A Nice Rational Equation from Crux Mathematicorum
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- Опубликовано: 7 фев 2025
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This was fun to solve. My aha moment was seeing that I could add 12x and make a perfect fourth power. Then it turned nicely into two similar quadratics which were not fun to solve.
indeed a very nice problem
Nice!
At 7:52 when you want to solve the quartic equation
x⁴ − 4x³ − 6x² − 4x + 1 = 0
you should _not_ start by depressing this equation, since this will only complicate things and is redundant when using Ferrari's method for solving quartic equations. First we bring all terms of a lower than the third degree over to the right hand side, which gives
x⁴ − 4x³ = 6x² + 4x − 1
Observe that at the left hand side we have x⁴ = (x²)² and 4x³ = 2·(x²)·(2x) so we can complete the square at the left hand side by adding (2x)² = 4x² to both sides since (x²)² − 2·(x²)·(2x) + (2x)² = (x² − 2x)². So, adding 4x² to both sides we get
(x² − 2x)² = 10x² + 4x − 1
Now, if we take any number k and add 2k(x² − 2x) + k² = 2kx² − 4kx + k² to both sides, then the left hand side will _remain_ a perfect square regardless of the value of k, because at the left hand side we then have (x² − 2x)² + 2·(x² − 2x)·k + k² = (x² − 2x + k)². So, adding 2k(x² − 2x) + k² = 2kx² − 4kx + k² to both sides we get
(x² − 2x + k)² = (10 + 2k)x² + (4 − 4k)x + (k² − 1)
Since the left hand side remains a perfect square regardless of the value of k, we are now free to select a value of k that will make the right hand side a perfect square as well. However, there is no need to calculate the discriminant of the quadratic in x at the right hand side and set this discriminant equal to zero and then solve the resulting cubic equation in k. It is easy to see that for k = 1 both the linear term (4 − 4k)x and the constant term (k² − 1) vanish, leaving only the quadratic term (10 + 2k)x² which equals 12x² for k = 1. So, with k = 1 we get
(x² − 2x + 1)² = 12x²
and since x² − 2x + 1 = (x − 1)² this can be written as
(x − 1)⁴ = 12x²
which is of course exactly the equation you obtained at 5:14.
Wow! This is pretty good. Thanks for sharing 😍
@@SyberMath Also note that when you depressed the quartic by substituting x = y + 1 you obtained
y⁴ = 12y² + 24y + 12
at 8:18 after bringing over all terms of a lower than the third degree to the right hand side. You then claim that adding 2ky² + k² to both sides didn't give you anything nice. Really?
Take a closer look at the right hand side of your equation in y. If we take out the common factor 12 we have
y⁴ = 12(y² + 2y + 1)
or
y⁴ = 12(y + 1)²
So, the right hand side actually _already is a perfect square_ and there is therefore no need to add anything to both sides to make the right hand side into a perfect square. Also note that if we back substitute y = x − 1 in this equation we get (x − 1)⁴ = 12x² which is exactly the equation obtained by applying Ferrari's method to the original quartic in x.
But let's suppose we don't notice that the right hand side of y⁴ = 12y² + 24y + 12 already is a perfect square and proceed by adding 2ky² + k² to both sides of the equation in y, then we obtain
(y² + k)² = (2k + 12)y² + 24y + (k² + 12)
The quadratic in y at the right hand side will be a perfect square, that is, the square of a linear polynomial in y, if and only if its discriminant is zero, which is the case if and only if k satisfies
24² − 4(2k + 12)(k² + 12) = 0
which reduces to
k³ + 6k² + 12k = 0
Do you see what I see? This cubic equation in k has no constant term, so k = 0 is a solution. This makes perfect sense, because, as we already saw, the right hand side 12y² + 24y + 12 already is a perfect square, so there is no need to add anything to both sides. Of course, adding 2ky² + k² to both sides with k = 0 amounts to adding nothing at all to both sides.
k=0, since 12y^2+24y+12 = (2 sqrt(3) (y+1))^2
That's great. I didn't have a clue about this one, and you made it easy - method 2.
Glad to hear that!
After noting that 1 is not a solution to the equation, we can put x=y+1/y-1 and get the quadratic equation 3y⁴-6y²-1=0. By solving this equation with the unknown y, we deduce the solutions to the equation with the unknown x.