Whenever solving a quartic, the first thing to do is to depress it by removing the cubic term with a substitution of x = t - b/4 where b is the coefficient of the cubic term (assuming the quartic term has coefficient 1). In this case, that substitution basically solves the problem, since it produces a biquadratic (a quadratic in x^2) which is easily solved. It even factors, no need to complete the square.
@@XJWill1 Well I don't think so. It is a matter of choosing between substituting x = t − ¹⁄₂ in x⁴ + 2x³ − x = 6 to get (t − ¹⁄₂)⁴ + 2(t − ¹⁄₂)³ − (t − ¹⁄₂) = 6 and expanding this to get t⁴ − ³⁄₂t² − ⁹¹⁄₁₆ = 0 or rewriting the equation as (x² + x)² = x² + x + 6 and adding 2k(x² + x) + k² = 2kx² + 2kx + k² to both sides to get (x² + x + k)² = (2k + 1)x² + (2k + 1)x + (k² + 6) and noting right away that with k = −¹⁄₂ the equation reduces to (x² + x − ¹⁄₂)² = (⁵⁄₂)² This is a lot less work compared to expanding (t − ¹⁄₂)⁴ and (t − ¹⁄₂)³ and besides, this immediately gives two quadratics x² + x − 3 = 0 and x² + x + 2 = 0 with integer coefficients. Solving the biquadratic t⁴ − ³⁄₂t² − ⁹¹⁄₁₆ = 0 is not difficult of course if you see that this factors as (t² − ¹³⁄₄)(t² + ⁷⁄₄) = 0 but this does not outweigh the effort of expanding (t − ¹⁄₂)⁴ and (t − ¹⁄₂)³ and collecting like terms. Also, as I noted in my main comment on this video, it is enough to note that (x² + x)² = x² + x + 6 can be written as (x² + x)² − 2·(x² + x)·(¹⁄₂) = 6 and to add (¹⁄₂)² = ¹⁄₄ to both sides to get (x² + x − ¹⁄₂)² = (⁵⁄₂)²
@@NadiehFan You are welcome to your fringe opinion. Almost everyone else can see that a simple substitution to remove the cubic term is quicker and easier. But you do you.
the equation can be manipulated to: (x^2 + x - 1/2)^2 = 25/4 giving 2 quadratic equations viz. x^2 + x -1/2 = +/- 5/2. Solving these gives the 4 distinct roots (2 real and 2 complex).
The discussion at 4:00 about completing x⁴ + 2x³ into a perfect square is a bit more complicated than it needed to be, because if we write this as (x²)² + 2·(x²)·x and compare this with a² + 2·a·b + b² = (a + b)² it is immediately clear that we need to add x² to get x⁴ + 2x³ + x² = (x²)² + 2·(x²)·x + x² = (x² + x)² You missed a nice opportunity to demonstrate that Ferrari's method for solving quartics essentially boils down to completing the square _twice._ When at 6:40 you have arrived at (x² + x)² = x² + x + 6 you recognize the common expression x² + x at the left hand side and the right hand side and infer that this calls for a substitution. That is an option, but note that we can bring x² + x over to the left hand side to get (x² + x)² − (x² + x) = 6 which we can write as (x² + x)² − 2·(x² + x)·(¹⁄₂) = 6 Do you see what I see? If we compare the left hand side with the identity a² − 2·a·b + b² = (a + b)² we immediately see that we can _again_ complete the left hand side into a perfect square by adding (¹⁄₂)² = ¹⁄₄ to both sides, which gives (x² + x)² − 2·(x² + x)·(¹⁄₂) + (¹⁄₂)² = 6 + ¹⁄₄ which can be written as (x² + x − ¹⁄₂)² = ²⁵⁄₄ or (x² + x − ¹⁄₂)² = (⁵⁄₂)² and applying the equal squares property A² = B² ⟺ A = B ⋁ A = −B (no, _not_ taking square roots) which says that squares of two quantities are equal _if and only if_ the quantities themselves are either equal or each others opposite this gives x² + x − ¹⁄₂ = ⁵⁄₂ ⋁ x² + x − ¹⁄₂ = −⁵⁄₂ x² + x = 3 ⋁ x² + x = −2 (x + ¹⁄₂)² = ¹³⁄₄ ⋁ (x + ¹⁄₂)² = −⁷⁄₄ (x + ¹⁄₂)² = (¹⁄₂√13)² ⋁ (x + ¹⁄₂)² = (i·¹⁄₂√7)² x + ¹⁄₂ = ¹⁄₂√13 ⋁ x + ¹⁄₂ = −¹⁄₂√13 ⋁ x + ¹⁄₂ = i·¹⁄₂√7 ⋁ x + ¹⁄₂ = −i·¹⁄₂√7 x = −¹⁄₂ + ¹⁄₂√13 ⋁ x = −¹⁄₂ − ¹⁄₂√13 ⋁ x = −¹⁄₂ + i·¹⁄₂√7 ⋁ x = −¹⁄₂ − i·¹⁄₂√7 All in all we have used completing the square four times and used the equal squares property three times to solve the quartic 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 Of course, we can also use Ferrari's method proper to solve the quartic equation x⁴ + 2x³ − x = 6 _without_ first depressing the quartic equation (that is, getting rid of the cubic term by means of a substitution). After bringing over all terms of the second and lower degrees to the right hand side and then completing the square at the left hand side, in this case by adding x² to both sides, we have (x² + x)² = x² + x + 6 Now, if we take any number k and add 2·k·(x² + x) + k² = 2kx² + 2kx + k² to both sides we have (x² + x)² + 2·k·(x² + x) + k² = x² + x + 6 + 2kx² + 2kx + k² which can be written as (x² + x + k)² = (2k + 1)x² + (2k + 1)x + (k² + 6) We have now completed the square at the left hand side a second time, and the left hand side will remain a perfect square regardless of the value of k, so we are now free to choose k in such a way that the quadratic in x at the right hand side will also become a perfect square. Generally, a quadratic polynomial ax² + bx + c with a ≠ 0 will be a perfect square, that is, the square (px + q)² of a linear polynomial px + q with p ≠ 0, if this quadratic has two coinciding and therefore identical zeros, which is the case if and only if the _discriminant_ b² − 4ac of the quadratic is equal to zero. For the quadratic in x at the right hand side of our equation we have a = 2k + 1, b = 2k + 1, c = k² + 6 so we will have b² − 4ac = 0 if k satisfies (2k + 1)² − 4(2k + 1)(k² + 6) = 0. This is a cubic equation in k, but we do not need to solve this equation. Since a = b = 2k + 1 we can immediately see that b² − 4ac = 0 if 2k + 1 = 0, that is, if k = −¹⁄₂. In fact, with this value for k, the right hand side of our equation no longer is a polynomial in x since it then reduces to a constant ²⁵⁄₄ = (⁵⁄₂)² so our equation becomes (x² + x − ¹⁄₂)² = (⁵⁄₂)² and this is of course the exact same equation we have already solved. Finally, your remark at 3:20 about a general formula to solve quintics. In spite of your insistence that it doesn't exist it actually does exist, but it _is_ true that no general formulas can exist which express the zeros of polynomials of degree five or higher algebraically with radicals in terms of their coefficients. I'm almost tempted here to say that you should be careful what you wish for because you may just get it, but I would recommend that you start by studying Felix Klein, _Vorlesungen über das Ikosaeder und die Auflösung der Gleichungen vom fünften Grade._ If you don't read German, don't worry, because there is an English translation of this book _Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree_ which is still available as a Dover reprint. You can also find the original German edition of the book legally for free on the internet archive since it is long out of copyright, and the English translation can also be borrowed digitally there. If you want to do a video on a nontrivial but algebraically solvable quintic you may want to take a look at x⁵ + 5x³ + 5x − 1 = 0 which can be solved in at least two different ways.
I got to the answer in a more haphazard way, but still managed to do it on my lunch break (without skipping lunch!), so I'm pretty pleased with myself. Starting with x⁴ + 2x³ -x = 6, I first tried to factor the left-hand side in a useful way. After some trial-and-error, I figured out that (x³-1) (x+2) was equal to x⁴ + 2x³ - x - 2. This means that (x³-1) (x+2) = 4. And because x³-1 is the difference of 2 cubes, the equation can further be factored as (x-1) (x²+x+1) (x+2) = 4 Now, multiplying (x-1) and (x+2), you arrive at (x²+x+1) (x²+x-2) = 4. Then, if you posit that t = x² + x, the equation becomes (t+1) (t-2) = 4, and finally t² - t - 2 = 4, or t² - t - 6 = 0. From there you proceed as in Sybermath's 2nd method (from 8:00) and I have nothing meaningful to add. (Not that any of the above was all that meaningful to begin with.)
Whenever solving a quartic, the first thing to do is to depress it by removing the cubic term with a substitution of x = t - b/4 where b is the coefficient of the cubic term (assuming the quartic term has coefficient 1).
In this case, that substitution basically solves the problem, since it produces a biquadratic (a quadratic in x^2) which is easily solved. It even factors, no need to complete the square.
Actually I never depress a quartic because that is unnecessary when using Ferrari's method.
@@NadiehFan Then you would waste your effort on equations like this one.
@@XJWill1 Well I don't think so. It is a matter of choosing between substituting
x = t − ¹⁄₂
in
x⁴ + 2x³ − x = 6
to get
(t − ¹⁄₂)⁴ + 2(t − ¹⁄₂)³ − (t − ¹⁄₂) = 6
and expanding this to get
t⁴ − ³⁄₂t² − ⁹¹⁄₁₆ = 0
or rewriting the equation as
(x² + x)² = x² + x + 6
and adding 2k(x² + x) + k² = 2kx² + 2kx + k² to both sides to get
(x² + x + k)² = (2k + 1)x² + (2k + 1)x + (k² + 6)
and noting right away that with k = −¹⁄₂ the equation reduces to
(x² + x − ¹⁄₂)² = (⁵⁄₂)²
This is a lot less work compared to expanding (t − ¹⁄₂)⁴ and (t − ¹⁄₂)³ and besides, this immediately gives two quadratics x² + x − 3 = 0 and x² + x + 2 = 0 with integer coefficients. Solving the biquadratic t⁴ − ³⁄₂t² − ⁹¹⁄₁₆ = 0 is not difficult of course if you see that this factors as (t² − ¹³⁄₄)(t² + ⁷⁄₄) = 0 but this does not outweigh the effort of expanding (t − ¹⁄₂)⁴ and (t − ¹⁄₂)³ and collecting like terms. Also, as I noted in my main comment on this video, it is enough to note that
(x² + x)² = x² + x + 6
can be written as
(x² + x)² − 2·(x² + x)·(¹⁄₂) = 6
and to add (¹⁄₂)² = ¹⁄₄
to both sides to get
(x² + x − ¹⁄₂)² = (⁵⁄₂)²
@@NadiehFan You are welcome to your fringe opinion. Almost everyone else can see that a simple substitution to remove the cubic term is quicker and easier. But you do you.
Thank you for this! I've been working on Masayoshi's ellipse puzzle and have had to resort to an online calculator to solve the quartic.
I love this. The contrived are some of the best!
the equation can be manipulated to: (x^2 + x - 1/2)^2 = 25/4 giving 2 quadratic equations viz. x^2 + x -1/2 = +/- 5/2. Solving these gives the 4 distinct roots (2 real and 2 complex).
All quintics are solvable! It's just that not by radicals! Some by radicals others by alternate means!
Nooooo! 🤪
@@SyberMath: Yesssss!
Nice - that was challenging, but eventually I got there!
Excellent!
The discussion at 4:00 about completing x⁴ + 2x³ into a perfect square is a bit more complicated than it needed to be, because if we write this as
(x²)² + 2·(x²)·x
and compare this with a² + 2·a·b + b² = (a + b)² it is immediately clear that we need to add x² to get
x⁴ + 2x³ + x² = (x²)² + 2·(x²)·x + x² = (x² + x)²
You missed a nice opportunity to demonstrate that Ferrari's method for solving quartics essentially boils down to completing the square _twice._
When at 6:40 you have arrived at
(x² + x)² = x² + x + 6
you recognize the common expression x² + x at the left hand side and the right hand side and infer that this calls for a substitution. That is an option, but note that we can bring x² + x over to the left hand side to get
(x² + x)² − (x² + x) = 6
which we can write as
(x² + x)² − 2·(x² + x)·(¹⁄₂) = 6
Do you see what I see? If we compare the left hand side with the identity a² − 2·a·b + b² = (a + b)² we immediately see that we can _again_ complete the left hand side into a perfect square by adding (¹⁄₂)² = ¹⁄₄ to both sides, which gives
(x² + x)² − 2·(x² + x)·(¹⁄₂) + (¹⁄₂)² = 6 + ¹⁄₄
which can be written as
(x² + x − ¹⁄₂)² = ²⁵⁄₄
or
(x² + x − ¹⁄₂)² = (⁵⁄₂)²
and applying the equal squares property A² = B² ⟺ A = B ⋁ A = −B (no, _not_ taking square roots) which says that squares of two quantities are equal _if and only if_ the quantities themselves are either equal or each others opposite this gives
x² + x − ¹⁄₂ = ⁵⁄₂ ⋁ x² + x − ¹⁄₂ = −⁵⁄₂
x² + x = 3 ⋁ x² + x = −2
(x + ¹⁄₂)² = ¹³⁄₄ ⋁ (x + ¹⁄₂)² = −⁷⁄₄
(x + ¹⁄₂)² = (¹⁄₂√13)² ⋁ (x + ¹⁄₂)² = (i·¹⁄₂√7)²
x + ¹⁄₂ = ¹⁄₂√13 ⋁ x + ¹⁄₂ = −¹⁄₂√13 ⋁ x + ¹⁄₂ = i·¹⁄₂√7 ⋁ x + ¹⁄₂ = −i·¹⁄₂√7
x = −¹⁄₂ + ¹⁄₂√13 ⋁ x = −¹⁄₂ − ¹⁄₂√13 ⋁ x = −¹⁄₂ + i·¹⁄₂√7 ⋁ x = −¹⁄₂ − i·¹⁄₂√7
All in all we have used completing the square four times and used the equal squares property three times to solve the quartic 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
Of course, we can also use Ferrari's method proper to solve the quartic equation
x⁴ + 2x³ − x = 6
_without_ first depressing the quartic equation (that is, getting rid of the cubic term by means of a substitution). After bringing over all terms of the second and lower degrees to the right hand side and then completing the square at the left hand side, in this case by adding x² to both sides, we have
(x² + x)² = x² + x + 6
Now, if we take any number k and add 2·k·(x² + x) + k² = 2kx² + 2kx + k² to both sides we have
(x² + x)² + 2·k·(x² + x) + k² = x² + x + 6 + 2kx² + 2kx + k²
which can be written as
(x² + x + k)² = (2k + 1)x² + (2k + 1)x + (k² + 6)
We have now completed the square at the left hand side a second time, and the left hand side will remain a perfect square regardless of the value of k, so we are now free to choose k in such a way that the quadratic in x at the right hand side will also become a perfect square.
Generally, a quadratic polynomial ax² + bx + c with a ≠ 0 will be a perfect square, that is, the square (px + q)² of a linear polynomial px + q with p ≠ 0, if this quadratic has two coinciding and therefore identical zeros, which is the case if and only if the _discriminant_ b² − 4ac of the quadratic is equal to zero. For the quadratic in x at the right hand side of our equation we have a = 2k + 1, b = 2k + 1, c = k² + 6 so we will have b² − 4ac = 0 if k satisfies (2k + 1)² − 4(2k + 1)(k² + 6) = 0.
This is a cubic equation in k, but we do not need to solve this equation. Since a = b = 2k + 1 we can immediately see that b² − 4ac = 0 if 2k + 1 = 0, that is, if k = −¹⁄₂. In fact, with this value for k, the right hand side of our equation no longer is a polynomial in x since it then reduces to a constant ²⁵⁄₄ = (⁵⁄₂)² so our equation becomes
(x² + x − ¹⁄₂)² = (⁵⁄₂)²
and this is of course the exact same equation we have already solved.
Finally, your remark at 3:20 about a general formula to solve quintics. In spite of your insistence that it doesn't exist it actually does exist, but it _is_ true that no general formulas can exist which express the zeros of polynomials of degree five or higher algebraically with radicals in terms of their coefficients. I'm almost tempted here to say that you should be careful what you wish for because you may just get it, but I would recommend that you start by studying Felix Klein, _Vorlesungen über das Ikosaeder und die Auflösung der Gleichungen vom fünften Grade._ If you don't read German, don't worry, because there is an English translation of this book _Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree_ which is still available as a Dover reprint. You can also find the original German edition of the book legally for free on the internet archive since it is long out of copyright, and the English translation can also be borrowed digitally there.
If you want to do a video on a nontrivial but algebraically solvable quintic you may want to take a look at
x⁵ + 5x³ + 5x − 1 = 0
which can be solved in at least two different ways.
- 2 one of...?
I got to the answer in a more haphazard way, but still managed to do it on my lunch break (without skipping lunch!), so I'm pretty pleased with myself. Starting with x⁴ + 2x³ -x = 6, I first tried to factor the left-hand side in a useful way. After some trial-and-error, I figured out that
(x³-1) (x+2) was equal to x⁴ + 2x³ - x - 2.
This means that (x³-1) (x+2) = 4.
And because x³-1 is the difference of 2 cubes, the equation can further be factored as (x-1) (x²+x+1) (x+2) = 4
Now, multiplying (x-1) and (x+2), you arrive at (x²+x+1) (x²+x-2) = 4.
Then, if you posit that t = x² + x, the equation becomes (t+1) (t-2) = 4, and finally
t² - t - 2 = 4, or t² - t - 6 = 0.
From there you proceed as in Sybermath's 2nd method (from 8:00) and I have nothing meaningful to add.
(Not that any of the above was all that meaningful to begin with.)
This is very meaningful! Thank you for sharing 😍