A quartic equation solved in two ways! No quartic formula!!! x^4+4x-1=0

This video is about a quartic equation that can be solved in two ways. No formula!!!
Quartic equations, factoring, Algebra, x^4+4x-1=0

[SyberMath]

This problem received a lot of interest! Thanks to all who watched it. It's an interesting problem and there are different variations of it such as x^4+16x-12=0 or x^4+36x-45=0.

[Megathescientist]

you could search for a difference of two squares by adding and subtracting the quantity (2x^2+1). In this way you find out that
x^4+4x-1=(x^2+1)^2-2(x-1)^2

[mimathematics]

I'm a Japanese student
I always enjoy watching this channel's video
I'm very glad because I could solve this problem by myself!

[djvalentedochp]

I'm proud to say I solved it myself just factoring exactly as the second method. The problems you bring to the videos are tough!

[SyberMath]

After watching the video one more time, I realized that I could've used a different approach for the first method. using ax and -ax instead of ax and cx since there are no terms with x^3. 😄
I should also add the fact that u^3+4u-16 can also be factored this way:
and it's obvious that u^2+2u+8=0 does not have any real solutions so the only real root from here is u=2!

[mrisouza]

I used a similar method as that one used for solving the quartics:
x^4 = 1 - 4x
x^4 + yx^2 + y^2/4 = 1+y^2 - 4x + yx^2
(x^2 + y/2)^2 = yx^2 - 4x + (1+y^2 ), for a convenient y.

Convenient here means to turn the right-hand-side expression into a "perfect square". As delta = 16-y^3-4y = (y-2)(y^2+16y+8) = 0 leads to y = 2 as a solution, we can write:

(x^2 + 1)^2 = 2x^2 - 4x + 2 = 2(x-1)^2
hence (x^2+1)^2 = 2(x-1)^2 and now the problem is reduced to solving quadratics.

[farhansadik]

old video, nice to see how much you've improved these days, quality wise and problem solving wise!

[thedarkknight]

I believe when such type of equations are given to you the best method would be hit and trial to find just one root

And after that with help of factorisation you can resolve other roots. But this question no hit and trial would work so sir your way of solving was quite impressive

[math_person]

I was able to solve it with the second method :) But with a slight variation. We got a^2 = 2b^2, so a = sqrt(2) * b and a = -sqrt(2) * b are the two solutions.

[kennethstevenson]

Thank you for revealing math manipulation that I have not mastered!

[RatanDas-verq]

The first method is complicated but it is very useful to solve these equations.I learn a new method.Thank you!!

[seegeeaye]

the equation is written as x(x^3 + 4) = 1, the LHS is a V-shaped curve crossing the x-axis at 0 and -- cubic rt of 4, by considering the RHS y =1, we know there are two real solutions: one is slightly less than -- cubic rt of 4, the other is little more than 0.

[jbreezy]

This is great for introducing the integration technique of partial fractions

[math_person]

In the first method, you should prove why factoring the equation into two arbitrary quadratic equations works. I think you should have started with f(x) = (x - a) * (x - b) * (x - c) * (x - d), which is the same as [(x - a) * (x - b)] * [(x - c) * (x - d)], so basically its a product of two quadratic equations.

[pierregillet]

With (x^2+1)^2-2*(x-1)^2=0
you have (x^2+1)^2=2*(x-1)^2
then abs(x^2+1)=sqrt(2)*abs(x-1)
then x^2+1=sqrt(2)*(x-1) or x^2+1=sqrt(2)*(1-x)
Only the term wih (1-x) is the quadratic equation with the two real roots (approx. 0.2490 and -1.6633)

[michelkhoury]

Very nice solution. I solved this by using some tricky manners like adding and subtracting some terms in order to factorising the polynomial

[xfefbkc]

I am a Junior high school 9 grade in Taiwan I find you due to this video . I am so excited about your
solved methods . Your video is sosososo excellent .
I hope I can learn a lot from your channel.

[WallaceChan]

let a, b, c, d be roots of x⁴+4x-1=0 then (x-a)(x-b)(x-c)(x-d)=0 By collecting coefficient of like terms of the polynomial, we get 4 equations of 4 unknowns a, b, c, d by solving the 4 equations we have the roots are the same roots of the 2 quadratic equations (x²-√2x+√2)=0 and (x²+√2x+1-√2)=0 two roots for each quadratic equation. To check whether they are real roots, use the discriminant δ=b²-4ac in the quadratic equation ax²+bx+c=0 or (x-α)(x-β)=0 to check
NB: α+β=-b/a αβ=c/a

[konstantinjoukovski]

In the first method, when you wrote down first X, you made a mistake, writing + instead of minus, so both of your roots appeared to be complex, and in fact the first one is real.

[holyshit]

Method of undetermined coefficients is quite general but i suggest to get rid of cubic term before we use it
because it will be less computations to solve this system of equations