I Solved A Quartic Equation in Two Ways

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[@NadiehFan]

In fact the second method is not as contrived as you make it out to be, it can be used to solve any quartic, even quartics that _do_ have a cubic term. If we add 2kx² + k² to both sides of
x⁴ = 5x² + 6x + 5
we have
x⁴ + 2kx² + k² = 5x² + 6x + 5 + 2kx² + k²
which is
(x² + k)² = (2k + 5)x² + 6x + (k² + 5)
Now the left hand side is a perfect square regardless of the value of k, but the quadratic in x at the right hand side will be a perfect square if and only if its discriminant is zero. So, in order to have a perfect square on both sides of our quartic equation k must satisfy
36 − 4(2k + 5)(k² + 5) = 0
or
(2k + 5)(k² + 5) = 9
This is a cubic equation in k which we could solve formally, but since the quartic equation we need to solve is a competition problem and therefore likely to have a nice factorization into two quadratics with integer coefficients we should start by looking for integer values of k that satisfy the condition for the right hand side of our quartic to become a perfect square. If that fails, you should also look for integer multiples of ½ (think about that).
Clearly, if k is an integer, then the product (2k + 5)(k² + 5) can only equal 9 if we have k² + 5 = 9 and 2k + 5 = 1 which implies k = −2. With this choice for k the equation
(x² + k)² = (2k + 5)x² + 6x + (k² + 5)
becomes
(x² − 2)² = x² + 6x + 9
which gives
(x² − 2)² = (x + 3)²

[@SyberMath]

Very nice! Thank you ❤️

[@bosorot]

4:10 . Does not need a cubic equation . Just use the rational root theorem, b=5 and c=-1 or b=-5 and c= 1 . Only the later is the answer.

[@mcwulf25]

If you only knew how I feel when I see you make a mistake. And how the world feels safe again when you spot it three lines later!!!

[@kwiky5643]

😂

[@SyberMath]

I don’t know what to say! 😁🤪

[@luggepytt]

Third method:
Add x²+1 to both sides. Then we get some nice numbers:
x⁴+x²+1 = 6x²+6x+6
Hunch: 6x²+6x+6 is divisible by x²+x+1. It feels like x⁴+x²+1 is too.
Let’s try if we can construct x⁴+x²+1 from x²+x+1.
Start with x²+x+1
Add x²(x²+x+1), and we get
x⁴+x³+2x²+x+1
Now subtract x(x²+x+1), and we get
x⁴+x²+1
So we have proved that;
x⁴+x²+1 = (x²-x+1)(x²+x+1)
Now, we can factor both sides in our equation
x⁴+x²+1 = 6x²+6x+6
like this:
(x²-x+1)(x²+x+1) = 6(x²+x+1)
Move everything to the left hand side:
(x²-x-5)(x²+x+1) = 0
Finally, we just have two quadratics to solve:
(x²-x-5) = 0
or
(x²+x+1) = 0
which is trivial.

[@mcwulf25]

Genius!
I did what Syber did in method 2.

[@NadiehFan]

Nice. Note that the zeros of x² + x + 1 are the complex cube roots of unity since (x − 1)(x² + x + 1) = x³ − 1. If ω is either of the complex cube roots of 1 we therefore have ω² + ω + 1 = 0 which implies ω⁴ + ω² + 1 = ω + ω² + 1 = 0 since ω⁴ = ω³·ω = 1·ω = ω. So, each of the zeros of x² + x + 1 is a zero of x⁴ + x² + 1 which implies that x² + x + 1 is a factor of x⁴ + x² + 1. Similarly, x² + x + 1 is a factor of e.g. x⁸ + x⁴ + 1 and x¹⁰ + x⁵ + 1 and generally of any trinomial xᵐ + xⁿ + 1 whenever m ≡ 2 (mod 3) and n ≡ 1 (mod 3) or vice versa.

[@SyberMath]

pretty good

[@giuseppemalaguti435]

Risulta semplicemente x^4-4x^2+4=x^2+6x+9....

[@9허공]

subtracting x both sides, x^4 - x = 5x^2 + 5x + 5
=> x(x^3 - 1) = x(x - 1)(x^2 + x + 1) = 5(x^2 + x + 1)
=> (x^2 + x + 1) (x^2 - x - 5) = 0

[@scottleung9587]

Got em all!

[@bobbyheffley4955]

Method 1: Descartes method
Method 2: Ferrari method

[@adamsmithson486]

Pozdrawiam serdecznie i życzę miłego dnia

[@rakenzarnsworld2]

x^4-5x^2+6x=5
x = 1 -> 1-5+6=2
x = 2 -> 16-20+12=8
Therefore, 1 < x < 2
x = -1 -> 1-5-6=-10
x = -2 -> 16-20-12=-16
x = -3 -> 81-45-30=6
Therefore, -3 < x < -2

[@sendai-shimin]

After a little thinking...
I did factorize forcibly.
(x^2 + x + 1)(x^2 - x -5)

[@bobbyheffley4955]

The solutions to the left factor are the complex cube roots of 1.

[@broytingaravsol]

so easy one

[@SyberMath]

why?

[@broytingaravsol]

@@SyberMath shifting an x in RHS to LHS leads to 2 quadratic formulae and the solutions