 1)² - 4.[(2λ + 50).(λ² - 600)] → then, Δ = 0
1 - 4.[(2λ + 50).(λ² - 600)] = 0
(2λ + 50).(λ² - 600) = 1/4
2.(λ + 25).(λ² - 600) = 1/4
(λ + 25).(λ² - 600) = 1/8
λ³ - 600λ + 25λ² - 15000 = 1/8
λ³ + 25λ² - 600λ - 15000 - (1/8) = 0
λ³ + 25λ² - 600λ - (120001/8) = 0
λ = - 49/2
Restart
(z² + λ)² - [z².(2λ + 50) - z + (λ² - 600)] = 0 → where: λ = - 49/2
[z² + (- 49/2)]² - [z².(2.{- 49/2} + 50) - z + ({- 49/2}² - 600)] = 0
[z² - (49/2)]² - [z².(- 49 + 50) - z + ({2401/4} - 600)] = 0
[z² - (49/2)]² - [z² - z + (1/4)] = 0 ← we can recognize a square
[z² - (49/2)]² - [z - (1/2)]² = 0 → recall: a² - b² = (a + b).(a - b)
[z² - (49/2) + z - (1/2)].[z² - (49/2) - z + (1/2)] = 0
(z² + z - 25).(z² - z - 24) = 0
First case: (z² + z - 25) = 0
z² + z - 25 = 0
Δ = 1 - (4 * - 25) = 101
z = (- 1 ± √101)/2 → recall: x = z - 5
x = [(- 1 ± √101)/2] - 5
x = (- 10 - 1 ± √101)/2
x = (- 11 ± √101)/2
First solution: x = (- 11 + √101)/2 ← rejected because the condition
Second solution: x = (- 11 - √101)/2
Second case: (z² - z - 24) = 0
z² + z - 24 = 0
Δ = 1 - (4 * - 24) = 97
z = (1 ± √97)/2 → recall: x = z - 5
x = [(1 ± √97)/2] - 5
x = (- 10 + 1 ± √97)/2
x = (- 9 ± √97)/2
Third solution: x = (- 9 + √97)/2
Fourth solution: x = (- 9 - √97)/2 ← rejected because the condition
Solution = { (- 11 - √101)/2 ; (- 9 + √97)/2 }
Nice solution.
The equation √(25-y)=y²-25 can be solved also as follows:
Let t=25: √(t-y)=y²-t
After squaring we get the quadratic in t: t²-(2y²+1)t+y⁴+y=0 with delta = (2y²+1)²-4(y⁴+y)=(2y-1)² and two roots:
t=y²+y or t=y²-y+1.
After back substitution t=25 we get two quadratics in y: y²+y-25=0 and y²-y-24=0.