Solving the Quartic x^4-16x-12=0

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #PolynomialEquations

SIMILAR PROBLEMS

PLAYLISTS 🎵 :

[MrLidless]

Should use Descartes’ formula for a depressed quartic (ie no x^3 term), only since there isn’t a x^2 term either it is much easier, and should perhaps be called a manically depressed quartic.

[seegeeaye]

The equation is rewritten as x(x^3 – 16) = 12. The LHS is a concave up curve crossing the x-axis at 0 and cubic root of 16, by considering y = 12, the two real solutions (the intersecting points) : one is a little more than cubic root of 16, the other a little less than 0.

[laurentthais]

This is a special case of Galois factorization (symmetric group S4) which leads to a cubic equation for the z^2 complex variable.

[yoav]

I can not solve x^4-16x-12=0, but i can solve x^2+2x+6=0 and x^2-2x-2=0😀

[DrQuatsch]

I basically tried the first method, but with - c instead of + c, so I had bc = 12 instead of -12. Just a preference.
It got me b - c - a^2 = 0, a(b + c) = 16 and bc = 12. Here I just took the time to see if integer factors of 12 would add up to a factor of 16, which 2 and 6 did. 2 x 6 = 12, and 2 + 6 = 8, which is a factor of 16. So if it worked out like that, a has to be 2, which I used in the first equation. 6 - 2 - 2^2 = 0, which checked out, where b had to be 6 and c had to be 2. So I had my factorisation into two quadratics.
The only logic I used here is that these type of problems usually have a nice way to factor the quartic down to a nicer to solve equation. This doesn't generally work, but it cannot hurt to check it first. Saved me a couple of minutes of writing.

[NadiehFan]

I do not understand why you keep solving quartic equations using the cumbersome method of factoring a depressed quartic into two quadratics using the method of indetermined coefficients, i.e. your first method, which is really just Descartes' method. You spend a full 7 minutes on setting up and solving your system in a, b and c. Your second method here is not really a method, because it is not applicable to an arbitrary quartic and depends on trial and error or on prior knowledge of a factorisation of your quartic into quadratic factors.

A really simple way to solve this equation is using an adaptation of Ferrari's method. First bring all terms which do not contain x⁴ or x³ over to the right hand side, which gives

x⁴ = 16x + 12

where the left hand side x⁴ = (x²)² is a square but the right hand side is not. Since (x² + t)² = x⁴ + 2tx² + t² the left hand side will remain a square if we add 2tx² + t² to both sides and this gives

(x² + t)² = 2tx² + 16x + (t² + 12)

Now, the left hand side is a square for any t, and the quadratic in x at the right hand side can also be made into a square if we choose t in such a way that this quadratic has two equal zeros. This is the case if the discriminant of the quadratic is zero, that is, if

16² − 4·2t·(t² + 12) = 0

Dividing both sides by 8 this simplifies to

32 − t(t² + 12) = 0

It is easy to see that t = 2 is a solution of this cubic equation in t. With t = 2, our equation becomes

(x² + 2)² = 4x² + 16x + 16

which can be written as

(x² + 2)² = (2x + 4)²

which is exactly what you got with your second 'method'. Of course, the solution of the equation is now easy. We can either bring over the square from the right hand side to the left hand side and apply the difference of two squares identity to get a factorisation of the original quartic x⁴ − 16x − 12 into (x² − 2x − 2)(x² + 2x + 6) or we can apply the principle that A² = B² is equivalent with A = B or A = −B to obtain x² + 2 = 2x + 4 ∨ x² + 2 = −2x − 4. Obviously these approaches are equivalent since

A² = B² ⇔ A² − B² = 0 ⇔ (A − B)(A + B) = 0 ⇔ A − B = 0 ∨ A + B = 0 ⇔ A = B ∨ A = −B

[SuperYoonHo]

thank you so much! these videos are really great and also encouraged me to do math olympiad problems and has worked out fine with the help from you and the thinking brain of mine. i think everyone would appreciate if you made a book. really. maybe you should try mixing up the ingrediants in your math videos starting from the first video to the latest one. that is this one. if you do it please let me know. i will be honoured to have such book. thank you once more. bye

[qwang]

Such an equation can only be solved if the relation among coefficients is perfect, just like we factor a quadratic equation.
As -12 = -2 x 6. We can try out luck with -2 and 6 (or -3 or 4).
(x^2 + ax + 6)(x^2 -ax - 2). It turns out a = 2 works. Done.
...

[MrLidless]

This has never been posted on the Internet, so this is the first:

A manically depressed quartic, x⁴ + px + q = 0 can be factored into a biquadratic (x² + ax + b)(x² - ax + c) = 0, each of which can then be easily solved, where:

a = √{³√[p²/2 + √(p⁴/4 - 64q³/27)]+ ³√[p²/2 - √ (p⁴/4 - 64q³/27)]}
b = (a² - p/a)/2
c = (a² + p/a)/2

[akssumusic]

Interesting problem idea:
You have the graph y=1/x
How many radians does it take to rotate the equation clockwise about the origin for it to intersect the point (2, 0)

[PunmasterSTP]

Solving the quartic? More like "Super-cool math, isn't it?" Thanks for sharing!

[elkincampos]

First this polynomial has no rational roots. if a root is integer number then a^4=16*a+12=4*(4*a+3). Thus 2 divides a^4. It follows 2 divides a. Therefore 2^4 divides a^4. But 2^3 doesn't divide 4*(4*a+3) contradition.

[srividhyamoorthy]

Nice video I now have to go to school cuz I was on holidays all these days i missed few vids but love your vids

[ilana]

thank you dear teacher! two remarks:
explanation is very fast.
it is difficult to see the notes on the board and this interferes with understanding. the previous videos were devoid of these flaws ..
With gratitude and respect!

[mahdiali]

Sir you are best teacher but we need more video

[gregwochlik]

On this one, I cheated. I re-activated my graphing calculator bought in the year 1999. I got it to graph it for me first, to check for real solutions. Then I got it to give the decimal roots to me.
I have also made a discovery in Python (programming language). In the "numpy" module, there is a "Polynomial" class. It gives you the roots by calling a method.

[moeberry]

I just used the quartic formula it was much easier that way.

[gregedmondson]

Not well explained for students learning. Tutor goes too fast without explaining each step.

[holyshit]

I run my program for numerical eigenvalues and it was easy to read values from decimal expansion

If you want to write program or record video on calculating eigenvalues numerically i suggest
Calculate Hessenberg reduction via Householder reflections
but you should write / show how to multiply by Householder matrices from both left and right
It can be done more efficently than inner product of each entry but nobody shows that
QR decomposition for Hessenberg matrix - it doesn't matter if you use rotations or reflections
but rotations are a little bit easier to derive
Clever choice of shift
For me it wolud be enough to use Wilkinson shift or doubly implicit shift - in case of complex eigenvalues
Deflation - maybe if our QR decomposition works on blocks of matrix instead of whole n by n matrix it will be easier to
write deflation without increasing space complexity