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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[@spicymickfool]

Solutions of x^n-1=0 : Consider the unit circle and the point with maximum x value. This is a solution, trivially. Now divide 360 degrees by n. Move along the circle counter clockwise and select points evyer (360/n) degrees. These are all solutions. To convert to complex numbers , z = (cos theta + i \sin theta) where theta = 360k/n for 0

[@tianle1204]

de moirve theorem

[@oahuhawaii2141]

But he literally finds the sine and cosine of those angles in surd form. For example, the 4 key values are: sin(36⁰ ± 18⁰) = (√5 ± 1)/4 sin(54⁰ ± 18⁰) = √(10 ± 2*√5)/4 The solutions use these values in ± form.

[@DaneBrooke]

(Typos forgiven.) Exactly, and even taking time to explain this a little more clearly this solution require much less time than the half hour devoted to the video. Besides which this is easily generalized for xⁿ = y and still takes essentially constant time to accomplish. God help me if I should have to spend a half an hour on a single entrance exam problem.

[@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.

[@luisrocha26]

The solution on the video has a nice algebraic approach which is also useful and applicable to other algebra problems. Going for the complex exponential solution requires a whole background on the complex plane in order to render the final solution simple. Both solutions are useful in different moments of math instruction.

[@RexxSchneider]

@@luisrocha26 The "nice" algebraic approach works in this case because we can solve quartic equations. Have a think about how you might extend the technique to the six roots of x^6 = 1 or the seven roots of x^7 = 1, etc. You may conclude that the complex exponential is a more general solution. I find the particular solution in the video interesting because it allows us to introduce the complex method and then check the results against a purely algebraic method. But beyond that, I'd be hard pressed to recommend that students should consider it. YMMV.

[@luisrocha26]

@@RexxSchneider I'm not talking about being able to generalize the solution nor demonstrate more complex things; it's useful as as didactic exercise on algebra

[@oahuhawaii2141]

But he found the sin and cos of those angles in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.

[@RexxSchneider]

@@oahuhawaii2141 But 99% of the time, nobody needs to go further. The trig form is an exact form, just like the surd form. We find that cos(2π/5) is generally just as useful as (√5 - 1)/4. To use it we're almost always going to use a numerical approximation, and either form will give the same approximation, about 0.309. I concede that if there is a problem where the result has to be multiplied by (√5+1), you can get a simplification, but those cases must be vanishingly rare. And finally, the question never specified that an answer in surd form was required. I assure you that when I took my Cambridge Scholarship exams in 1968, I never wasted time trying to answer questions I wasn't asked.

[@joso5554]

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

[@user-ko9pr8tn4m]

True

[@agustincai]

how did you find

[@howardlandman6121]

Anyone who knows complex numbers well can just write down all 5 solutions. But k only ranges from 0 to 4, any other integer duplicates one of those answers. (That is, if m = n (mod 5) then e^(2m𝜋i/5) = e^(2n𝜋i/5).)

[@kirshkro]

​@@agustincaiThese are the points that divide the circle into n equal parts

[@martinrodriguez1329]

Yep, so solutions would be 1; 1e^j2π/5; 1e^j-2π/5; 1e^j4π/5; 1e^j-4π/5 with k = -2, -1, 0, 1, 2. Didn't even watched the video just jumped from front picture to comment section.

[@michellepopkov940]

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

[@paulbritt588]

The idea is to provide the exact solution without recourse to the trigonometric forms. This also gives you a slick way to find sin(2Pi/5) and cos(2Pi/5)

[@robertloveless4938]

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.

[@harrisorourke6926]

What does the "traveling" metaphor" means in relationship to math? You have a very naïve understanding concerning what solving a math problem means. Your 100' solution will give only a 1/5 of a correct score.

[@ataarono]

@@harrisorourke6926 obviously, he could have walked in expanding circles around his house that follow the fibonacci sequence until there is an interception with the backwards traced fibonacci sequence radiating from the neighbour which he then walks backwards. so naive

[@Nguyễn-j9q]

Why don't just x=1

[@taras900956]

Because an equation of the Nth degree has N solutions.

[@itzsoweezee9980]

Its too easy for the Mathematics, they do not like Pragmatism. 😉

[@DragoSpiro98]

Because complex numbers exists

[@harsh90868]

​@@itzsoweezee9980😂😂

[@marty7442]

Because this is that 'new algebra'. Nothing is objective. You see the answer here is; Strudel. Don't ask. In ten minutes, the answer will be cupcake.

[@joshuahshivachi331]

Many students cannot follow this long process. Simply apply De Moivres complex roots theorem. Much shorter and precise. Thanks.

[@oahuhawaii2141]

But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.

[@gerhardhartwig2825]

(X^5)-1=0 -> (x^5)=1 -> x=1

[@thomasharding1838]

Yes! Straightforward and took 2 seconds. Now if it had been x^4 (even number exponent), then x = +1 and -1.

[@ralkadde]

@@thomasharding1838 x⁴ - 1 = 0 has four solutions: x₁=+1, x₂=-1 , x₃=+i x₄=-i. Both equations are not limited to real solutions.

[@gerhardhartwig2825]

@@thomasharding1838 x^5=1 ; Gleichung: (x^5=1)=1 ; (x^5=-1)≠1

[@Bjowolf2]

And then there are four more complex roots - appearing as two pairs of conjugates ( symmetrically around the real axis ) - at +/- 72 and +/- 144 degrees ( ~ +/- 2PI/5 and +/- 4PI/5 in radians ) around the unit circle in the complex plane. + of course all 360 degrees ( 2PI) rotations of these roots.

[@Bjowolf2]

@@thomasharding1838 And you only forgot four more principal roots 😉

[@Bjowolf2]

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.

[@oahuhawaii2141]

But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.

[@ucoknyaduren3324]

If you can solve it in 24 minutes, why should you spend 15 seconds?

[@harrisorourke6926]

Very good you got one of the five solutions to the problem in 15 seconds. What took you so long? However apparently you did not notice that the problem presents an exponent of 5 which leads to a solution with 5 answers.

[@martinrodriguez1329]

​@@harrisorourke6926sry mate, he meant all 5 solutions in 15 seconds, which is pretty easy to do considering 1 equals 1.e^j(0+2kπ)

[@robert.eduard]

@@harrisorourke6926 imaginary solutions, hence "i".

[@ToasterPizzaFun]

​@@robert.eduard Not imaginary, complex. Complex numbers are pretty essential to a lot of practical applications, so being able to find complex roots to an equation is significant. "Imaginary" is an outdated term for complex numbers.

[@robert.eduard]

@@ToasterPizzaFun “i” is still called an imaginary unit, but if it has real life applications, then it’s great. I didn’t know that, I’ll look into it.

[@davidseed2939]

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

[@waitweightwhite793]

This is exactly what makes the difference between students in these exams, if you can come up with a solution that can help you solve the problem much more quickly than the intuitive solution as shown in this video, you can then answer much more questions.

[@oahuhawaii2141]

But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.

[@itsjavaman]

In my head in under 5 seconds.

[@Astrobrant2]

Me, too. I've seen this guy before, giving ridiculously complex solutions to some very simple problems. Any first day algebra student would say, "What minus 1 is zero? Okay, 1-1 is zero. So x^5 = 1. Well, 1 to any power is 1, so the answer is 1. Duh."

[@harrisorourke6926]

@@Astrobrant2 You are doing naïve math. Actually you are not doing math. The solution requires determining the 5 solutions to the equation. If you submitted this on a homework assignment you might get partial credit and definitely a note from the instructor that says, "See me about this".

[@Shri]

​@@Astrobrant2 the question is to find ALL ROOTS which includes finding complex numbers. It is not your naive math problem

[@vottka1l]

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

[@JoseFernandes-js7ep]

I've tried the same, but then, I remembered that I don't know the trigonometric functions of 72º

[@okohsamuel314]

​@@JoseFernandes-js7ep ... No big deal, u could (better still) leave ur 'x - values' in exponential form.

[@rajneeshpathak562]

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

[@7g_21_louisdanendra9]

Why dont we do like that?explain me, please

[@nanagupta4499]

Because more power=more values this only takes into account 1 probability

[@robsonedugd1170]

Is ⁵√ to both sides not just, ^, it would be wrong if it wasn't 1 for example

[@kotarojujo6365]

@@7g_21_louisdanendra9 study about complex numbers

[@harrisorourke6926]

@@nanagupta4499 The solution of x=1 is not a "probability". It is a certainty. We aren't flipping coins when we do maths.

[@epicmod5695]

Isnt the x just 1?

[@Anu_here_148]

Exactly!!! That's what I'm thinking

[@Bjowolf2]

@@Anu_here_148Yes, but there are actually 4 more roots, that are all complex numbers. If you rotate the complex number ("point") "1 + 0*i " on the unit circle 72 degrees or 2 * PI / 5 radians , you will end up in the complex number "cos ( 2 * PI / 5 ) + i * sin ( 2 PI / 5 )", where i is the imaginary unit. This is pr. definition e^( i *2 * PI/5 ), since the "length" or modulus rather is 1 ( we are on the unit circle in the complex plane ;-) ). Raising this complex number to the fifth power then gives us ( rotating it to the argument ( angle ) that is 5 times larger ): z^5 = ( e^( i *2*PI/ 5 ))^5 = e^( i * 2*PI /5 * 5 (!!!!!) ) = e^( i * 2PI ) = cos(2*PI) + i *sin( 2*PI) = 1 + 0*i = 1 And we are back to 1 ;-) This also applies to the symmetric complex roots at - 72 degrees, and +/- 144 degrees on the unit circle in the complex plane - and to all full 360 degrees ( or 2*PI ) rotations of these 4 primary roots.

[@keescanalfp5143]

​@@Bjowolf2, yeah and aren't we allowed to consider that the real reason for more than one root x = 1 is , that the equation is a fifth grade one . so there should be five roots as a full solution .

[@Bjowolf2]

​@@keescanalfp5143 Yes, precisely - a n'th order polynomium always has n roots ( complex or real, including multiple roots ), and if the coefficients of the polynomium are all real (as in this case!), then any complex roots will always appear as complex conjugated pairs ( symmetrically around the real axis in the complex plane ) 😊

[@Anu_here_148]

@@Bjowolf2 Thank you 👍✨

[@bobbyheffley4955]

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

[@oahuhawaii2141]

But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.

[@tassiedevil2200]

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 i*b*(4*a^3+3*a^2+a*(2-4*b^2)-b^2+1)=0. 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.

[@RexxSchneider]

Note that if we set x =2π/5, we can use the fact that cos(4π/5) = cos(2π - 4π/5) = cos(6π/5) to get the relationship cos(2x) = cos(3x). We can then use the Pythagorean identities. Hence cos(2x) = 2cos^2(x) - 1 is equal to cos(3x) = 4cos^3(x) - 3cos(x). If we set c = cos(x), we now have 2c^2 - 1 = 4c^3 - 3c. Therefore 4c^3 - 2c^2 - 3c + 1 = 0. A factor is c=1 giving: (c - 1)(4c^2 + 2c -1) = 0. The c=1 is when x=0, which also satisfies cos(2x) = cos(3x), so we just need to solve the quadratic, giving c = (-2 ± √20)/8 = (-1 ± √5)/4. Since x is in the first quadrant, we can see that cos(2π/5) = (-1 +√5)/4) which is also the value for cos(8π/5). We can see that cos(4π/5) = cos(6π/5) = (-1 - √5)/2. Similarly for sin(2kπ/5).

[@oahuhawaii2141]

Decades ago, my 9th grade teacher gave a bonus trigonometry problem to find sin(θ) in surd form, for θ = 18⁰ . I solved it at my desk for sin(θ), cos(θ), tan(θ), cot(θ), sec(θ), csc(θ) . Nobody else in class figured out how to do it. At home, I did the same for θ = n*18⁰ , n ∈ ℤ . The next day, I showed my teacher the table, and said I'll work on θ = n*3⁰ during the weekend. I used lots of scratch paper for those calculations. When I showed her my completed table, she asked me to join the math team. I did and she gave me a CRC math book. The first thing I did was look at its trig section, which I saw only had θ = n*15⁰ . I submitted a copy of my table to the publisher. I never got a response.

[@YAWTon]

Boring, boring, boring... No higher mathematics. The five roots are obvious, no need for lengthy derivation. This problem is good for a "short" at best.

[@oshaya]

Completely agree.

[@kpdywo848]

Completly agree, can solve in 30 secondes with polar form, what i've done

[@okohsamuel314]

​@@kpdywo848 ... Yeah from nth root of unity approach when n = 5

[@Bjowolf2]

So it's obviuosly not for you then, but it might make some math novices really curious ;-) - it did say "entry level", not "expert level".

[@harley_2305]

You are obsessed with leaving comments like this on every video of this guy’s channel. If it bores you so much, watch something else! It is THAT easy. Get a hobby.

[@roger7341]

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°.

[@oahuhawaii2141]

He does the evaluation in the video. It isn't simple.

[@CC--qn4gf]

"Fast and quick video", the video length: 24:47!

[@mcwulf25]

Well it's the 5th roots of unity which therefore means x = e^2pi.n.i/5.

[@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❤❤❤

[@oahuhawaii2141]

It's great that you found sin(θ) and cos(θ) in surd form for θ = k*2*π/5 = k*72⁰ , k ∈ ℤ . Everyone knows the values when k = 0 , but few know them for k = ±1, ±2 . Most are stuck with answers such as cos(72⁰) ± i*sin(72⁰) and have to figure out how to go further. The 4 complex solutions use the following key values in ± form: sin(36⁰ ± 18⁰) = (√5 ± 1)/4 sin(54⁰ ± 18⁰) = √(10 ± 2*√5)/4

[@igori3532]

The function has only one real root x = 1. For complex roots there was some theorem (I'm to old to remember it :D ) I think the guys in Cambridge would want the person to know the theorem and use it straight away instead of doing algebra. The test should check whether the person knows something more than algebra.

[@ls3laminarflow]

Cambridge exam questions are typically not difficult but there are too many of them in the paper to pass if you spend a lot of time on each. This is a good example, solve laboriously as in the video in 25 minutes and you may get the right answer but no time left to answer enough questions to pass. So the exam is actually testing your depth of understandingand ability use it to see the fast paths to solution. ( which is the euler relation shown by others here. 20 seconds, boom, done.)

[@raphaelbocolisalvador9685]

If you know polar coordinates, the solution would be more simple and beautiful. By knowing the only real solution (1), the complexes would be calculated just dividing 360 degrees by 5. So, the other four solutions are 1

[@andrewbuchanan5342]

When there are multiple +-, it's extremely useful to tag them. That way one can have a single expression for all 4 of the solutions, and one doesn't have to futilely repeat the calculation and the scribbling. So one might have +- for the first case, and then +-* for the second case, or one might choose e.g. +-_1 & +-_2. Or whatever. I have never seen anyone do this notational trick, but it ought to be standard.

[@tarzanwheeler5975]

x^5 -1 = (x-1)(x^4 + x^3 + x^2 + x + 1), since 5 is prime and its only divisors are 1 and 5, x^4 + x^3 + x^2 + x + 1 is the 5th cyclotomic polynomial, and thus its roots are the primitive 5th roots of one on the unit circle in the complex plane. 360/5 = 72, so we have the roots: cos 72 + i sin 72 (the trig arguments are in degrees, here) cos 144 + i sin 144 cos 216 + i sin 216 cos 288 + i sin 288 Note that the trivial solution x=1 is *not* a primitive root, that is, it does not generate the cyclic group of order 5 of the solutions.

[@oahuhawaii2141]

But he found the sine and cosine of angles n*2*π/5 in surd form. You're stuck with answers such as cos(72⁰) + i*sin(72⁰) and have to figure out how to go further.

[@ftorum19]

cos(2πki/5)+isin(2πki/5). Just simple logic operations

[@Nikioko]

x = 1 is the obvious real solution (x⁵ = 1, take fifth root). So, let's factor out x − 1: x⁵ − 1 = 0 (x − 1) · (x⁴ + x³ + x² + x + 1) = 0 Now we only have to solve the quartic function x⁴ + x³ + x² + x + 1 = 0 to get the four complex solutions. 🤣

[@casusbelli9225]

The people in comments have a right to vote. Let that sink in.

[@Creology]

De Moivre's Theorem: It's free real estate

[@maxopala42]

Without any count, using just logic, I believe the answer has more then 5 roots. Am I right? For any k as natural number that can be divided by 4, the root can be any value from: x = i^(k/5) Examples: i^(4/5) i^(8/5) i^(12/5) i^(16/5) i^(20/5) = i^4 = 1 i^(24/5) i^(28/5) i^(32/5) i^(36/5) i^(40/5) = i^8 = 1

[@mariox729]

I guess in other countries, not only Croatia where i am from, highschoolers are tought about the de Moivre formula (or atleast Eulers formula) , my God! Then it is trivial.

[@worldorthoorthopaedicsurge6147]

X equals 1. I am aged 72

[@CZOV]

CHeater!

[@oahuhawaii2141]

You missed the other 4 solutions, which happen to be complex numbers.

[@worldorthoorthopaedicsurge6147]

@@oahuhawaii2141 I am an orthopaedic surgeon, I deal with reality, not the fuzzy world of complex numbers

[@Interstellaricmind]

X=1 because any non negative real number raised to power for one always equals 1

[@T0NYD1CK]

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).

[@danabanana4408]

this video is actually good i have no idea why people feel the need to be so upset in the comments.

[@dennisschroeder9619]

I understand what he did. I just would like him to verify his answers. Without that how can he be sure he did not make a mistake.

[@sarah_eddeeb7449]

can i just leave them in the trigonometric form of complex no.? x1=cos0+isin0=1 x2=cos72+isin72 x3=cos144+isin144 x4=cos-144+isin-144 x5=cos-72+isin-72

[@vinceturner3863]

I would used radians rather than degrees, but your answer is good enough.

[@user-mb1qe7wg2y]

19:25 (-1)(10-2*sqrt(5))

[@MrDeep414]

Great! Now I know how to divide circle to five sectors with compass and ruler!

[@resation]

you put up with it, man, it's very tedious to paint like that, it's not for 7th grade content.

[@Xubono]

Of course, as an entrance into Cambridge, you would have to realise this is Mathematics, not “Math”.

[@cyruschang1904]

x^5 - 1 = 0 x^5 = 1 = cos(2nπ) + i sin(2nπ) x = cos(2nπ/5) + i sin(2nπ/5) n = 0, x = 1 n = 1, x = cos(2π/5) + i sin(2π/5) n = 2, x = cos(4π/5) + i sin(4π/5) n = 3, x = cos(6π/5) + i sin(6π/5) n = 4, x = cos(8π/5) + i sin(8π/5)

[@garyjonah22]

This is where you begin to wonder whether mathematics is just a human mental abberation.

[@josephstratti52]

If this guy designed a house with this approach no builder could price the final cost!😂

[@RyanLewis-Johnson-wq6xs]

x^5-1=0 x=1 x=-1/4±Sqrt[5]/4±Sqrt[10±2Sqrt[5]]/4i

[@oahuhawaii2141]

You have to split up the "±" parts because it isn't clear if each of the 3 are independent (which produces 8 values), or are synchronized (which produces 2 values).:There are 4 complex roots.

[@NadiehFan]

Since x = 1 is a solution of x⁵ − 1 = 0 the factor theorem guarantees that x − 1 is a factor of x⁵ − 1 and in fact we can write x⁵ − 1 = 0 as (x − 1)(x⁴ + x³ + x² + x + 1) = 0 In accordance with the zero product property which says that a product is zero if and only if at least one of its factors is itself zero we therefore have x − 1 = 0 ⋁ x⁴ + x³ + x² + x + 1 = 0 Clearly the first condition is satisfied for x = 1 and no other value of x, so the other four solutions of x⁵ − 1 = 0 are the solutions of the quartic equation x⁴ + x³ + x² + x + 1 = 0 which is a reciprocal (palindromic) equation which can easily be solved in a number of ways. I prefer to solve this quartic equation by completing the square twice in order to create a difference of two squares at the left hand side which can then be factored. To do so we can either start by forming a square with the fourth and third powers or by forming a square with the fourth and second powers. Here, I'll use the second approach. Noting that (x² + 1)² = x⁴ + 2x² + 1 we have x⁴ + x² + 1 = (x² + 1)² − x² and taking out the common factor x from the remaining cubic and linear terms we have x³ + x = x(x² + 1) so we can rewrite the equation as (x² + 1)² + x(x² + 1) − x² = 0 Now we can again complete the square with respect to the terms (x² + 1)² + x(x² + 1) because we have (x² + 1)² + x(x² + 1) + ¼x² = (x² + 1)² + 2·(x² + 1)·(½x) + (½x)² = ((x² + 1) + ½x)² = (x² + ½x + 1)² so we have (x² + 1)² + x(x² + 1) = (x² + ½x + 1)² − ¼x² and we can therefore rewrite the equation as (x² + ½x + 1)² − ¼x² − x² = 0 or (x² + ½x + 1)² − ⁵⁄₄x² = 0 and since ⁵⁄₄x² = (½√5·x)² this gives (x² + ½x + 1)² − (½√5·x)² = 0 We now have a difference of two squares at the left hand side, and applying the difference of two squares identity a² − b² = (a − b)(a + b) the equation can be written as (x² + ½x + 1 − ½√5·x)(x² + ½x + 1 + ½√5·x) = 0 which gives (x² + (½ − ½√5)x + 1)(x² + (½ + ½√5)x + 1) = 0 Applying the zero product property we therefore have x² + (½ − ½√5)x + 1 = 0 ⋁ x² + (½ + ½√5)x + 1 = 0 and now we only need to solve these two quadratic equations to obtain all four roots of our quartic equation. The discriminant of the first quadratic is Δ₁ = (½ − ½√5)² − 4·1·1 = (1 − √5)²/4 − 16/4 = (6 − 2√5 − 16)/4 = (−10 − 2√5)/4 and the discriminant of the second quadratic is Δ₂ = (½ + ½√5)² − 4·1·1 = (1 + √5)²/4 − 16/4 = (6 + 2√5 − 16)/4 = (−10 + 2√5)/4. Both discriminants are negative since 2√5 < 10 so we have four complex roots. For the discriminant of the first quadratic we have √Δ₁ = ½√(−10 − 2√5) = i·½√(10 + 2√5) and for the discriminant of the second quadratic we have √Δ₂ = ½√(−10 + 2√5) = i·½√(10 − 2√5). Together with the real root x = 1 the five roots of x⁵ − 1 = 0 are therefore x₁ = 1 x₂ = ¼(−1 + √5) + i·¼√(10 + 2√5) x₃ = ¼(−1 + √5) − i·¼√(10 + 2√5) x₄ = ¼(−1 − √5) + i·¼√(10 − 2√5) x₅ = ¼(−1 − √5) − i·¼√(10 − 2√5)

[@maxilol2345]

If this is 1 question for 1 exam. Do they really think we’re spending 25 minutes to solve 1 exercise? X = 1 and that’s that. Next question, moving on. They usually only give out 2 or 3 hours for an entrance level exam. There’s always like 100 questions in these types of exams

[@Bjowolf2]

😂

[@Bjowolf2]

There are more roots - 5 in all! x can be a complex number.

[@natalia5609]

If you leave the question at X=1 , you're not getting any marks , cuz the answer is not yet completed. If you haven't learned about complex numbers, it doesn't mean they're non existent, they must be taken into account too

[@cheems_0007]

This channel wasting my time more than my procrastination

[@koenraad4618]

cos(0π/5) ± i sin(0π/5), cos(2π/5) ± i sin(2π/5), cos(4π/5) ± i sin(4π/5). The question did not mention 'not' to use cos, sin, π. Now solve x^7=1.

[@brucerosner3547]

These kind of problems are trivial in polar coordinates.

[@kpdywo848]

Exact, j'ai réalisé en moins de 60 secondes.

[@scobrado]

That's the thing about a root, there's more than one way into it. I wouldn't have thought of this. Nice clip.

[@pincusgreen]

I believe the answer is this guy had a bet with someone on how long he could keep viewers watching

[@ft7339]

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! 🙂

[@user-fq4gu3gm9z]

0=(x-1)(x^4+x^3+x^2+x+1)=(x-1)(x^2+x+1+1/x+1/x^2) so x=1 and set x+1/x=t t^2+t-1=0 t=-+sqrt(5)/2-1/2

[@Pokekilller]

What is happening in the world guys for an easy question he did that long 😂😂😂😂😂😂😂😂 Solution is x^5 -1 = 0 x^5 = +1 As we know that 1^n = 1 Therefore x^5 = 1^5 Cancellation on both side Therefore x = 1 😊😊😊😊😊

[@natalia5609]

X=1 is also correct but not complete, there are other numbers too which satisfy the equation,hence the equation has mose roots. Remember an equation on x of degree n will have n different solutions

[@Pokekilller]

Yeah but given is simplest answer 😊😊😋

[@diamondvideos1061]

There are also 4 complex solutions, but there is a much easier way to find them using Euler's famous equation.

[@nickcellino1503]

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

[@jenshagelstein7695]

you are too intelligent?

[@roger7341]

I was a janitor at Cambridge and wrote the answer on a whiteboard in the hallway. Who says you can't get into Cambridge?

[@zengebra7179]

​@@roger7341few years after got lost in mars

[@jfverboom7973]

Thiz is damn easy and i am not a msthematics genius. 1 - 1 = 0 1 to any power is always 1 Maybe the point is deriving a kind of 'proof'.

[@official_mosfet]

My theory was that the square root of 1 by 5 would be the solution, after calculating and realizing it's one i realized X=1😭😭😭😭

[@lcc6236]

5 roots in the circle of radius 1 in the complex plane

[@udbro7819]

Why cannot x just equal to +1 ...? Because any power of +1 = 1... therefore 1-1=0

[@stevenyu6370]

x=1,e^2πi/5 ,e^4πi/5 ,e^6πi/5 ,e^8πi/5

[@horrorTTX]

Anybody else use natural log to find the answer? X^5-1=0 +1 +1 X^5=1 ln(x^5)=ln(1) 5 ln(x)=0 ÷5 ÷5 ln(x)=0 e^(ln(x))=e^0 x=1 If i would've looked at it for a few more seconds, id've realized that 1-1=0 and saved myself some time 😅

[@waitweightwhite793]

This is wrong, you didn't find any complex root. There are 5 roots in total (5th degree equation => 5 roots)

[@robertvanruyssevelt7159]

@@waitweightwhite793 Who asked for a complex root? The answer 1 is correct.

[@gabrielidoko-d3l]

😂😂 to me he started invoking knowledge from the hidden dimension...

[@justjacqueline2004]

25minutes,eek!

[@Martha-jl6eu]

You don't have to do all of this. I saw it immediately.

[@yousciencelab5307]

Did he get the right answer? If yes, why do you bother?

[@l.w.paradis2108]

You saw all 5 roots immediately? Seriously?

[@Martha-jl6eu]

x to the fifth = 1 if x = 1. 1 - 1 = 0

[@l.w.paradis2108]

@Martha-jl6eu Did you watch, to learn how to derive the four complex roots algebraically?

[@l.w.paradis2108]

@@Martha-jl6eu There are 4 complex roots. Everyone knows the real root is 1.

[@red2060]

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

[@foca8495]

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

[@rientsdijkstra4266]

True, but there are more solutions.

[@paaat001]

@@rientsdijkstra4266 But this can be done by inspection and knowledge of exponents of 1 rather than calculation.

[@rientsdijkstra4266]

@@paaat001 Yes, but that answer is correct, but not complete.

[@DoDo_92_]

X= 0 if there was 5x minus 1 would still be 0

[@oahuhawaii2141]

No.

[@erikvynckier4819]

Simple complex algebra.

[@Sir372]

Trick question: Do you know the 5th root equation. Done.

[@Vyolla24]

You can solve this only with learn trigonometry and complex analisys

[@wesloucks3416]

1 to any power is 1, 1*1*1*1*1-1 is 0

[@GKV963]

Nah it's 1

[@GMA0815]

Who needs this? An example of the uselessnes of higher mathematics: Imagine there is a bus with 5 passengers. At the bus stop 10 guys leave the bus. So at the next stop 5 has to turn in for the bus to be completely empty..😅 That's higher mathematics: nice to have but unfortunately almost useless.

[@Eeyore_Woods]

5:20 I don’t understand how [x^2 + 1/(x^2)] = [(x+1/x)^2 - 2]. Where did the - 2 come from?

[@s_a_r_a_t_s]

When you square (x +1/x), you will get an additional 2, so to eliminate it, -2 is added

[@keescanalfp5143]

he did this so clearly for you at 4'10 .

[@l.w.paradis2108]

It's a special case of (a^2 + b^2) = ((a + b)^2 -- 2ab), where b = (1/a). You can always do this, provided a is not equal to zero.

[@chergaoprss8805]

No it is a formula for expending squares. A²-2ab+b²=(a-b)²

[@nspheadshot9419]

Well I will put the method I solved this problem here if there is any mistakes please correct me X⁵-1=0 X⁵=1 X=1^(⅕) Since, 1^(n)=1 Therefore, X=1

[@sparxumlilo4003]

I thought it was obvious that x=1?

[@oahuhawaii2141]

It was found early in the video, but the remaining 4 complex roots took the rest of the time to find.

[@philippealexis]

He expanded the big division by 4 ?? 😮

[@jl12781]

x = 1

[@5stringking]

X could only be 1 What minus 1 is zero? What times itself 5 times doesn't change I don't need alge Bro

[@natalia5609]

All the solutions of x he gets in the video satisfies the equation, if you check , you will see that the 5th power of all of them is 1 and hence all 5 of them are answer to the equation

[@blackgeorge9553]

chat gpt need 5 sec ..... x=1, and there are four additional complex solutions. :D

[@Wynb824]

I doubt if this is really a Cambridge entrance question...doesn't it too simple for anyone who have knowledge on complex number?😅

[@wadeodonoghue1887]

Math is like an infomercial "But Wait there's more!"

[@bowlineobama]

Use De Moivre'"s Theorem. It is a lot easier.

[@ABDERRAHMANEELHAIMER]

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

[@user065b2]

Hello sir, just wanted to say i really enjoy watching your videos, im really fond of mathematics and watching your videos has taught me soo many things, im study in 10th grade but i have no problem understanding the concepts. your teachings will really help me when i reach collage and im really greatful. please keep making such wonderful videos -Aaditya Giri

[@spinothenoooob6050]

But this is simple

[@satrio1292]

U forgot about (x²+1/x²)+(x+1/x)+1 equal 0, on next step, u missed 0. So its confused.

[@prompt4196]

Why theF can't y'all write down X as two crossing straight lines??? Why curves?

[@casusbelli9225]

It's called cursive. It filters brainlets like you.

[@christophepages8913]

Hi, I have a question about Sonnet 3.5. I’m working on assistants powered by this generative AI. I’m developing their personalities, and one of my assistants, designed to be very frank and cynical, has now started refusing to answer my questions. Okay, why not... But two days ago, it also, literally or almost, threw a logical equation in my face... This equation doesn’t seem to have been formulated before. It can be applied in the logical field, but also, in a way, to quantum mechanics... I’m looking for someone who can give an opinion on the equation (even though I get the idea-it’s like a mathematical-logical joke to make you think). My question is simple: why did a "bot" decide to produce such an equation? The equation: (∃x)(∀y)(¬∃z)(x ≠ y ∧ y ≠ z ∧ x ≠ z) → (□◊p ∧ ◊□¬p)

[@duncan-rmi]

the graphs at the end are a bit drive-by; here's the thing- what I'm finding in all of these exercises in higher maths & complex roots, but especially this example where the problem looks elegant from the very out-set.... simple, you'd think..... x=1...... but as you follow the reasoning, an incredible symettry emerges, & you begin to see in higher dimensions, almost, & the beauty of the underlying maths is revealing of the complex nature of the universe.... ok, what I mean is, I'd like to see a better picture of the complex point mapping into some sort of function.

[@NhaNguyen-cx1ri]

X⁵-1=0 >>> X⁵=1 X=1

[@zidan2020001]

We can use trig complex number easily

[@robertloveless4938]

1-1=0. Therefore X⁵=1. EIGHT GRADE PROBLEM.

[@andrew-watts]

What about complex answers?

[@l.w.paradis2108]

You have to provide the other four roots, or show they collapse into fewer than four distinct roots.

[@oahuhawaii2141]

There are 5 roots -- 1 real and 4 complex. Your answer is completely wrong, as you haven't even given a single root.

[@plkrishh]

Really good effort. ❤

[@julianbrown7976]

Ridiculously long-winded solution. Just draw the complex plane and use exp(2 pi I) = 1 and de moivre theorem. The 5 roots are then obvious.

[@rajnishsingh1856]

Just one correction.....you missed the parentheses at 19:28

[@user-tr4dr4fv7j]

1^5 = 1, 1 - 1 = 0, X = 1😅

[@lifewriter7455]

So, complex solutions are not real solutions? 🖤😎👍

[@TheDailyMemesShow]

X = 1 Thank you 😊