Harvard University Admission Interview Tricks | Find the Value of k=?

Harvard University Admission Interview Tricks | Find the Value of k=?

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Explore the interesting world of exponential equations with varying bases and unlock the power of exponentiation! In this educational video, we delve into the concept of exponential equations where the base is not a constant value. Learn the fundamentals, discover the properties, and master solving equations with different bases, all explained clearly and concisely. Join us on this exponential journey and level up your mathematical prowess!

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[lynnrathbun]

Six equally spaced points on the circumference of the circle of radius 2 in the complex plane. 2e^(i*n*(pi)/3 ) n=0 to 6

[tedpancoast]

As my favorite teacher used to say, “isn’t it intuitively obvious?”

[jiangchuYT]

Using de Moivre's formula, the answer is trivially 2cis(k\pi/3), 0<=k<=5.

[igpoo]

Hours is the best way of doing it. You could also see the solutions of +/-2, and divide k power 6 - 2 power 6 by k-2, then by k+2 and solve the remaing quyartic

[MADUCHUKWUDI-ev]

Excellent. In advance level, you are required to show exhaustive solutions for K.

Unlike ordinary level in which only real solutions are asked for.

[AndorMatus]

Great example, how to make science from the fact 2=2.

[oahuhawaii]

k⁶ = 2⁶
k = 2*cis (2*π*n/6), n integer
= 2*cis(n*π/3), n = 0, ±1, ±2, 3
= ±2, 1±i*√3, -1±i*√3

[bakulchandramajumder]

As it is an equation of degree 6, it has 6 solutions..needs to find all six roots...2 are real and 4 imaginary/complex roots.

[billrandle]

Solve the equation :
k⁵=2⁶.
Rearrange the equation to give :
(k/2)⁶ - 1 = 0
Let z =k/2, so we have:
z⁶ - 1 = 0
This is a polynomial equation of degree 6 and so
It has six roots. zᵣ. (r = 0, 1, 2, 3, 4, 5).

The 6 roots are:
zᵣ= cos(2rπ/6) + isin(2rπ/6)
r = 0, 1, 2, 3, 4
So the 6 roots are
±1, 1/2 ± i√(3/2), - 1/2 ± i√(3/2) .

Recall, z=k/2

thus k=±2, 1± i√3, - 1±i√3.

,

[AbouTaim-Lille]

Dévide by 2⁶ and then put Z=e^2πi Θ= x/2 so that e^2πiΘ = 1= e^2πik.
Then u substitute k for 0, 1, ...5 and the rest is calculous.

[Manganyiful]

Why are you not applying the exponential rule instead of bitting behind the bush. Answer is -2 or 2 a exp n = b exp n, therefore a=b

[yiutungwong]

π = 2 in the Riemann Paradox and Sphere Geometry System Incorporated...

In English and Deutsch ...
University 🎓🏫 Universal ♾️💍🔐 As same as Universe...
That's means π= 2 in the Universe...

So K can be Solved for 2, -2, π and -π (Real Number)
+ 4 Complex Number Solution

[MrPandaJJ]

Divide both sides by 2^6
Sub x = k/2
Then it becomes a 6th roots of unity problem
and the rest follows

I don't think Harvard would assume the candidate not knowing Euler's formula...

[robertloveless]

Ever heard of Ocham's razor?? PERFECT example

[yiutungwong]

4 Complex Number in That Solution...
And K = 2 and (-2)... (Real Number)

[carlosjimenez]

Mucho más sencillo: son las seis raíces sextas de 64, todas de módulo 2 y argumentos, 0, 60, 120, 180, 240 y 300, como bien dice g414nk0.

[gtnk]

Las 6 soluciones (2 reales, 4 complejas) forman los vértices de un hexágono regular.

[Roger-xghh]

Whatttt... I knew that the answer was 2 and -2 in three second... It is just to see the ecuation

[lynnrathbun]

Do it in your head in 15 seconds in complex polar coordinates

[PSRao_]

This shows the standard of Harvard University.. such a simple problem.. I am sure the same person takes 1 hour to solve a=b/2, while b=10, what is a=?