This equation can be solved much more elegantly (1) without having to mess with fractions, (2) without substitutions, and (3) without ending up with unelegant expressions for the roots which still need to have their denominators rationalized. 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 x²(x² + 1) to eliminate both fractions (which is allowed since x ≠ 0 and x ≠ −1) we get (x + 1)² − x² = x²(x + 1)² I ultimately want to create squares on both sides, which is why I start by bringing x² over to the right hand side to get (x + 1)² = x²(x + 1)² + x² This doesn't seem like a step forward because we first had a perfect square on the right hand side but not on the left hand side, and now we have a perfect square on the left hand side but not on the right hand side. But the reason I do this is because we can incorporate the term x² into the product x²(x + 1)² since they both have a common factor. How? First, expanding (x + 1)² we have (x + 1)² = x²(x² + 2x + 1) + x² and taking out the common factor x² at the right hand side this gives (x + 1)² = x²((x² + 2x + 1) + 1) which is (x + 1)² = x²(x² + 2x + 2) We now have a product of two quantities x² and x² + 2x + 2 which we can turn into a difference of squares. The average (arithmetic mean) of x² and x² + 2x + 2 is x² + x + 1 and half the difference between x² + 2x + 2 and x² is x + 1 so we have x² = (x² + x + 1) − (x + 1) and x² + 2x + 2 = (x² + x + 1) + (x + 1) so we can rewrite the equation as (x + 1)² = ((x² + x + 1) − (x + 1))((x² + x + 1) + (x + 1)) and applying the difference of two squares identity (a − b)(a + b) = a² − b² this can be written as (x + 1)² = (x² + x + 1)² − (x + 1)² and bringing the term (x + 1)² from the right hand side over to the left hand side this gives 2(x + 1)² = (x² + x + 1)² We now have a square on both sides of our equation because 2(x + 1)² = (√2)²(x + 1)² = (√2(x + 1))² = (√2·x + √2)² so we have (√2·x + √2)² = (x² + x + 1)² For convenience sake, because I want the square of the quadratic polynomial on the left hand side and the square of the linear polynomial on the right hand side, I'll swap both sides of the equation so we now have (x² + x + 1)² = (√2·x + √2)² We can now proceed in two ways which are fundamentally the same. One way is to bring the square from the right hand side over to the left hand side to create a difference of two squares on the left hand side whereas the right hand side becomes zero. We can then factor the left hand side into two quadratic polynomials using the difference of two squares identity a² − b² = (a − b)(a + b) and apply the zero product property which says that a product is zero if and only if at least one of its factors is itself zero. The other way is to use the equivalence A² = B² ⟺ A = B ⋁ A = −B In English: if the squares of two quantities are equal, then these quantities are either equal _or_ each others opposite, and vice versa. The two ways are fundamentally the same because this equivalence 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 Now, returning to our equation and applying the equivalence A² = B² ⟺ A = B ⋁ A = −B we have x² + x + 1 = √2·x + √2 ⋁ x² + x + 1 = −√2·x − √2 which gives x² + (1 − √2)x + (1 − √2) = 0 ⋁ x² + (1 + √2)x + (1 + √2) = 0 and now we only need to solve these two quadratic equations using the quadratic formula to obtain all four roots of our equation. The discriminant of the first quadratic is Δ = (1 − √2)² − 4·1·(1 − √2) = 3 − 2√2 − 4 + 4√2 = 2√2 − 1 and the discriminant of the second quadratic is Δ = (1 + √2)² − 4·1·(1 + √2) = 3 + 2√2 − 4 − 4√2 = −2√2 − 1 The discriminant of the first quadratic is positive and the discriminant of the second quadratic is negative, so we have two real and two (conjugate) complex solutions. For the first quadratic we have √Δ = √(2√2 − 1) but for the second quadratic we have √Δ = √(−2√2 − 1) = i√(2√2 + 1) and so we have the solutions x = ½(−1 + √2 + √(2√2 − 1)) x = ½(−1 + √2 − √(2√2 − 1)) x = ½(−1 − √2 + i√(2√2 + 1)) x = ½(−1 − √2 − i√(2√2 + 1))
Nice solution. Other approach: substitution t=1/(x+1) leads to symmetric quartic: t^4-2t^3+t^2-2t+1=0 which can be solved by dividing by t^2 and another substition y=t+1/t. The new equation y^2-2y-1=0 with two roots y=1+/-sqrt(2) leads to two quadratics t^2-(1+/-sqrt(2))t+1=0. Finally we get two real and two complex roots for t and consequently for x.
This equation can be solved much more elegantly (1) without having to mess with fractions, (2) without substitutions, and (3) without ending up with unelegant expressions for the roots which still need to have their denominators rationalized. 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 x²(x² + 1) to eliminate both fractions (which is allowed since x ≠ 0 and x ≠ −1) we get
(x + 1)² − x² = x²(x + 1)²
I ultimately want to create squares on both sides, which is why I start by bringing x² over to the right hand side to get
(x + 1)² = x²(x + 1)² + x²
This doesn't seem like a step forward because we first had a perfect square on the right hand side but not on the left hand side, and now we have a perfect square on the left hand side but not on the right hand side. But the reason I do this is because we can incorporate the term x² into the product x²(x + 1)² since they both have a common factor. How? First, expanding (x + 1)² we have
(x + 1)² = x²(x² + 2x + 1) + x²
and taking out the common factor x² at the right hand side this gives
(x + 1)² = x²((x² + 2x + 1) + 1)
which is
(x + 1)² = x²(x² + 2x + 2)
We now have a product of two quantities x² and x² + 2x + 2 which we can turn into a difference of squares. The average (arithmetic mean) of x² and x² + 2x + 2 is x² + x + 1 and half the difference between x² + 2x + 2 and x² is x + 1 so we have x² = (x² + x + 1) − (x + 1) and x² + 2x + 2 = (x² + x + 1) + (x + 1) so we can rewrite the equation as
(x + 1)² = ((x² + x + 1) − (x + 1))((x² + x + 1) + (x + 1))
and applying the difference of two squares identity (a − b)(a + b) = a² − b² this can be written as
(x + 1)² = (x² + x + 1)² − (x + 1)²
and bringing the term (x + 1)² from the right hand side over to the left hand side this gives
2(x + 1)² = (x² + x + 1)²
We now have a square on both sides of our equation because 2(x + 1)² = (√2)²(x + 1)² = (√2(x + 1))² = (√2·x + √2)² so we have
(√2·x + √2)² = (x² + x + 1)²
For convenience sake, because I want the square of the quadratic polynomial on the left hand side and the square of the linear polynomial on the right hand side, I'll swap both sides of the equation so we now have
(x² + x + 1)² = (√2·x + √2)²
We can now proceed in two ways which are fundamentally the same. One way is to bring the square from the right hand side over to the left hand side to create a difference of two squares on the left hand side whereas the right hand side becomes zero. We can then factor the left hand side into two quadratic polynomials using the difference of two squares identity a² − b² = (a − b)(a + b) and apply the zero product property which says that a product is zero if and only if at least one of its factors is itself zero. The other way is to use the equivalence
A² = B² ⟺ A = B ⋁ A = −B
In English: if the squares of two quantities are equal, then these quantities are either equal _or_ each others opposite, and vice versa. The two ways are fundamentally the same because this equivalence 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
Now, returning to our equation and applying the equivalence A² = B² ⟺ A = B ⋁ A = −B we have
x² + x + 1 = √2·x + √2 ⋁ x² + x + 1 = −√2·x − √2
which gives
x² + (1 − √2)x + (1 − √2) = 0 ⋁ x² + (1 + √2)x + (1 + √2) = 0
and now we only need to solve these two quadratic equations using the quadratic formula to obtain all four roots of our equation. The discriminant of the first quadratic is
Δ = (1 − √2)² − 4·1·(1 − √2) = 3 − 2√2 − 4 + 4√2 = 2√2 − 1
and the discriminant of the second quadratic is
Δ = (1 + √2)² − 4·1·(1 + √2) = 3 + 2√2 − 4 − 4√2 = −2√2 − 1
The discriminant of the first quadratic is positive and the discriminant of the second quadratic is negative, so we have two real and two (conjugate) complex solutions. For the first quadratic we have √Δ = √(2√2 − 1) but for the second quadratic we have √Δ = √(−2√2 − 1) = i√(2√2 + 1) and so we have the solutions
x = ½(−1 + √2 + √(2√2 − 1))
x = ½(−1 + √2 − √(2√2 − 1))
x = ½(−1 − √2 + i√(2√2 + 1))
x = ½(−1 − √2 − i√(2√2 + 1))
😍
A both brilliant and elegant solution and very well explained!
Wow man, you are also really good at writing!
Nice solution.
Other approach: substitution t=1/(x+1) leads to symmetric quartic: t^4-2t^3+t^2-2t+1=0 which can be solved by dividing by t^2 and another substition y=t+1/t.
The new equation y^2-2y-1=0 with two roots y=1+/-sqrt(2) leads to two quadratics t^2-(1+/-sqrt(2))t+1=0.
Finally we get two real and two complex roots for t and consequently for x.
Thank you!
Nice solution, thanks
You are welcome
mind blowing
Substitution is awesome, isn’t it? 😍
A mind is required for one to be "blown"
This problem is a monster...
A little bit 😁
Are you interested in solving our math Olimpiad ( Algeria 🇩🇿)
انا مهتم انا من اليمن 🇾🇪 عربي نفسك ياجزائري❤❤
Sure. Can you provide some links?
i did the same, but started with multiplying both sides by x²
Nice job!
Thanks!
👍
asnwer=1 isit
You put "Math Olympiad" in description.
Would you take an extra 2 minutes to put which Olympiad, Year, and problem number on thumbnail?
Thanks
Not sure which year but I know it's from Russia.
There's another method : treat the left side as difference of 2 squares and then simplify.