We note by inspection that x=-1 is a solution, so we introduce y=x+1 and multiply the equation by (x+2)^2 = (y+1)^2 to get (y-1)^2[(y+1)^2 + 4] =(y^2-1)^2 + 4(y-1)^2 = 5(y+1)^2. Multiplying out and collecting terms gives (y^3 -3y-18)y = 0 By search we find that the third order polynomial has a zero at y=3. The three zeros, a,b,c of this polynomial obey the Vieta relations a+b+c=0, abc=18. We have found (say) c=3, so the remaining two satisfy a+b =-3, ab=6. Hence they are zeros of the polynomial (y^2+3y+6). We have found the integer factorization y(y-3)(y^2+3y+6) = (x+1)(x-2)(x^2+5x+10) = 0.
The equation to solve is x² + (2x/(x + 2))² = 5 This equation can be solved elegantly (1) without having to mess with fractions and (2) without substitutions. The key here is to use a technique which I demonstrated earlier, i.e. converting a product of quantities into a difference of squares. If we start by multiplying both sides by (x + 2)² to eliminate the fraction (which is allowed since x ≠ −2) we get x²(x + 2)² + 4x² = 5(x + 2)² I want to create a difference of two squares on the left hand side, and to do so I first incorporate the term 4x² at the left hand side into the product x²(x + 2)². This can be done since they both have a common factor x². Expanding (x + 2)² at the left hand side we have x²(x² + 4x + 4) + 4x² = 5(x + 2)² and taking out the common factor x² at the left hand side this gives x²((x² + 4x + 4) + 4) = 5(x + 2)² which is x²(x² + 4x + 8) = 5(x + 2)² We now have a product of two quantities x² and x² + 4x + 8 at the left hand side which we can turn into a difference of squares. The _average_ (arithmetic mean) of x² and x² + 4x + 8 is half their sum which is x² + 2x + 4 and the difference between x² + 4x + 8 and x² is 4x + 8 = 4(x + 2) so _half their difference_ is 2(x + 2). So, we have x² = (x² + 2x + 4) − 2(x + 2) and x² + 4x + 8 = (x² + 2x + 4) + 2(x + 2) and we can therefore rewrite the equation as ((x² + 2x + 4) − 2(x + 2))((x² + 2x + 4) + 2(x + 2)) = 5(x + 2)² and applying the difference of two squares identity (a − b)(a + b) = a² − b² to the left hand side this can be written as (x² + 2x + 4)² − 4(x + 2)² = 5(x + 2)² and bringing the term 5(x + 2)² from the right hand side over to the left hand side this gives (x² + 2x + 4)² − 9(x + 2)² = 0 We now again have a difference of two squares at the left hand side whereas the right hand side is zero. Again applying the difference of two squares identity a² − b² = (a − b)(a + b) to the left hand side this gives (x² + 2x + 4 − 3(x + 2))(x² + 2x + 4 + 3(x + 2)) = 0 which is (x² − x − 2)(x² + 5x + 10) = 0 and applying the zero product property which says that a product is zero _if and only if_ at least one of its factors is itself zero we get x² − x − 2 = 0 ⋁ x² + 5x + 10 = 0 Factoring the first quadratic and completing the square of the second quadratic this gives (x − 2)(x + 1) = 0 ⋁ (x + ⁵⁄₂)² = −¹⁵⁄₄ Here we can again apply the zero product property to the first quadratic. For the second quadratic we can note that −¹⁵⁄₄ = (i·¹⁄₂√15)² and apply the equal squares property. This says that squares of two quantities are equal _if and only if_ the quantities themselves are either equal or each others opposite. Using these properties we find x = 2 ⋁ x = −1 ⋁ x = −⁵⁄₂ + i·¹⁄₂√15 ⋁ x = −⁵⁄₂ − i·¹⁄₂√15 which completes the solution of the equation. Note that applying the zero product property A·B = 0 ⟺ A = 0 ⋁ B = 0 and applying the equal squares property A² = B² ⟺ A = B ⋁ A = −B to solve a quadratic equation are fundamentally the same approaches. This is so because the equal squares property is a simple consequence of the difference of two squares identity and the zero product property, since we have A² = B² ⟺ A² − B² = 0 ⟺ (A − B)(A + B) = 0 ⟺ A − B = 0 ⋁ A + B = 0 ⟺ A = B ⋁ A = −B
We note by inspection that x=-1 is a solution, so we introduce y=x+1 and multiply the equation by (x+2)^2 = (y+1)^2 to get
(y-1)^2[(y+1)^2 + 4] =(y^2-1)^2 + 4(y-1)^2 = 5(y+1)^2.
Multiplying out and collecting terms gives
(y^3 -3y-18)y = 0
By search we find that the third order polynomial has a zero at y=3. The three zeros, a,b,c of this polynomial obey the Vieta relations a+b+c=0, abc=18. We have found (say) c=3, so the remaining two satisfy a+b =-3, ab=6. Hence they are zeros of the polynomial (y^2+3y+6). We have found the integer factorization
y(y-3)(y^2+3y+6) = (x+1)(x-2)(x^2+5x+10) = 0.
The equation to solve is
x² + (2x/(x + 2))² = 5
This equation can be solved elegantly (1) without having to mess with fractions and (2) without substitutions. The key here is to use a technique which I demonstrated earlier, i.e. converting a product of quantities into a difference of squares.
If we start by multiplying both sides by (x + 2)² to eliminate the fraction (which is allowed since x ≠ −2) we get
x²(x + 2)² + 4x² = 5(x + 2)²
I want to create a difference of two squares on the left hand side, and to do so I first incorporate the term 4x² at the left hand side into the product x²(x + 2)². This can be done since they both have a common factor x². Expanding (x + 2)² at the left hand side we have
x²(x² + 4x + 4) + 4x² = 5(x + 2)²
and taking out the common factor x² at the left hand side this gives
x²((x² + 4x + 4) + 4) = 5(x + 2)²
which is
x²(x² + 4x + 8) = 5(x + 2)²
We now have a product of two quantities x² and x² + 4x + 8 at the left hand side which we can turn into a difference of squares. The _average_ (arithmetic mean) of x² and x² + 4x + 8 is half their sum which is x² + 2x + 4 and the difference between x² + 4x + 8 and x² is 4x + 8 = 4(x + 2) so _half their difference_ is 2(x + 2). So, we have x² = (x² + 2x + 4) − 2(x + 2) and x² + 4x + 8 = (x² + 2x + 4) + 2(x + 2) and we can therefore rewrite the equation as
((x² + 2x + 4) − 2(x + 2))((x² + 2x + 4) + 2(x + 2)) = 5(x + 2)²
and applying the difference of two squares identity (a − b)(a + b) = a² − b² to the left hand side this can be written as
(x² + 2x + 4)² − 4(x + 2)² = 5(x + 2)²
and bringing the term 5(x + 2)² from the right hand side over to the left hand side this gives
(x² + 2x + 4)² − 9(x + 2)² = 0
We now again have a difference of two squares at the left hand side whereas the right hand side is zero. Again applying the difference of two squares identity a² − b² = (a − b)(a + b) to the left hand side this gives
(x² + 2x + 4 − 3(x + 2))(x² + 2x + 4 + 3(x + 2)) = 0
which is
(x² − x − 2)(x² + 5x + 10) = 0
and applying the zero product property which says that a product is zero _if and only if_ at least one of its factors is itself zero we get
x² − x − 2 = 0 ⋁ x² + 5x + 10 = 0
Factoring the first quadratic and completing the square of the second quadratic this gives
(x − 2)(x + 1) = 0 ⋁ (x + ⁵⁄₂)² = −¹⁵⁄₄
Here we can again apply the zero product property to the first quadratic. For the second quadratic we can note that −¹⁵⁄₄ = (i·¹⁄₂√15)² and apply the equal squares property. This says that squares of two quantities are equal _if and only if_ the quantities themselves are either equal or each others opposite. Using these properties we find
x = 2 ⋁ x = −1 ⋁ x = −⁵⁄₂ + i·¹⁄₂√15 ⋁ x = −⁵⁄₂ − i·¹⁄₂√15
which completes the solution of the equation.
Note that applying the zero product property A·B = 0 ⟺ A = 0 ⋁ B = 0 and applying the equal squares property A² = B² ⟺ A = B ⋁ A = −B to solve a quadratic equation are fundamentally the same approaches. This is so because the equal squares property is a simple consequence of the difference of two squares identity and the zero product property, since we have
A² = B² ⟺ A² − B² = 0 ⟺ (A − B)(A + B) = 0 ⟺ A − B = 0 ⋁ A + B = 0 ⟺ A = B ⋁ A = −B
A = x
B= 2x/(x+2)
A - B = AB/2
Now A^2+B^2 = 5
(A-B)^2 + 2 AB = 5
(AB)^2 + 8 AB -20= 0
AB = 2 , -10
Case 1
x^2-x-2 = 0
x= - 1 , 2
Case 2
x^2 + 5x+10 = 0
x = (-5+√15 i)/2 , (- 5-√15 i)/2
x² + [2x/(x + 2)]² = 5
x² + [4x²/(x + 2)²] = 5
[x².(x + 2)² + 4x²]/(x + 2)² = 5
x².(x + 2)² + 4x² = 5.(x + 2)²
x².(x² + 4x + 4) + 4x² = 5.(x² + 4x + 4)
x⁴ + 4x³ + 4x² + 4x² = 5x² + 20x + 20
x⁴ + 4x³ + 3x² - 20x - 20 = 0 → the aim is to eliminate terms to the power 3
Let: x = z - (b/4a) → where:
b is the coefficient for x³, in our case: 4
a is the coefficient for x⁴, in our case: 1
x⁴ + 4x³ + 3x² - 20x - 20 = 0 → let: x = z - (4/4) → x = z - 1
(z - 1)⁴ + 4.(z - 1)³ + 3.(z - 1)² - 20.(z - 1) - 20 = 0
(z - 1)².(z - 1)² + 4.(z - 1)².(z - 1) + 3.(z² - 2z + 1) - 20z + 20 - 20 = 0
(z² - 2z + 1).(z² - 2z + 1) + 4.(z² - 2z + 1).(z - 1) + 3z² - 6z + 3 - 20z = 0
z⁴ - 2z³ + z² - 2z³ + 4z² - 2z + z² - 2z + 1 + 4z³ - 4z² - 8z² + 8z + 4z - 4 + 3z² - 6z + 3 - 20z = 0
z⁴ - 3z² - 18z = 0
z.(z³ - 3z - 18) = 0
First case: z = 0
Recall: x = z - 1
→ x = - 1
Second case: (z³ - 3z - 18) = 0
z³ - 3z - 18 = 0 → let: z = u + v
(u + v)³ - 3.(u + v) - 18 = 0
(u + v)².(u + v) - 3.(u + v) - 18 = 0
(u² + 2uv + v²).(u + v) - 3.(u + v) - 18 = 0
u³ + u²v + 2u²v + 2uv² + uv² + v³ - 3.(u + v) - 18 = 0
u³ + v³ + 3u²v + 3uv² - 3.(u + v) - 18 = 0
u³ + v³ + 3uv.(u + v) - 3.(u + v) - 18 = 0
u³ + v³ + (u + v).(3uv - 3) - 18 = 0 → suppose that: (3uv - 3) = 0 ← equaton (1)
u³ + v³ + (u + v).(0) - 18 = 0
u³ + v³ - 18 = 0 ← equaton (2)
(1): (3uv - 3) = 0
(1): 3uv = 3
(1): uv = 1
(1): u³v³ = 1 ← this is the poduct P
(2): u³ + v³ - 18 = 0
(2): u³ + v³ = 18 ← this is the sum S
u³ & v³ are the solution of the equation: a² - Sa + P = 0
a² - 18a + 1 = 0
Δ = 18² - 4 = 320 = 5 * 64
a = (18 ± 8√5)/2
a = 9 ± 4√5
u³ = 9 + 4√5 → u = (9 + 4√5)^(1/3)
v³ = 9 - 4√5 → v = (9 + 4√5)^(1/3)
Recall: z = u + v
z = (9 + 4√5)^(1/3) + (9 + 4√5)^(1/3)
z = 3
Recall: x = z - 1
→ x = 2
Restart
x⁴ + 4x³ + 3x² - 20x - 20 = 0 → we've just seen that 2 is a root, so we can factorize (x - 2)
(x - 2).(x³ + ax² + bx + 10) = 0 → you expand
x⁴ + ax³ + bx² + 10x - 2x³ - 2ax² - 2bx - 20 = 0 → you group
x⁴ + x³.(a - 2) + x².(b - 2a) + x.(10 - 2b) - 20 = 0 → you compare with: x⁴ + 4x³ + 3x² - 20x - 20 = 0
For x³ → (a - 2) = 4 → a = 6
For x² → (b - 2a) = 3 → b = 3 + 2a → b = 15
For x → (10 - 2b) = - 20 → 2b = 30 → b = 15
(x - 2).(x³ + ax² + bx + 10) = 0 → where: a = 6
(x - 2).(x³ + 6x² + bx + 10) = 0 → where: b = 15
(x - 2).(x³ + 6x² + 15x + 10) = 0 → you know that x = 2 is a solution, then we study the equation:
x³ + 6x² + 15x + 10 = 0 ← we can see that (- 1) is an obvious root, so we can factorize (x + 1)
(x + 1).(x² + ax + 10) = 0 → you expand
x³ + ax² + 10x + x² + ax + 10 = 0 → you group
x³ + x².(a + 1) + x.(10 + a) + 10 = 0 → you compare with: x³ + 6x² + 15x + 10 = 0
For x² → (a + 1) = 6 → a = 5
For x → (10 + a) = 15 → a = 5
Restart
(x + 1).(x² + ax + 10) = 0 → where: a = 5
(x + 1).(x² + 5x + 10) = 0 → you know that x = - 1 is a solution, then we study the equation:
x² + 5x + 10 = 0
Δ = 5² - (4 * 10) = 25 - 40 = - 15 = 15i²
x = (- 5 ± i√15)/2
Summarize:
x² + [2x/(x + 2)]² = 5
x⁴ + 4x³ + 3x² - 20x - 20 = 0
(x - 2).(x³ + 6x² + 15x + 10) = 0
(x - 2).(x + 1).(x² + 5x + 10) = 0
Solution = { - 1 ; 2 ; (- 5 - i√15)/2 ; (- 5 + i√15)/2 }
(x-2)(x+1)(x^2+5x+10)=0 ,
let t = x + 1,
x^2(x + 2)^2 - 5(x + 2)^2 + 4x^2 = 0
=> (t - 1)^2(t + 1)^2 - 5(t + 1)^2 + 4(t - 1)^2 = t^4 - 3t^2 - 18t = t(t - 3)(t^2 + 3t + 6) =0
=> t = { 0, 3, (-3 ± √15i)/2 } => x = { -1, 2, (-3 ± √15i)/2 }
X^2 + (2X)^2 /(X +2)^2 = 5 Обе стороны умножить на (X +2)^2
X^2 (X +2)^2 + (2X)^2 =5 (X+2)^2
X^2 (X^2 + 2×X×2 + 2^2) +4X^2 - 5(X^2 + 4X +4) =0
X^4 + 4X^3 +4X^2 + 4X^2 -5X^2 - 20X -20 = 0
4X^3 = 3X^3 + 1X^3
(X^4 +X^3) + (3X^3 +3X^2) -20X - 20 =0
X^3(X+1) +3X^2(X+1) -20(X+1) = 0
(X+1)(X^3 + 3X^2 - 20) = 0
X +1=0 ; X= -1
X^3 +3X^2 -20 = 0
X^2 (X+3) = 20; 20= 4×5
X^2 (X+3) = 4×5
X^2 = 4 X+3 = 5
X =2 X=2
X= -2 (не реальный ответ, т.к. при проверке на 0 делить нельзя)
Ответ: X =-1
X=2