Can You Pass Oxford University Entrance Exam?
ฝัง
- เผยแพร่เมื่อ 7 ก.ย. 2024
- What do you think about this question? If you're reading this ❤️. Have a great day!
Check out my latest video (Everything is possible in math): • Everything is possible...
Can You Pass Harvard's Entrance Exam?: • Can You Pass Harvard's...
Hello My Friend ! Welcome to my channel. I really appreciate it!
@higher_mathematics
#maths #math
![[LIVE] MCHOICE & MINT AWARDS 2024 at TRUE ICON HALL, ICONSIAM](http://i.ytimg.com/vi/wMewUeT5nbY/mqdefault.jpg)
x⁶- x³ = 2 is a 6ᵗʰ order polynomial, so there are 6 roots. Two complex roots are missing from the solution.
The two missing complex roots come from
x³ = 2
x₅,₆ = cbrt(2)e^(±2πi/3)
x₅ = cbrt(2)[-½ + i sqrt(3)/2]
x₆ = cbrt(2)[-½ - i sqrt(3)/2]
These roots can also come from factoring:
x³ - 2 = [x - cbrt(2)][x² + cbrt(2)x + cbrt(4)]
and obtaining the quadratic roots
x =
½[-cbrt(2)±i sqrt(3)•cbrt(2)]
There are 6 roots. Mr has not solved the equation x^3 - 2=0 with all the roots
You are correct, he did leave out the other two solutions for x^3=2, he only did the real solution
The equation being of order 6, 2 more roots in the complex plane are needed. Those come from x³=2, which has the solutions
x=2^(⅓)·e^(2πin/3) for n=0,±1.
The n=0 solution is in the video. The other 2 are
x=2^(-⅔)·(-1±i√3)
You can also get these roots from
0=x³-2=[x-2^(⅓)][x²+2^(⅓)·x+2^(⅔)].
The 2 extra roots come from the quadratic.
correct but that are complex solutions (commenting before watching the video)
@@seekingCK
and x3, x4 given above are not complex solutions ?
@@whoff59 true. i solved it myself and got -1, 2^1/3, (1±i√3)/2 and then wrote into polar form and got -(2^1/3)/(2) ± i((√3)(2^1/3))/2 as the 6 answers
x⁶- x³ = 2 is a 6ᵗʰ order polynomial, so there are 6 roots. Two complex roots are missing from the solution.
The two missing complex roots come from
x³ = 2
x₅,₆ = cbrt(2)e^(±2πi/3)
x₅ = cbrt(2)[-½ + i sqrt(3)/2]
x₆ = cbrt(2)[-½ - i sqrt(3)/2]
These roots can also come from factoring:
x³ - 2 = [x - cbrt(2)][x² + cbrt(2)x + cbrt(4)]
and obtaining the quadratic roots
x = ½[-cbrt(2)±isqrt(3)•cbrt(2)]
The missing roots come from his partial solution to x^3 = 2, which has 3 roots, not just x = 2^(1/3). You could get 3 roots by factoring x^3-2=0 (after writing 2 in the form c = (2^(1/3))^3, so it becomes x^3 - c^3=0 and factors as (x-c)(x^2+xc+c^2) = 0. Solving the quadratic part is annoying, but straightforward.
Higher Mathematics doesn't usually miss stuff like this, I watch a lot of his videos. I noticed the two missing roots right after he got the first one, but figured he would come back to them. As someone else pointed out, the remaining quadratic is messy - it has cube-root coefficient/constants - but a careful use of parentheses and the rules of indices and you can harvest them. I used polynomial long division (x^3 - 2) divided by [ x - 2^(1/3) ] to get X^2 + [ 2^(1/3) ] X - [ 2^(1/3) ]^2 = 0 and used the quadratic formula from there. to get the two remaining +/- roots.
Highly doubt this is an Oxford Entrance exam question. Maybe if we open the solution space to the complex numbers.
We have a quadratic equation in x³, so we have 6 solutions in the complex space.
A= x³; A²-A=2 or A(A-1)=2 has two real solutions: A=2 and A=-1
x³=-1 has one real solution '-1 and two complex solutions -1/2 +/- i V3/2 (= cos a + i sin a, where a = 2pi/3)
x³= 2 has the same solutions, each time multiplied by - 2^(1/3)
Why does your Argand Diagram show 6 roots and you find only 4? A 6th order polynomial has 6 roots. I'm afraid you would not be reading Maths at Oxford!
1 has 3 cube roots: 1, -1/2 +/- irt3/2. cube rt 2 is each of these multiplied by the basic cube rt of 2, 1.2599.
cube rt of -1 : -1, 1/2 +/- irt3/2. Then they will consider marking your next answer. Hard luck!
x = -1 ± √3i are solutions of this equation too.
Why the roundabout process of factoring t^2 - t - 2 = 0?
It very apparent that the factors are (t -2) and (t + 1), if you know how to factor, not figure.
Who agrees with me? This guy is the Bob Ross of math! Love your videos. It’s a 6th exp so it should have six solutions. In the beginning where X^3=2 shouldn’t you have another complex set. Keep the videos coming sir.
Thank you for your support! Have a great day! Much love and respect!❤❤❤ What do you think about this math question?
The problem is very easy. Are you sure it is an OU entry exam question? And your solution is incomplete: x^3-2=0 has three solutions, one real, two complex. Otherwise, the video has the length (10 minutes) and quality that I expect from Higher Mathematics.
@higher_mathematics Can you please (pretty please with a pi pie on top! lol) remake this video, or, at least edit in another part, where you find the 2 missing roots. Thank you! Otherwise, the video is great!
Hmm, you forgot to solve all the roots of the first partial solution the same way you did for the second one
A 6-degree polynomial has 6 roots. You only found 4.
Thanks for teaching us this math proceeding. I learned the substitution to T variable. 😊😊😊
Life is too short to calculate , just say -1
Should we not have 6 values for X as the equation has x^6? I will appreciate your feedback👍👍 enjoy your videos.
There are 6 roots. Mr has not solved the equation x^3 - 2=0 with all the roots
@@kirillbarkov2 I thought I was going mad. Thanks👍👍
@@kirillbarkov2 one more enquiry, most of the values for X do not satisfy the equation, am I correct? Only solution is 2^1/3
@@kevinseptember2917 2 real solutions are -1 & 2^(1/3) and 4 Complex
@@kevinseptember2917 and 2^1/3 and -1 are the only Real solutions
Hello Mr, where are 5,6 roots. You forgot
I hear the call for the missing:
"Roots, bloody roots!"
- Sepultura
I would have never imagined, that there are more than two solutions, as I would have considered only the "Real Number Space". How can we be sure, that there are no more solutions in polydimensional spaces ?
Nice Explanation Sir ❤
(x^6)^2 ➖ (x^3)^2 = {x^36 ➖ x^9}= x^27 x^3^3 (x ➖ 3x+3)
Ehat is the step from t(t-2)+(t-2) to (t-2)(t+1)?
Factoring (t-2) as a common factor
just used calculator wont it be around 1.26
Thx a lot
my brain stop at x3 (lol)
Fail to enter oxford university 😂😂😂
i ❤ Mathematics
3:12 ...can some1 explain more
What he did was take the factors for his created equations. By changing -t into -2t+t it enabled this.
(t^2-2t)+(t-2) He could take the "t" factor out of the first equation and the implied 1 out of the second which could have been written as ((1)t-(1)2). without changing the second equation. This could also be written as ((1)(t-2)) if that would be easier to see. I hope this makes more sense now 🙏
Детские примеры )
You talk too much. Too much!
x^6-x^3=2
(x^3)^2-x^3-2=0
x^3 = t
t^2 - t - 2=0
t1=2
t2= - 1
x^3=-1
(x+1)(x^3+1). x^2 - x +1
x^3 + x^2
- x^2 +1
- x^2 - x
x+1
x+1
0
(x+1)(x^2-x+1) = 0
x1=-1
x2=(1 + isqrt3) /2
x3= (1 - isqrt3) /2
x^3 = 2
(x-cbrt 2) (x^3-2) x^2 + cbrt2 *x + cbrt4
x^3 - cbrt2 * x^2
cbrt2*x^2 - 2
cbrt2*x^2 - cbrt4 * x
cbrt 4 *x - 2
cbrt4 * x - 2
0
(x-cbrt 2)( x^2 + cbrt2*x + cbrt 4) =0
x4= cbrt 2
x5 = (-cbrt2 + sqrt( - 3cbrt4))/2
x5= (-cbrt2 + i*sqrt(3cbrt4))/2
x6= (-cbrt2 - i*sqrt(3cbrt4))/2