Solving 7x^5+x+8 without Using the Quintic Formula

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#ChallengingMathProblems #PolynomialEquations

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[haider]

I like how the title says "without using the quintic formula" 😂

[rnld]

Ahh yes I love the totally real quintic formula you chose not to use because it definitely exists.

[RajeshYadav-nbwp]

Using Rene Descartes rule of signs, we can say that it has atmost 1 negative real root(-1) and 0 positive real root. *So it has only 1 real root.*

[damianbla]

04:20 Can you factor this quartic:
7x^4 - 7x^3 + 7x^2 - 7x + 8
into two quadratics?
(Ax^2 + Bx + C) * (Dx^2 + Ex + F)
(in other words, solve for A, B, C, D, E, F).

[pwmiles]

I have an approach to solving the quartic
P(x) = x^4 - x^3 + x^2 -x + 8/7 = 0

Viz., write
P(x) = (x^2 - x/2 + b)^2 - Q(x)
Q(x) = (2*b - 3/4)*x^2 + (1 - b)*x + b^2 - 8/7

The idea is to make Q(x) a perfect square. This happens if the discriminator is zero i.e. 
-8*b^3 + 4*b^2 + (50/7)*b - 17/7 = 0

Put b = c+1/6 and you get the depressed cubic
c^3 - (41/42)*c + 55/378 = 0

You could use Cardan's formula to solve for c and hence b and then write Q=R^2 with R linear in x. We have 
x^2 - x/2 + b = +/-R(x)

Rearrange and solve the quadratic for x. Bit fiddly to actually do though... EDIT: I checked it numerically and it works

[michaelz]

All you need to say at the end is: We have f'(x) = 35x^4 + 1 > 0 for all x so f is increasing. So f(-1) = 0 means f(x) > 0 for x > -1 and f(x) < 0 for x < -1. So there are no other zeroes.

[DangRenBo]

I wonder why you choose not to use a factoring method like synthetic division, which is faster. Real question. No shade being thrown.

[carloshuertas]

Hey SyberMath! As soon as I looked at this math problem, I found out that x=-1 in my head. Thanks a lot!

[HoSza]

No need for calculus, because M = x4-x3+x2-x+1 = (x2+1)2-x(x2+1)+x2+1 = (x2+1)(x2-x+1) > 0. Then 7M+1 > 0, so there will be no more real roots.

[alainrogez]

-1 is an answer. It is the only real root because the derivative is 35x^4+1>0. The other 4 roots have to 2 conjugate roots.

[dihey]

At 7:25 you say it will intersect in the x-axis only once because its always increasing but thats kinda false since the function can be always increasing and have no roots. Like e^x for example.

[janakkathayat]

Thanks sir I am your waiting for video so Thanks sir I am your regular viewer sir so next video come quickly sir your video chanel is most important for mathematics lover so sir I am some note understand this equation so sir I sir my problem do solved sir your amazing for trick i am say thanks sir I hope your another come quickly video sir

[rakenzarnsworld]

This problem is pratically impossible to solve without using computer program.

[yuxycww]

Левая часть равенства - монотонно возрастающая функция . Правая- константа. По теореме о корне данное уравнение имеет не более одного корня. Подстановкой находим, что х= -1.

[fdh]

Could be interesting to mention the limits of the polynom at - oo and +oo
Furthermore to the fact that the polynom is increasing
Thanks a lot for your awsome exercices

[scorchable]

Notice that at x=0, while it is not a root of the equation, it is an interesting point since it is an inflection point. The second derivative test will show that too. :)

[tonyhaddad]

Hello syber, my approach :
7x^5 + x + 8 = 0
first by the rational root theorem
we have (-1) is a solution
so (x+1) is a factor
(x+1)(7x^4 - 7x^3 + 7x^2 - 7x + 8) = 0
7(x^4 - x^3 + x^2 - x + 8/7 ) = 0
Divide 7 both side
x^4 - x^3 + x^2 - x + 8/7 > =
min (x^4 - x^3) + min(x^2-x) + 8/7
By derivitive :
= (3/4)^4 - (3/4)^3 +(1/2)^2 - (1/2)+8/7 > 0
so the only real solution is (-1)

[aoughlissouhil]

Nice video . I really like it.
👍
I wanna ask you a question if you don't mind .
How can we prove that the Fibonacci sequence is greater than or equal to n
F_n≥n

[moeberry]

Descartes rule of sign shows only 1 negative real solution and the others are 2 pairs of complex conjugates.

[raffaellabrunetti]

The only real soluzion x=-1 is easy to find. To demonstrate it is the only one NOT using derivatives: drow a rough graph of f(x)=7x^5 (a simple power-function, similar to x^3, increasing for all x) and compare with f(x)=-x-8, a decreasing straight line. Both graphs are unlimitated and represent continuos finctions existing for every value in R. In this situation they must have one and ONE ONLY common point... Sorry for my English :)