A Special Nonic Equation | Math Olympiads

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[tanishkgoyal]

3:34 we could have factorised it by taking y common and applying y²-a² and taking y-a from the 2 terms

[copernic]

Another way is by guessing √6 as a solution of the cubic equation, then factorizing as such:
(y-√6)(y²+y√6-1)=0
which can easily be solved.

[BabisKoutroulis]

There's an easier way to find the roots: y^3 - 6y - y +\|6 =y (y^2 -( \|6)^2) - (y-\|6) = 0... The rest is obvious.

[demenion]

similar to the problem with the imaginary constant, we can simply use algebra here as well to make our lives easier. since the cubic equation only contains odd powers of y and the constant is just a multiple of sqrt(6), we know that we can rationalize the equation by defining y=sqrt(6)*u. then 6u³-7u+1=0 which is obviously solved by u=1 and the other two solutions follow from the quadratic formula.

[goldfing]

This "inverting" of the equation is very nice, but I have to say that in this case, you can easily see at 3:00 already, that the cubic equation y^3 - (a^2 + 1)y + a = 0 has a solution y = a (which you only got much later at 6:45, after lots of calculations :-), because if you plug in y = a, you get

a^3 - (a^2 + 1)a + a =
a^3 - (a^3 + a) + a =
a^3 - a^3 - a + a
= 0.

So you can apply polynomial divison:

(y^3 - (a^2 + 1)y + a) : (y - a) = y^2 + ay - 1
- (y^3 - ay^2)

ay^2 - a^2y - y + a
- (ay^2 - a^2y)

-y + a
- (-y + a)

0

The two remaining solutions are the roots of y^2 + ay - 1 = 0
y_2, 3 = (-a +- sqrt(a^2 + 4))/2
which are both real because the discriminant is always positive (assuming a is real).

Since a = sqrt(6) and a^2 = 6 and a^2 + 4 = 10, we get
y1 = a = sqrt(6)
y_2, 3 = (-sqrt(6) +- sqrt(10))/2

and finally
x1 = cubrt(sqrt(6)) = sixth root of 6
x_2, 3 = cubrt(y_2, 3) = cubrt((-sqrt(6) +- sqrt(10))/2).

[andylee]

Cool trick! I've been enjoying the ones where you treat the equation as quadratic in a variable (or in this case a constant) I would not have thought of.

[nasrullahhusnan]

The solution of y³-7y+sqrt(6)=0 can be obtained sing Cardano formula. Alternatively the solution can be found by factorization.
LHS of y³-7y+sqrt(6)=0 can be factored as outlined by Babis Koutroulis y³-6y-y+sqrt(6)=0
y(y²-6)-[y-sqrt(6)]=0
As y²-6=[y+sqrt(6)][y-sqrt(6)] then y(y²-6)-[y-sqrt(6)]=0 becomes
[y{y+sqrt(6)}-1][y-sqrt(6)]=0
Hence y=sqrt(6) --> x=6^⅙
y{y+sqrt(6)}-1=0
is a quadratic equation in y. The solution can be obtained using well known formula for solving a quadratic equation.
General method foractorization may also be used:
(y²+ay+b)(y+p)=0
y³+(a+p)y²+(ap+b)y+bp=0
Thus a+p=0, ap+b=-7, bp=sqrt(6)
It will turn out that a=sqrt(6), b=-1, and p=-sqrt(6). We come up with

[goldfing]

Hehe, "Italien guys". There were lots of them during 16th century playing a role: Luca Pacioli, Scipione del Ferro, Antonio Maria del Fior, Niccolo Tartaglia, Geronimo Cardano, Ludovico Ferrari, Raffaele Bombelli :-)

[paultoutounji]

Please put more equations of this type.❤

[dariosilva]

You didnt check the solutions, is that because the substitution was cubic, not quadratic, and therefore no false roots?

[ezzatabdo]

Is great feedback follow more than one motivation, thanks sir.

[rodantero]

NIce one.... at 3:52 you could have group first two terms and the last two terms... y(y^2-a^2) - (y-a)=0 ===> y(y-a)(y+a) - (y-a)=0 ===> (y-a)[y(y+a)-1]=0

[scottleung]

Got 'em all (well, the real ones anyway)!

[DonRedmond-jkhj]

For someone who loves substitution why didn't you make a sub to get the sqrt(6) out of the eqn. Then you'd just be dealing with an eqn with integer coeffs.

[kacemtoubal]

I think you have to set a definition domain to the equation first, because may you could get a rejected solution.
Please could you give us a new idea of sum of sequences
Thank you for your effort, and keep up

[shivanshnigam]

Even if we had a nonic formula it would be practically impossible to do it on just an iPad or whatever device you use😂

[abdoshaat]

Where's the other 6 roots it's nine degree equation

[Hobbitangle]

Substitution x³=y/√6 results in
y³/6√6-y•7/√6+√6=0
y³-42•y+36=0
what can be easily factorized:
y³-42•y+36=(y-a)•(y²+a•y-b)
solving the equation system
a•b=36
a²-b=42
with the divisor candidates method:
a=±{1;2;3;4;6;9;12;36}
The fifth attempt gives the result
1²±36≠42
2²±18≠42
3²±12≠42
4²±9≠42
6²+6=42 ✓

so a=b=-6
y³-42•y+36=(y+6)•(y²-6•y+6)=0
The first root y=a=-6

y²-6•y+6=0
(y-3)²=3²-6=3

The second and the third roots:
y=3±√3

x=3_√(y•√6)
The first root
x=3_√(-6•√6)=-√6

The second and the third roots:


Rest of the roots are complex and can be found multiplying the known real roots by
-½±i√3/2

[ClaudioJoseBruno]

How can I solve that?
3×4^x-56^x+9^x=0

[groscolisdery]

let u=x^3
the complexity of the equation is reduced
u^3-7u+✓6=0
then I will work that on paper and I will come back tomorrow:⁠, ⁠-⁠)