Cross multiply: (x^2 + x + 1)^2 = x^2 (3x^2 + x + 1). Let y^2 = 2x^2 + x + 1, so (y^2 - x^2)^2 = x^2 (x^2 + y^2). Expand: (y^4 + x^4 - 2 x^2 y^2) = (x^4 + x^2 y^2). Subtract rhs: y^4 - 3x^2 y^2 = 0. Factor: y^2 (y^2 - 3x^2) = 0. This equation can be solved by either y^2 = 3x^2 or y^2 = 0, from which we find that either 2x^2 + x + 1 = 3x^2, or 2x^2 + x + 1 = 0. The first quadratic becomes x^2 - x - 1 = 0, which is solved by x = (1 +/- sqrt(5))/2, and the second quadratic is solved by x = (-1 +/- sqrt(7)i )/4.
At 1:59 you have a quartic equation with integer coefficients which is 2x⁴ − x³ − 2x² − 2x − 1 = 0 As you discovered, converting this equation into a monic equation and then depressing it is not really the way to go. I have two recommendations for you and for everyone else trying to solve quartics from math contests which most probably have a nice factorization into two quadratics with integer coefficients: 1. _Do not_ convert a non-monic quartic with integer coefficients into a monic quartic if this results in non-integer coefficients. 2. _Do not_ depress a quartic which has a cubic term. So, how do we solve this non-monic non-depressed quartic with integer coefficients? First, multiply both sides by _four times the coefficient of the leading (quartic) term_ which is 4·2 = 8. This gives 16x⁴ − 8x³ − 16x² − 16x − 8 = 0 Now bring all terms of the second and lower degrees over to the right hand side, which gives 16x⁴ − 8x³ = 16x² + 16x + 8 Observe that at the left hand side we now have 16x⁴ = (4x²)² which is the square of 4x² whereas 8x³ = 2·4x²·x is twice the product of 4x² and x. Therefore, using the identity a² − 2·a·b + b² = (a − b)², we can complete the square at the left hand side by adding x² to both sides which gives 16x⁴ − 8x³ + x² = 17x² + 16x + 8 which can be written as (4x² − x)² = 17x² + 16x + 8 Now the left hand side is a perfect square but the right hand side is not. To make the right hand side into a perfect square as well we can add something to both sides, but we need to do that in such a way that the left hand side will remain a perfect square. If we take any number k and add 2k(4x² − x) + k² = 8kx² − 2kx + k² to both sides then the left hand side will remain a perfect square regardless of the value of k, because, in accordance with the identity a² + 2·a·b + b² = (a + b)², we will then have (4x² − x)² + 2·(4x² − x)·k + k² = (4x² − x + k)² at the left hand side. So, adding 2k(4x² − x) + k² = 8kx² − 2kx + k² to both sides we get (4x² − x)² + 2k(4x² − x) + k² = 17x² + 16x + 8 + 8kx² − 2kx + k² which can be written as (4x² − x + k)² = (17 + 8k)x² + (16 − 2k)x + (8 + k²) Since the left hand side remains a perfect square regardless of the value of k, we are now free to choose k in such a way that the right hand side will also become a perfect square. But how can we do that? The right hand side of our equation is a quadratic in x, and a quadratic ax² + bx + c will be a perfect square (that is, the square of a linear polynomial in x _or_ the square of a constant) _if and only if_ its discriminant b² − 4ac is zero. Here we have a = 17 + 8k, b = 16 − 2k, c = 8 + k², so the right hand side of our equation will be a perfect square if and only if k satisfies (16 − 2k)² − 4(17 + 8k)(8 + k²) = 0 This is a cubic equation in k, which is generally referred to as the cubic _resolvent_ of our quartic equation. But here is my next recommendation: 3. _Do not_ attempt to solve a cubic resolvent of a quartic equation from a math contest formally unless this is absolutely necessary. Quartics with integer coefficients from math contests usually have a nice factorization into two quadratics with integer coefficients, and we can take advantage of this. Since we started out by multiplying our quartic by four times the leading coefficient, this means that there should exist an _integer value of k_ which makes the right hand side (17 + 8k)x² + (16 − 8k)x + (8 + k²) of our quartic equation the square of a linear polynomial mx + n with _integer coefficients m and n_ if our quartic does indeed have a factorization into two quadratics with integer coefficients. So, since (mx + n)² = m²x² + 2mnx + n², we only need to check integer values of k which make the coeficient 17 + 8k = m² of x² the square of an integer and check if this same integer value of k _also_ makes the constant term 8 + k² = n² the square of an integer. Moreover, here we only need to check integer values of k which make 17 + 8k the square of an _odd_ integer because 17 + 8k cannot be the square of an even integer for any integer k. First, we try 17 + 8k = 1² = 1 which is true for k = −2 but this will not make the constant term 8 + k² = 8 + (−2)² = 12 the square of an integer. Next, we try 17 + 8k = 3² = 9 which is true for k = −1 and with this value of k the constant term 8 + k² = 8 + (−1)² = 9 = 3² is indeed the square of an integer, as required. So, we select k = −1 and with this value of k our quartic equation (4x² − x + k)² = (17 + 8k)x² + (16 − 2k)x + (8 + k²) becomes (4x² − x − 1)² = 9x² + 18x + 9 which can be written as (4x² − x − 1)² = (3x + 3)² Bringing over the square from the right hand side to the left hand side this gives (4x² − x − 1)² − (3x + 3)² = 0 and applying the difference of two squares identity a² − b² = (a − b)(a + b) to the left hand side this gives (4x² − 4x − 4)(4x² + 2x + 2) = 0 Note that the first quadratic factor is 4 times a quadratic with integer coefficients and that the second quadratic factor is 2 times a quadratic with integer coefficients. This is quite understandable because we started out by multiplying both sides of our original quartic by 4·2 = 8. Dividing both sides by 4·2 = 8 we get (x² − x − 1)(2x² + x + 1) = 0 which is therefore an integer factorization of our original quartic 2x⁴ − x³ − 2x² − 2x − 1 = 0. Applying the zero product property A·B = 0 ⟺ A = 0 ⋁ B = 0 this gives x² − x − 1 = 0 ⋁ 2x² + x + 1 = 0 and now we only need to solve these two quadratics to obtain all four solutions of our quartic equation. With regard to your second method I note that when at 5:03 you have (x² + x + 1)² = x²(3x² + x + 1) we do not need to distribute the right hand side and then divide both sides by x² to proceed. If we substitute x² + x + 1 = t we can write the equation as t² = x²(t + 2x²) which is a quadratic in t which can be written as t² − x²t − 2x⁴ = 0 Note that this quadratic in t is easy to factor, because −2x² and x² are two quantities with sum −x² and product −2x⁴ so we get (t − 2x²)(t + x²) = 0 and therefore t = 2x² ⋁ t = −x² and since t = x² + x + 1 this gives x² + x + 1 = 2x² ⋁ x² + x + 1 = −x² and so x² − x − 1 = 0 ⋁ 2x² + x + 1 = 0 which are the exact same quadratics obtained with the first method. If you are brave enough you could even dispense with a substitution altogether and proceed using factoring by grouping as follows (x² + x + 1)² = x²(3x² + x + 1) (x² + x + 1)² = x²(x² + x + 1) + 2x⁴ (x² + x + 1)² − x²(x² + x + 1) − 2x⁴ = 0 (x² + x + 1)² − 2x²(x² + x + 1) + x²(x² + x + 1) − 2x⁴ = 0 (x² + x + 1)(x² + x + 1 − 2x²) + x²(x² + x + 1 − 2x²) = 0 (x² + x + 1 − 2x²)(x² + x + 1 + x²) = 0 (−x² + x + 1)(2x² + x + 1) = 0 (x² − x − 1)(2x² + x + 1) = 0 x² − x − 1 = 0 ⋁ 2x² + x + 1 = 0 which again gives the exact same quadratics.
In Soviet Russia, math problem solves you.
You won
This reminds me of something the Khrushchev-pig character said in Walt Kelly's Pogo.
Cross multiply: (x^2 + x + 1)^2 = x^2 (3x^2 + x + 1). Let y^2 = 2x^2 + x + 1, so (y^2 - x^2)^2 = x^2 (x^2 + y^2). Expand: (y^4 + x^4 - 2 x^2 y^2) = (x^4 + x^2 y^2). Subtract rhs: y^4 - 3x^2 y^2 = 0. Factor: y^2 (y^2 - 3x^2) = 0. This equation can be solved by either y^2 = 3x^2 or y^2 = 0, from which we find that either 2x^2 + x + 1 = 3x^2, or 2x^2 + x + 1 = 0. The first quadratic becomes x^2 - x - 1 = 0, which is solved by x = (1 +/- sqrt(5))/2, and the second quadratic is solved by x = (-1 +/- sqrt(7)i )/4.
Ok lets t= x^2+x+1 solving eqn gives t= 2x^2;-x^2 gives the same solns instant.
At 1:59 you have a quartic equation with integer coefficients which is
2x⁴ − x³ − 2x² − 2x − 1 = 0
As you discovered, converting this equation into a monic equation and then depressing it is not really the way to go. I have two recommendations for you and for everyone else trying to solve quartics from math contests which most probably have a nice factorization into two quadratics with integer coefficients:
1. _Do not_ convert a non-monic quartic with integer coefficients into a monic quartic if this results in non-integer coefficients.
2. _Do not_ depress a quartic which has a cubic term.
So, how do we solve this non-monic non-depressed quartic with integer coefficients? First, multiply both sides by _four times the coefficient of the leading (quartic) term_ which is 4·2 = 8. This gives
16x⁴ − 8x³ − 16x² − 16x − 8 = 0
Now bring all terms of the second and lower degrees over to the right hand side, which gives
16x⁴ − 8x³ = 16x² + 16x + 8
Observe that at the left hand side we now have 16x⁴ = (4x²)² which is the square of 4x² whereas 8x³ = 2·4x²·x is twice the product of 4x² and x. Therefore, using the identity a² − 2·a·b + b² = (a − b)², we can complete the square at the left hand side by adding x² to both sides which gives
16x⁴ − 8x³ + x² = 17x² + 16x + 8
which can be written as
(4x² − x)² = 17x² + 16x + 8
Now the left hand side is a perfect square but the right hand side is not. To make the right hand side into a perfect square as well we can add something to both sides, but we need to do that in such a way that the left hand side will remain a perfect square. If we take any number k and add 2k(4x² − x) + k² = 8kx² − 2kx + k² to both sides then the left hand side will remain a perfect square regardless of the value of k, because, in accordance with the identity a² + 2·a·b + b² = (a + b)², we will then have (4x² − x)² + 2·(4x² − x)·k + k² = (4x² − x + k)² at the left hand side.
So, adding 2k(4x² − x) + k² = 8kx² − 2kx + k² to both sides we get
(4x² − x)² + 2k(4x² − x) + k² = 17x² + 16x + 8 + 8kx² − 2kx + k²
which can be written as
(4x² − x + k)² = (17 + 8k)x² + (16 − 2k)x + (8 + k²)
Since the left hand side remains a perfect square regardless of the value of k, we are now free to choose k in such a way that the right hand side will also become a perfect square. But how can we do that? The right hand side of our equation is a quadratic in x, and a quadratic ax² + bx + c will be a perfect square (that is, the square of a linear polynomial in x _or_ the square of a constant) _if and only if_ its discriminant b² − 4ac is zero.
Here we have a = 17 + 8k, b = 16 − 2k, c = 8 + k², so the right hand side of our equation will be a perfect square if and only if k satisfies
(16 − 2k)² − 4(17 + 8k)(8 + k²) = 0
This is a cubic equation in k, which is generally referred to as the cubic _resolvent_ of our quartic equation. But here is my next recommendation:
3. _Do not_ attempt to solve a cubic resolvent of a quartic equation from a math contest formally unless this is absolutely necessary.
Quartics with integer coefficients from math contests usually have a nice factorization into two quadratics with integer coefficients, and we can take advantage of this. Since we started out by multiplying our quartic by four times the leading coefficient, this means that there should exist an _integer value of k_ which makes the right hand side (17 + 8k)x² + (16 − 8k)x + (8 + k²) of our quartic equation the square of a linear polynomial mx + n with _integer coefficients m and n_ if our quartic does indeed have a factorization into two quadratics with integer coefficients.
So, since (mx + n)² = m²x² + 2mnx + n², we only need to check integer values of k which make the coeficient 17 + 8k = m² of x² the square of an integer and check if this same integer value of k _also_ makes the constant term 8 + k² = n² the square of an integer. Moreover, here we only need to check integer values of k which make 17 + 8k the square of an _odd_ integer because 17 + 8k cannot be the square of an even integer for any integer k.
First, we try 17 + 8k = 1² = 1 which is true for k = −2 but this will not make the constant term 8 + k² = 8 + (−2)² = 12 the square of an integer. Next, we try 17 + 8k = 3² = 9 which is true for k = −1 and with this value of k the constant term 8 + k² = 8 + (−1)² = 9 = 3² is indeed the square of an integer, as required. So, we select k = −1 and with this value of k our quartic equation
(4x² − x + k)² = (17 + 8k)x² + (16 − 2k)x + (8 + k²)
becomes
(4x² − x − 1)² = 9x² + 18x + 9
which can be written as
(4x² − x − 1)² = (3x + 3)²
Bringing over the square from the right hand side to the left hand side this gives
(4x² − x − 1)² − (3x + 3)² = 0
and applying the difference of two squares identity a² − b² = (a − b)(a + b) to the left hand side this gives
(4x² − 4x − 4)(4x² + 2x + 2) = 0
Note that the first quadratic factor is 4 times a quadratic with integer coefficients and that the second quadratic factor is 2 times a quadratic with integer coefficients. This is quite understandable because we started out by multiplying both sides of our original quartic by 4·2 = 8. Dividing both sides by 4·2 = 8 we get
(x² − x − 1)(2x² + x + 1) = 0
which is therefore an integer factorization of our original quartic 2x⁴ − x³ − 2x² − 2x − 1 = 0. Applying the zero product property A·B = 0 ⟺ A = 0 ⋁ B = 0 this gives
x² − x − 1 = 0 ⋁ 2x² + x + 1 = 0
and now we only need to solve these two quadratics to obtain all four solutions of our quartic equation.
With regard to your second method I note that when at 5:03 you have
(x² + x + 1)² = x²(3x² + x + 1)
we do not need to distribute the right hand side and then divide both sides by x² to proceed. If we substitute
x² + x + 1 = t
we can write the equation as
t² = x²(t + 2x²)
which is a quadratic in t which can be written as
t² − x²t − 2x⁴ = 0
Note that this quadratic in t is easy to factor, because −2x² and x² are two quantities with sum −x² and product −2x⁴ so we get
(t − 2x²)(t + x²) = 0
and therefore
t = 2x² ⋁ t = −x²
and since t = x² + x + 1 this gives
x² + x + 1 = 2x² ⋁ x² + x + 1 = −x²
and so
x² − x − 1 = 0 ⋁ 2x² + x + 1 = 0
which are the exact same quadratics obtained with the first method. If you are brave enough you could even dispense with a substitution altogether and proceed using factoring by grouping as follows
(x² + x + 1)² = x²(3x² + x + 1)
(x² + x + 1)² = x²(x² + x + 1) + 2x⁴
(x² + x + 1)² − x²(x² + x + 1) − 2x⁴ = 0
(x² + x + 1)² − 2x²(x² + x + 1) + x²(x² + x + 1) − 2x⁴ = 0
(x² + x + 1)(x² + x + 1 − 2x²) + x²(x² + x + 1 − 2x²) = 0
(x² + x + 1 − 2x²)(x² + x + 1 + x²) = 0
(−x² + x + 1)(2x² + x + 1) = 0
(x² − x − 1)(2x² + x + 1) = 0
x² − x − 1 = 0 ⋁ 2x² + x + 1 = 0
which again gives the exact same quadratics.
Superb! 😍😍
Substituting for x+1 works too.
poser x^2=a et x+1=b et continuer c est tres facile après
Request: x^18-(x^6-x^2)^3-(x^5-1)^3=1, find all real and complex solutions
the 2a always gets you 😏
my try
(x+1+1/x)² =lhs = xx+x+1 +x+1+1/x +1+1/x +1/xx = xx+1+1/xx + 2(x+1 +1/x)
3x²+x+1=rhs
(x^2 + x + 1)^2
----------------------- = x^2
3x^2 + x + 1
Let t = x^2 + x + 1 [1]
t^2 = x^2(t + 2x^2)
2x^4 + tx^2 - t^2 = 0
Solve for x^2
x^2 = (-t +/- sqrt(t^2 + 8t^2))/4
x^2 = (-t +/- 3t)/4
x^2 = t/2 or x^2 = -t
If x^2 = t/2
t = 2x^2
Subs. in [1]
x^2 - x - 1 = 0
x = (-1 +/- sqrt(5))/2
If x^2 = -t
t = -x^2
Subs. in [1]
2x^2 + x + 1 = 0
x = (-1 +/- i sqrt(7))/4 (as per Wolfram Alpha)
very good!
a^2/(2b+a)=b =>a^2=2ab+b^2 the rest is trivial t=a/b and so on
Nice
I’m confused, after I cross multiplied I got a^2=2b^2+ab not 2ab+b^2
Nice!
Thanks!
We need more and more tricks master bye bye
(x^2+x+1)^2=(x^2+x+1)x^2+2x^2*x^2..(x^2+x+1)(x+1)=2x^4...2x^4-x^3-2x^2-x-1=0...x=-0,836..x=1,518...mah
Who are You ? You ! No, not Me, You. I'm You....😂😂😂😂😂😂 From Rush Hour.
😁
problem
(x² + x + 1)² / (3 x² + x + 1) = x²
Expand.
x⁴ + 2x³ + 3x² + 2x + 1 = 3x⁴ + x³ + x²
2x⁴ - x³ - 2x² - 2x - 1 = 0
Multiply by 8.
16x⁴ - 8x³ - 16x² - 16x - 8 = 0
Let
t/2 = x
t⁴ - t³- 4t²- 8t - 8 = 0
Let
t = y + 1/4
y⁴ - 35y²/8 - 81y/8 - 2627/256 = 0
Multiply by 256.
256 y⁴-1120y² - 2592 y - 2627 = 0
Let
y = u/4
u⁴- 70 u²- 648 u - 71•37 = 0
Factor (no cubic term) with variable a an unknown.
(u² +au - 71) (u² - au + 37) = 0
u⁴ + u² (-34-a²) + u 108 a = 0
Matching the coefficients:
-34-a² = -70
a² = 36
108 a = -648
a = -648/108
= -6
Factors are
(u² - 6u - 71) (u² + 6u + 37) = 0
By zero product property, the left factor, right factor:
u = 3 ± 4√5, u = -3 ± 2 i √7
Back substitution.
y = u/4
y = 3 /4 ± √5, y = -3/4 ± i √7 / 2
t = y + 1/4
t = 1 ± √5, t = -1/2 ± i √7 / 2
t/2 = x
x = 1 /2 ± √5 / 2, x = -1/4 ± i √7 / 4
answer
x ∈ { 1 /2 - √5 / 2, 1 /2 + √5 / 2,
-1/4 - i√7 / 4, -1/4 + i√7 / 4 }