x⁴-2x³-6x²+16x-8=0 x³(x-2)-2(3x²-8x+4)=0 x³(x-2)-2(x-2)(3x-2)=0 (x-2)(x³-6x+4)=0 either x=2 or x³-4x-2x+4=0 x(x²-4)-2(x-2)=0 x(x-2)(x+2)-2(x-2)=0 (x-2)(x²+2x-2)=0 either x=2 or x=-1±√3
If you want to graph it, take eqn 1 and factorise it (x-1)(y-1) = 1 This iz a rectangular hyperbola centred on (1,1) The solutions are where it intersects with the circle in eqn 2.
first notice that 2 is an obvious solution, and d/dx((x^2-8)(x-1)^2+x^2)=2(2x^3 - 3 x^2 - 6 x + 8) is also 0 for x=2 so 2 is a root of order 2 : (x^2-8)(x-1)^2+x^2=(x-2)^2(x^2+ax+b) for some a and b, identifying the coefficients yield a=2, b=-2, and the two remaining roots are easy to compute (sqrt(3)-1 and -1-sqrt(3))
Other approach - completing the square: x²+(x/(x-1))²+2(x)(x/(x-1))-2(x)(x/(x-1)) = 8 (x+x/(x-1))²-2x²/(x-1) = 8 ((x²+x-x)/(x-1))²-2x²/(x-1) = 8 (x²/(x-1))²-2x²/(x-1)-8 = 0 We get the quadratic in x²/(x-1) with two roots: x²/(x-1)=4 or x²/(x-1)=-2 i.e. x²-4x+4=0 or x²+2x-2=0. Finally we get 4 real roots: x=2 (double), x=-1+√3, x=-1-√3
In the quartic expansion x^4 - 2x^3 - 6x"2 + 16x - 8 = 0 put x = 2u 16u^4 - 16u^3 - 24u^2 + 32u - 8 = 0 2u^4 - 2u^3 - 3u"2 + 4u - 1 = 1 [1] Sum of coefficients is 0 so u=1, x=2 is a solution. Since x/(x-1) is involutive i.e. self-inverse, if x is a solution so is x/(x-1). With x=2, x/(x-1)=2. So x=2 is a repeated root, likewise u=1. Dividing [1] by (u-1)^2=(u^2 - 2u + 1) leaves 2u^2 + 2u - 1 = 0 u = (-2 +/- sqrt(4+8))/4 u = (-1 +/- sqrt(3))/2 x = -1 +/- sqrt(3)
x^4-2x^3-6x^2+16x-8=0..x^4-16-2x^3-6x^2+16x+8=0..(x^4-16)-(2x^3+6x^2-16x-8)=0...(x^2+4)(x^2-4)-2(x-2)(x^2+5x+2)=0...quindi una prima soluzione è x=2...semplifico rimane una cubica x^3-6x+4=0..(x-2)(x^2+2x-2)=0...ancora x=2,inoltre x=-1+√3,x=-1-√3
These types of equations seem to be tremendously popular on TH-cam math channels. You already discussed this same equation a year ago, but unlike this time, you got it right then: th-cam.com/video/dS38tvwlPj4/w-d-xo.html My preferred method to solve this equation is to start by multiplying both sides by (x − 1)² to get rid of the fraction. Then we can proceed as follows: x²(x − 1)² + x² = 8(x − 1)² x²(x² − 2x + 1) + x² = 8(x − 1)² x²(x² − 2x + 2) = 8(x − 1)² (x² − x + 1 + (x − 1))(x² − x + 1 − (x − 1)) = 8(x − 1)² (x² − x + 1)² − (x − 1)² = 8(x − 1)² (x² − x + 1)² − 9(x − 1)² = 0 (x² − x + 1 − 3(x − 1))(x² − x + 1 + 3(x − 1)) = 0 (x² − 4x + 4)(x² + 2x − 2) = 0 x² − 4x + 4 = 0 ∨ x² + 2x − 2 = 0 (x − 2)² = 0 ∨ (x + 1)² − 3 = 0 x = 2 ∨ x = −1 + √3 ∨ x = −1 − √3 All that was needed was the difference of two squares identity, first to convert the product x²(x² − 2x + 2) at the left hand side into a difference of two squares (x² − x + 1)² − (x − 1)², and then, after reducing the right hand side to zero, once again to factor the left hand side (x² − x + 1)² − 9(x − 1)² into two quadratics. You can turn any product of two quantities into a difference of two squares by taking their average and rewriting the quantities as their average plus and minus half their difference. Here we had the product of x² and x² − 2x + 2. Their average is x² − x + 1 and their difference is 2x − 2 = 2(x − 1) so we have x² = (x² − x + 1) + (x − 1) and x² − 2x + 2 = (x² − x + 1) − (x − 1). Using the difference of two squares identity (a − b)(a + b) = a² − b² we can therefore rewrite the product x²(x² − 2x + 2) as (x² − x + 1)² − (x − 1)². With regard to your first method which, as so often, you didn't finish, there is _no need to depress a quartic in order to solve it_ using Ferrari's method. At 1:15 you have: x⁴ − 2x³ + 2x² = 8x² − 16x + 8 Subtract x² from both sides to make the left hand side into a perfect square which gives us (x² − x)² = 7x² − 16x + 8 Now we need to create a perfect square on the right hand side as well, but we need to do that in such a way that the left hand side will remain a perfect square. Taking any number k we can add 2k(x² − x) + k² to both sides, which gives us (x² − x)² + 2k(x² − x) + k² = 7x² − 16x + 8 + 2k(x² − x) + k² so we get (x² − x + k)² = (2k + 7)x² − (2k + 16)x + (k² + 8) The left hand side remains a perfect square regardless of the value of k, so we are now free to choose k in such a way that the right hand side will also become a perfect square. The quadratic in x at the right hand side will be a perfect square if its discriminant is zero, that is, if k satisfies (2k + 16)² − 4(2k + 7)(k² + 8) = 0. This is a cubic equation in k, but most of the time quartic equations with integer coefficients from competitions have a nice factorization into two quadratics with integer coefficients in which case we can save ourselves the effort of resolving the cubic equation in k formally. If our equation has a nice factorization into two quadratics with integer coefficients, then there must exist a value of k which makes the right hand side of our equation a square of a linear polynomial in x with an _integer coefficient_ of x, so the coefficient (2k + 7) of x² at the right hand side must then be the _square of an integer._ Also, the constant term (k² + 8) must then be the _square of a rational number_ (not necessarily the square of an integer since there are odd multiples of ½ for k which make 2k + 7 the square of an integer). So, we only need to test values of k which make 2k + 7 equal to 1, 4, 9 ..., that is, k = −3, −1½, 1... and see if this makes k² + 8 the square of a rational number (actually, the square of an integer multiple of ½). Then, we quickly find that k = 1 fits the bill because this makes 2k + 7 = 9 = 3² while k² + 8 = 9 = 3² is also the square of an integer. Choosing k = 1 our equation (x² − x + k)² = (2k + 7)x² − (2k + 16)x + (k² + 8) becomes (x² − x + 1)² = 9x² − 18x + 9 which is (x² − x + 1)² = (3x − 3)² Bringing over the square from the right hand side to the left hand side we have (x² − x + 1)² − (3x − 3)² = 0 and applying the difference of two squares identity to the left hand side this gives (x² − 4x + 4)(x² + 2x − 2) = 0 and the zero product property then implies that we have x² − 4x + 4 = 0 ∨ x² + 2x − 2 = 0 so we end up with the exact same quadratic equations as in my first method.
10:10 there is a mistake, the equation should be t^2+2t-2 = 0, and this one has 2 real solutions
A^2+B^2 = 8 where A= x , B = x/(x-1) hence A+B = AB
(A+B)^2 - 2 AB = 8
(AB)^2 -2 AB -8 = 0
(AB +2) (AB - 4) = 0
AB = - 2 , 4
x^2/(x-1) = - 2 , 4
x^2 + 2x - 2 = 0 , x^2 - 4x +4 = 0
x= -1+√3 , -1 - √3 , 2
x⁴-2x³-6x²+16x-8=0
x³(x-2)-2(3x²-8x+4)=0
x³(x-2)-2(x-2)(3x-2)=0
(x-2)(x³-6x+4)=0
either x=2 or
x³-4x-2x+4=0
x(x²-4)-2(x-2)=0
x(x-2)(x+2)-2(x-2)=0
(x-2)(x²+2x-2)=0
either x=2 or
x=-1±√3
*AMEN.*
X = 2 OR X = -1±V3 ,
That's what I got too!
If you want to graph it, take eqn 1 and factorise it
(x-1)(y-1) = 1
This iz a rectangular hyperbola centred on (1,1)
The solutions are where it intersects with the circle in eqn 2.
first notice that 2 is an obvious solution, and d/dx((x^2-8)(x-1)^2+x^2)=2(2x^3 - 3 x^2 - 6 x + 8) is also 0 for x=2 so 2 is a root of order 2 : (x^2-8)(x-1)^2+x^2=(x-2)^2(x^2+ax+b) for some a and b, identifying the coefficients yield a=2, b=-2, and the two remaining roots are easy to compute (sqrt(3)-1 and -1-sqrt(3))
Obvious solution x=2 : 2^2 + (2/2-1)^2 = 4 + 4 = 8
Multiply both sides by (x-1)^2 in order to remove the denominator (with x ≠1)➔
x^4 - 2*x^3 - 6*x^2 + 16*x - 8 = 0 ➔ polynomial division by x-2 ➔
x^4 - 2*x^3 - 6*x^2 + 16*x - 8 = (x-2)*(x^3- 6*x+4) = (x-2)*(x-2)*(x^2+2*x -2) = 0 ➔
x = 2 (double root) and x = -1 +- sqrt(3)
Other approach - completing the square:
x²+(x/(x-1))²+2(x)(x/(x-1))-2(x)(x/(x-1)) = 8
(x+x/(x-1))²-2x²/(x-1) = 8
((x²+x-x)/(x-1))²-2x²/(x-1) = 8
(x²/(x-1))²-2x²/(x-1)-8 = 0
We get the quadratic in x²/(x-1) with two roots: x²/(x-1)=4 or x²/(x-1)=-2 i.e. x²-4x+4=0 or x²+2x-2=0.
Finally we get 4 real roots: x=2 (double), x=-1+√3, x=-1-√3
I just looked at the problem and in 6 seconds came up with x = 2. I'll work a bit longer to see if other solutions exist.
Please check your microphone position. The tapping of your stylus on the screen sounds almost as loud as your voice. Thank you for your videos!
In the quartic expansion
x^4 - 2x^3 - 6x"2 + 16x - 8 = 0
put x = 2u
16u^4 - 16u^3 - 24u^2 + 32u - 8 = 0
2u^4 - 2u^3 - 3u"2 + 4u - 1 = 1 [1]
Sum of coefficients is 0 so u=1, x=2 is a solution.
Since x/(x-1) is involutive i.e. self-inverse, if x is a solution so is x/(x-1). With x=2, x/(x-1)=2. So x=2 is a repeated root, likewise u=1. Dividing [1] by (u-1)^2=(u^2 - 2u + 1) leaves
2u^2 + 2u - 1 = 0
u = (-2 +/- sqrt(4+8))/4
u = (-1 +/- sqrt(3))/2
x = -1 +/- sqrt(3)
(x-2)(x-2)(x^2+2x-2)=0 ,
x^4-2x^3-6x^2+16x-8=0..x^4-16-2x^3-6x^2+16x+8=0..(x^4-16)-(2x^3+6x^2-16x-8)=0...(x^2+4)(x^2-4)-2(x-2)(x^2+5x+2)=0...quindi una prima soluzione è x=2...semplifico rimane una cubica x^3-6x+4=0..(x-2)(x^2+2x-2)=0...ancora x=2,inoltre x=-1+√3,x=-1-√3
How nice 😊
I actually got three real solutions for x, one of them being a double root.
Is it weird that I could do this in my head in ~3 seconds?
The complex solutions too?
Crazy! I did this problem in my head with u sub and came up with {2, -1+-√3} for x.
I got via 4+ 4 = 8 .... Its done
X= 2
These types of equations seem to be tremendously popular on TH-cam math channels. You already discussed this same equation a year ago, but unlike this time, you got it right then:
th-cam.com/video/dS38tvwlPj4/w-d-xo.html
My preferred method to solve this equation is to start by multiplying both sides by (x − 1)² to get rid of the fraction. Then we can proceed as follows:
x²(x − 1)² + x² = 8(x − 1)²
x²(x² − 2x + 1) + x² = 8(x − 1)²
x²(x² − 2x + 2) = 8(x − 1)²
(x² − x + 1 + (x − 1))(x² − x + 1 − (x − 1)) = 8(x − 1)²
(x² − x + 1)² − (x − 1)² = 8(x − 1)²
(x² − x + 1)² − 9(x − 1)² = 0
(x² − x + 1 − 3(x − 1))(x² − x + 1 + 3(x − 1)) = 0
(x² − 4x + 4)(x² + 2x − 2) = 0
x² − 4x + 4 = 0 ∨ x² + 2x − 2 = 0
(x − 2)² = 0 ∨ (x + 1)² − 3 = 0
x = 2 ∨ x = −1 + √3 ∨ x = −1 − √3
All that was needed was the difference of two squares identity, first to convert the product x²(x² − 2x + 2) at the left hand side into a difference of two squares (x² − x + 1)² − (x − 1)², and then, after reducing the right hand side to zero, once again to factor the left hand side (x² − x + 1)² − 9(x − 1)² into two quadratics.
You can turn any product of two quantities into a difference of two squares by taking their average and rewriting the quantities as their average plus and minus half their difference. Here we had the product of x² and x² − 2x + 2. Their average is x² − x + 1 and their difference is 2x − 2 = 2(x − 1) so we have x² = (x² − x + 1) + (x − 1) and x² − 2x + 2 = (x² − x + 1) − (x − 1). Using the difference of two squares identity (a − b)(a + b) = a² − b² we can therefore rewrite the product x²(x² − 2x + 2) as (x² − x + 1)² − (x − 1)².
With regard to your first method which, as so often, you didn't finish, there is _no need to depress a quartic in order to solve it_ using Ferrari's method. At 1:15 you have:
x⁴ − 2x³ + 2x² = 8x² − 16x + 8
Subtract x² from both sides to make the left hand side into a perfect square which gives us
(x² − x)² = 7x² − 16x + 8
Now we need to create a perfect square on the right hand side as well, but we need to do that in such a way that the left hand side will remain a perfect square. Taking any number k we can add 2k(x² − x) + k² to both sides, which gives us
(x² − x)² + 2k(x² − x) + k² = 7x² − 16x + 8 + 2k(x² − x) + k²
so we get
(x² − x + k)² = (2k + 7)x² − (2k + 16)x + (k² + 8)
The left hand side remains a perfect square regardless of the value of k, so we are now free to choose k in such a way that the right hand side will also become a perfect square. The quadratic in x at the right hand side will be a perfect square if its discriminant is zero, that is, if k satisfies (2k + 16)² − 4(2k + 7)(k² + 8) = 0. This is a cubic equation in k, but most of the time quartic equations with integer coefficients from competitions have a nice factorization into two quadratics with integer coefficients in which case we can save ourselves the effort of resolving the cubic equation in k formally.
If our equation has a nice factorization into two quadratics with integer coefficients, then there must exist a value of k which makes the right hand side of our equation a square of a linear polynomial in x with an _integer coefficient_ of x, so the coefficient (2k + 7) of x² at the right hand side must then be the _square of an integer._ Also, the constant term (k² + 8) must then be the _square of a rational number_ (not necessarily the square of an integer since there are odd multiples of ½ for k which make 2k + 7 the square of an integer).
So, we only need to test values of k which make 2k + 7 equal to 1, 4, 9 ..., that is, k = −3, −1½, 1... and see if this makes k² + 8 the square of a rational number (actually, the square of an integer multiple of ½). Then, we quickly find that k = 1 fits the bill because this makes 2k + 7 = 9 = 3² while k² + 8 = 9 = 3² is also the square of an integer. Choosing k = 1 our equation
(x² − x + k)² = (2k + 7)x² − (2k + 16)x + (k² + 8)
becomes
(x² − x + 1)² = 9x² − 18x + 9
which is
(x² − x + 1)² = (3x − 3)²
Bringing over the square from the right hand side to the left hand side we have
(x² − x + 1)² − (3x − 3)² = 0
and applying the difference of two squares identity to the left hand side this gives
(x² − 4x + 4)(x² + 2x − 2) = 0
and the zero product property then implies that we have
x² − 4x + 4 = 0 ∨ x² + 2x − 2 = 0
so we end up with the exact same quadratic equations as in my first method.
The correct solution is ..
let x = u + 1
(u + 1)^2 + (u + 1 / u)^2 - 8 = 0
(u + 1)^2 + (u^-1 + 1)^2 - 8 = 0
u^2 + 2u + 1 + u^-2 + 2u^-1 + 1 - 8 = 0
(u^2 + 2 + u^-2) + (2u + 2u^-1) - 8 = 0
(u + u^-1)^2 + 2(u + u^-1) - 8 = 0
let (u + u^-1) = t
t^2 + 2t - 8 = 0
.
.
.
x = 2;
x = -1 + sqrt(3);
x = -1 - sqrt(3);
x ≠ 1, a = x/(x-1) => ax-a = x => x+a = ax and x²+y² = 8 = (x+a)²-2ax => 8+1 = (ax-1)²
ax = 1±3 => ax = 4 = a+x => x(4-x) = 4 => x²-4x +4 = 0 => x = 2 or ax = -2 = a+x => x(-2-x) = -2 => x² +2x -2 = 0 = (x+1)²-3 => x = -1±V3
There are only 3 solutions, all reals: x1=2 ; x2=-1+3^0.5 ; x3=-1-3^0.5.
No ! 4 solutions : x1 =2 is double root, x2=-1+3^0.5 ; x3=-1-3^0.5.
x = 2
problem
x² + [ x / ( x - 1 ) ]² = 8
Let
u = x - 1
x = u + 1
( u + 1 )² + [ ( u + 1 ) / u ]² = 8
Factor out ( u + 1 )²
( u + 1 )² [ 1 + 1 / (u²) ] = 8
( u + 1 )² ( u² + 1 ) / (u²) = 8
( u + 1 )² ( u² + 1 ) - 8 u² = 0
( u² + 1 ) ( u² + 2u + 1 ) - 8 u² = 0
Expand. Put into standard form.
u⁴ + 2 u³ + u² + u² + 2u + 1 -8 u² = 0
u⁴ + 2 u³- 6 u² + 2u + 1 = 0
Σ coeffs. = 0, u = 1 is a root, u-1 a factor.
u⁴ + 2 u³- 6 u² + 2u + 1 = 0
(u-1) u³ + 3 (u-1) u²-3(u-1) u-(u-1) = 0
(u-1) (u³ + 3 u²-3 u - 1) = 0
By zero-product property
u³ + 3 u²-3 u - 1 = 0
Σ coeffs. = 0, u = 1 is a root, u-1 a factor.
( u = 1 is a double root )
(u-1) u² + 4 (u-1) u + (u-1) = 0
(u-1) ( u² + 4 u + 1 ) = 0
By zero-product property
u² + 4 u + 1 = 0
By the quadratic formula
u = -2 ± √3
All the distinct u roots are then
u = 1
u = -2 - √3
u = -2 + √3
Back substitute x = u + 1
x = 2
x = -1 - √3
x = -1 + √3
answer
x ∈ { -1 -√3, -1 +√3, 2 }
Again!!