Can you solve this Cambridge Entrance Exam Question?

A great challenging math problem today x^5 - 1 = 0. All roots. What do you think about this question? If you're reading this ❤️.

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[joso]

Solutions (beyond x=1) are complex numbers :
e^(2k.Pi.i/5) with k being any integer.
Simple geometry in the complex plane.

[higher_mathematics]

Thank you for watching. A great question today x^5 - 1 = 0 (Finding all roots) Have a great day and take care! Wish you all the best in you life and career❤❤❤

[Bjowolf]

x1 = 1 ; x2, x3 = cos ( 2 * PI / 5 ) + / - i * sin ( 2 * PI / 5 ); x4, x5 = cos ( 4 * PI / 5 ) + / - i * sin (4 * PI / 5 ) - tthe primary values. + all 2 PI rotations of these roots.

Where 2 * PI / 5 ~ 72 degrees, 5 * 72 degrees = 360 degrees.

[ft]

Similarly, x^7 - 1 = 0 can be solved (you divide by x^3 instead of x^2), but you will get a 3rd degree equation for t (beginning with x^3 + 1/x^3 = t3 - 3t) like
t^3 + t^2 - 2t -1 = 0, which is difficult to solve but makeable! Of course, finding the solution by using the Euler formula is more elegant, but when I was in the 10th class Euler was not teached, so we had to workout everything algebraically! 🙂

[vottkal]

What? x = 1^(1/5) = exp(2ni*pi/5), n = 0; 1; 2; 3; 4 because other integers give the same numbers. So x_1 = 1; x_2to5 = exp(2ni*pi/5) = cos(n*72°) + i*sin(n*72°), n=1; 2; 3; 4

[robertloveless]

I've posted something like this before. You want to go to your neighbor's house, which is 100 feet to your west. You could walk 100' and arrive at your destination. OR you could travel East the entire diameter of the earth minus that 100'. ALL TOO OFTEN, the solutions shown take the long way to get there.

[rajneeshpathak]

.

x^5 - 1 = 0
=> x^5 = 0 + 1
=> x^5 = 1
=> x^5 = 1^5
=> x = 1

[RexxSchneider]

If you really intend to get in to Cambridge, you probably ought to know that exp(2kπi) = 1 where k ∈ ℤ.
Then x^5 = exp(2kπi), giving the five solutions as x = exp(2kπi/5) where k = { 0 ... 4 }. The trivial solution is when k=0, giving x = 1.
The principal root is therefore exp(2πi/5) = cos(2π/5) + i.sin(2π/5), and the other three are cos(4π/5) + i.sin(4π/5), cos(6π/5) + i.sin(6π/5) and cos(8π/5) + i.sin(8π/5).
The time you save (about 23 minutes, apparently) will allow you to do several more trivial questions.

[michellepopkov]

Polar plot. Radius =1. Rotate the vector 2pi/5. Project each stopping point with an X and Yi coordinate. Four complex roots and X=1. Done

[davidseed]

x^5=1=exp(i2πn)
hence
x=exp(i2πn/5) and using exp(iθ)=cis(θ)=cosθ+isinθ, for n=0..4
we have 5 solutions as required

[tassiedevil]

Depending on the time allotted for this, I would agree that the trig and complex exponential version "answers" for the complex roots might not get you far at Cambridge - as a mere demonstration of rote learning. Another approach is to notice that the roots are all unimodular - that is: taking x=a+i*b, (a & b Real), a^2+b^2=1. Substituting this in the quartic equation for x, we know both the Real and Imaginary parts must vanish. The imaginary part is As b is nonzero for roots of interest (we know from the monotonicity of x^5 there aren't other real roots), we can ignore the overall factor of b, so substituting in b^2=1 - a^2 elsewhere, and extracting a common factor of a (joining the b since a is nonzero too) leads to a quadratic for a : 4*a^2+2*a-1=0, roots of which are -1/4 +Sqrt[5]/4, and -1/4 - Sqrt[5]/4, and the corresponding Imaginary parts can be straightforwardly calculated (from Sqrt[1 - a^2]). Whether this is "cleverer" than the "completing the square" approach presented might be a matter of taste - a sharper eye there versus using more knowledge of the complex roots? Another connection with the presented solution is that for a unimodular complex number x, x +1/x is twice the real part, which is why the solutions above for a are half the presenter's "t" roots, indeed could use this instead of solving the second quadratic where the complex numbers first arise in the presentation. Of course for completeness one might wonder about the Real part of the x-quartic: after substituting for b^2 one gets a quartic for a - with four real roots, the pair above and two more (+1/Sqrt[2[, -1/sqrt[2]) that don't make the Imaginary part vanish - must confess that took the edge off a little.

[roger]

Simple. Draw a unit circle centered on the origin of the complex plane. Mark 5 points on the circle at 72° intervals starting on the real axis at x=1.
What do I think about the answer? All that is needed is to know how to evaluate sine and cosine of 72°, 144°, 216°, and 288°.

[bobbyheffley]

X=cos (2n*pi/5)+i*sin (2n*pi/5), with n=0, 1, 2, 3, and 4 for the principal values.

[streptococcuspneumoniae-ixve]

easy just used x = 1, 1*cis(360/5), 1*cis(720/5), 1*cis(1080/5), 1cis(1440/5) = 1, cis72, cis144, cis216, cis288 ; Deg. cisA = cos(A)+isin(A)

[nickcellino]

Well, I guess I'm not going to Cambridge.

[foca]

x⁵ - 1 = 0
x⁵ = 1
⁵✔(x⁵) = ⁵✔(1)
x = 1

[red]

Very nice !
( at 20:42, "=>" should be "=" )

[fqgugmz]

so x=1 and
set x+1/x=t t^2+t-1=0
t=-+sqrt(5)/2-1/2

[ABDERRAHMANEELHAIMER]

Using the exponential form and iterating through the first five numbers could have gotten this solved really quickly 🏍️🏍️

[TNYDCK]

x⁵ = 1 so 5 roots.
Angle between roots will be 360º / 5 = 72º.
So, answers are: (1, 0º), (1, 72º), (1, 144º), (1, 216º), (1, 288).