Well, k=5 is a solution by inspection. You could then take the full polynomial, k^3 - k^2 - 100 = 0 and divide (x-5) out of it and get a quadratic. That comes out x^2 + 4*x + 20 = 0, and the roots of that are -2 + i*4 and -2 - i*4. So 5 is the only real root.
@@TJ_104 Yes; it's a cubic equation and 5 is a root, so k-5 is a factor. Dividing it out leaves you with a quadratic. Of course you have to see that 5 is a root to start with - I'm just pretty good with numbers and knew that 5^3 is 125 and 5^2 is 25, and 125 - 25 is 100, so... there you go. It helps a lot that in problems like this on TH-cam the answer is almost always an integer, and usually a relatively small one. Not too many different things to try.
5 seconds to find the real root as k = 5. 5^3 = 125 and k^2 = 25 then 125 - 25 = 100. After that, divide by (k - 5) to get the quadratic and then the complex roots.
That was a fraudulent resolution of the exercise. As if by magic, 125 and 25 appear, both powers of 5, which is the solution. What a lack of seriousness of this channel!
Solving a general cubic equation is far too complicated for an admissions test so the applicant should assume that there is likely to be one solution that is easy to find. This is likely to be an integer and as k³ gets big very quickly, it's going to be a small integer. Given this thinking, it's pretty easy to find k=5.
It can be recived even esely: KKK - KK =100 KK*(K-1) =100 K*√(K-1)=10 10 can be obtained by multiplication 2*5, so if k=5, it is correct. In my daughter mathbook this type of equations often occurs.
You effectively noticed right at the beginning that k = 5 was a solution, so why not just long divide k^3 - k^2 - 100 by (k-5) to find the quadratic factor and then find the roots of that to finish the problem?
The first magic step, deciding to represent 100 as 125-25 (a cube - a square) is as hard as realizing that 5 is a solution by inspection, in which case, as others say, just divide by (k-5) and get the quadratic.
This was far too trivial: k*k*k has to be larger than 100, right? Well, 5 is the first integer for which this is true (8, 27, 64, 125, 216, 343,etc), and then it becomes obvious that 5 is in fact the solution. This should take 10 seconds max to solve without either a calculator or even a piece of paper.
I immediately guessed K=5 and that works. That extra DOT (for multiplication) was confusing, but then he wrote it again without the DOT. Anyway, if this is all it takes to get into Harvard, then I'll stick with my degree from CLEMSON UNIVERSITY.
I mean if you are just going to examine it and make an educated guess, then just say so, otherwise actually solve it using either the cubic formula, or the quadratic formula by factoring out (k-5).
@@superacademy247 You see immediately that x = 5 is a solution. Dividing the equation by (x-5) gives a quadratic equation that is trivial to solve, and you're done. Much shorter and less to write, with less risk of making mistakes.
I dont know mybe i am wrong but i see he wrote k.k.k.-k.k this is so wrong cuz you have multiplication dot right b4 minus which mean you multiply by one -k only proper equation should look like k*k*k*(-k)*k or i am wrong?
Instead of showing a long series of your algebra skills, it would be more intuitive if you discussed what approach you would take to arrive at your answer so that students would know what to think first when they encounter problems like this. The algebra behind is trivial
Hello Mr. Paper Killer! Here too much writing and talking again! Is it "very fancy" in very long way? When you realise that 100=125-25, k=5 is certain. But you like writing; often same thing over and over! Do you really need to write these? K-5=0 K=0+5 K=5 Lol!
It's not complex...I don't understand the complaints. In this example, rule is you do all the multi's then the subtraction. 5x5x5=125. 5x5=25. 125-25 = 100. Grow up.
sorry, but you are not presenting a solution. Instead you use the classic "trick" to guess and try a zero point of the cubic equation, and then do polynom division.
That's not maths, it's just guesswork. Five is a more or less obvious solution so you factor out (X-5) and you get a second degree equation. Nothing interesting here.
Sorry I don't know if you are a teacher or a student. First, please repeat x.x.x - x.x =100 ==> x.x.x= 100+x.x (from here we know for sure k>0 and this is the solution condition of the equation). When solving the equation, we only have one solution, which is 5. There is no addiction that satisfies K>0. So if you are a student, you should study again, and if you are a teacher, you should try to gain more knowledge and not be too stupid to get views or likes.
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Well, k=5 is a solution by inspection. You could then take the full polynomial, k^3 - k^2 - 100 = 0 and divide (x-5) out of it and get a quadratic. That comes out x^2 + 4*x + 20 = 0, and the roots of that are -2 + i*4 and -2 - i*4. So 5 is the only real root.
Démonstration moche!
Очередной пример иудацкого решения. При этом этот идиот десять раз переписыаает одно и то же... Ну тупой...
Dividing by k-5 is good idea. Now i finally understand why people do this^^
@@TJ_104 Yes; it's a cubic equation and 5 is a root, so k-5 is a factor. Dividing it out leaves you with a quadratic. Of course you have to see that 5 is a root to start with - I'm just pretty good with numbers and knew that 5^3 is 125 and 5^2 is 25, and 125 - 25 is 100, so... there you go. It helps a lot that in problems like this on TH-cam the answer is almost always an integer, and usually a relatively small one. Not too many different things to try.
k = 5 and two complex solutions.
k³ - k² - 100 = 0
(k - 5) (k² + 4k + 20) = 0
k₁ = 5
k₂,₃ = -2 ± √(4 - 20)
= -2 ± 4i
k₂ = -2 - 4i ∨ k₃ = -2 + 4i
アートラムイレーシスア
テーミと言う。
It took me 30 seconds, reasoning. K=5.
It took me 29sec
lol, took me 5 seconds.. but to be fair i'm teaching maths so i see 5^3=125 from time to time in other situations
@@TJ_104 I doubt it, life experience tells , that's a very long 5 seconds 😉
5 seconds to find the real root as k = 5. 5^3 = 125 and k^2 = 25 then 125 - 25 = 100. After that, divide by (k - 5) to get the quadratic and then the complex roots.
That was a fraudulent resolution of the exercise. As if by magic, 125 and 25 appear, both powers of 5, which is the solution. What a lack of seriousness of this channel!
I thought the same.
How would he approach:
K^3 - K^2 = 373.7854 ?
@@Russ--R
K = 373.7854 - 367.2146 =0
K = 5
Isn't a fraud, It is only different ways to do it.
Solving a general cubic equation is far too complicated for an admissions test so the applicant should assume that there is likely to be one solution that is easy to find. This is likely to be an integer and as k³ gets big very quickly, it's going to be a small integer. Given this thinking, it's pretty easy to find k=5.
No need for all the work. Use synthetic division to get the quadratic!!
It can be recived even esely:
KKK - KK =100
KK*(K-1) =100
K*√(K-1)=10
10 can be obtained by multiplication 2*5, so if k=5, it is correct.
In my daughter mathbook this type of equations often occurs.
You effectively noticed right at the beginning that k = 5 was a solution, so why not just long divide k^3 - k^2 - 100 by (k-5) to find the quadratic factor and then find the roots of that to finish the problem?
The first magic step, deciding to represent 100 as 125-25 (a cube - a square) is as hard as realizing that 5 is a solution by inspection, in which case, as others say, just divide by (k-5) and get the quadratic.
100 = 5^2×4 = 5^2(5-1)
k^3-k^2=k^2(k-1)
therefore
k=5
easy math
Factorise. k^2(k-1). Check with 2 or 3 naturals. 5 is solution: 25*4.
Please plot a graph for y=x^3 and y=x^2+100 and in how many places both the graphs cut each other. You can visualise the solution easily.
k^3 - k^2 = 100
k^3 - k^2 - 100 = 0
(k - 5)(k^2 + 4k + 20) = 0
k = 5, -2 +/- 4i
k^3 - k^2 = 100
F(x) k^3 = 3k^2
3k^2 - k^2 = 2k^2
F'(x) 2k^2 = 4k
4k = 100
k = 100 / 4
k = 25
25*3*2 - 25*2 = 100
150 - 50 = 100
This was far too trivial:
k*k*k has to be larger than 100, right? Well, 5 is the first integer for which this is true (8, 27, 64, 125, 216, 343,etc), and then it becomes obvious that 5 is in fact the solution. This should take 10 seconds max to solve without either a calculator or even a piece of paper.
You are ignoring the imaginary solutions.
@@billjones4159 Yes, and that is/was intentional. 🙂
k^3-k^2=k*k(k-1)=5*5*4 multipliers of 100 so k=5
You would better state at the beginning that you expect \(k\) to be complex, not an integer, for example.
I just look at it and PING (5.5.5)-(5.5)=100 125-25=100
I immediately guessed K=5 and that works. That extra DOT (for multiplication) was confusing, but then he wrote it again without the DOT. Anyway, if this is all it takes to get into Harvard, then I'll stick with my degree from CLEMSON UNIVERSITY.
Have you ever seen k = 5 + 0 and only after that conclude k = 5?!!! This person is not a teacher or math student. He makes math boring.
It's his voice.
When you decided that 100 was 125-25 (5^3 - 5^2), you gave the answer! K=5
Since 2:30
K^3 - K^2=5^3-5^2
K = 5
k³ - k² = 100
k².( k - 1 ) = 2². 5²
Le facteur: 5, apparaît évidant.
5².( 5 - 1) = 5². 4
( k³ - k² - 100 )/( x - 5 ) = x² + 4x + 20
Discriminant: 4² - 4.(1)(20) = -8²
x = -2 - 4.i ou x = -2 + 4.i
Seule solution réelle: x = 5
K=5 só de olhar
Coloquem valores que no sean evidentes
I mean if you are just going to examine it and make an educated guess, then just say so, otherwise actually solve it using either the cubic formula, or the quadratic formula by factoring out (k-5).
Nice
Thanks
Before the solution he wrote the wrong problem he wrote k³ × - k² = 100 and solved for k³ - k² = 100
Video is too short, i guess you can explain solution much slower and deeper
When submitting an exercise, you should precise the domain of definition for k : R, Z, C… If not, no solution could be found
I agree, and conventionally k would an integer.
Are you serious? This thing has a 14 minutes video 🤷🤷🤷..
5*5*5-5*5= 125-25=100
✅✅✅
k = 5 by inspection. QED. LOL.
Too much work. This is a very simple equation.
Why do you think so? I disagree.
@@superacademy247 You see immediately that x = 5 is a solution. Dividing the equation by (x-5) gives a quadratic equation that is trivial to solve, and you're done.
Much shorter and less to write, with less risk of making mistakes.
Okay 👌
My child solved this problem in 2 minutes
Did he ge the imaginary numbers.
10^10 2^5^2^5 1^1^2^1 2^1 (x ➖ 2x+1).
Geometric solution...K*k=1, k*k*k=6, 6-1=5,k=5
K.k.k -- k.k = 100
k.k.k--k.k -- 100 = 0
k.k.k --125 --k.k+25 = 0
k.k.k--5.5.5--(k.k --5.5) = 0
(k--5){k.k+5k+25--(k+5)}= 0
(k--5) = 0
then k = + 5//
The Order is: Defend, Be Strong, Be Proud, Be Pure.
Solution by inspection 5
I answer 5 , less than 5 sec.
Just from Possibilties, probability
K = 5, 5×5×5 - 5x5 = 100
Easy One...
FIVE, not negative 5 like the other Harvard test.
Cubic equation with a trivial solution k=5. Divide by (k-5) to obtai quadratic equation fir two additionl complex solutions.
I dont know mybe i am wrong but i see he wrote k.k.k.-k.k this is so wrong cuz you have multiplication dot right b4 minus which mean you multiply by one -k only proper equation should look like k*k*k*(-k)*k or i am wrong?
It was a typo. It should be ignored!
x=5. There I just guessed it and turns out I was right
5 by inspection
Instead of showing a long series of your algebra skills, it would be more intuitive if you discussed what approach you would take to arrive at your answer so that students would know what to think first when they encounter problems like this.
The algebra behind is trivial
5😊
Pour un niveau de l olympiade il faut aller plus vite
can we say
k.k.k-1=5×5×4,
hence, k=5.
thx
3k-2k=100
K= 100
You wrote two different equations:
k*k*k*-k*k=100
and
k*k*k-k*k=100
(see time stamp 0:24)
It was an error.
and apply Ruffini's rule?
K = 5 a simple inspección.
He should apply a formula just as we apply the formula for a quadratic equation. This solution is unsatisfactory and inadequate
Great job.
Thanks for the visit 💪💪💪
Where did you get the numbers from. 4:57
Pulled it out of his ass
Ur guess of 100 = 125 - 25 implies u knew k = 5 already. BOO.
K=5. 125-25
Crazy
My chat did it in 3 sec!
Ugly problem, ugly solution. There's a kind of beauty here.
75 % I was focused....
the last 25 % I just can't understand!⚠️
I think you complicated all unusefully!😢
К =5
Harvard? Realy? 5 sec.
K=100
need 5 sec, no pen and paper
5
125-25...
Is this video a joke?!
k^2 (k-1) = 100. Looking at it & thinking for 10 seconds... Solution: k = 5
This should be short. Unnecessary made it long.
K 1= 10 also K 2= 99
k=5
I just use Matlab symbolic…
😂😂😂😂
I did it in my head in 10 seconds, tops. This video is nonsensical.
Hello Mr. Paper Killer! Here too much writing and talking again! Is it "very fancy" in very long way? When you realise that 100=125-25, k=5 is certain. But you like writing; often same thing over and over!
Do you really need to write these?
K-5=0
K=0+5
K=5
Lol!
Yes 😂. Many people don't realize the beauty of Math is hidden in trivial things .
That's not a good solution. How do you get to 125-25 out of the blue
By inspection
テーミと言う。
It's not complex...I don't understand the complaints. In this example, rule is you do all the multi's then the subtraction. 5x5x5=125. 5x5=25. 125-25 = 100. Grow up.
Too much work
Kann man damit Geld verdienen? 😮
Very primitive solution. Badly explained,
Not as good as Eddie Woo.
Useless video.
sorry, but you are not presenting a solution. Instead you use the classic "trick" to guess and try a zero point of the cubic equation, and then do polynom division.
Which is a solution.
The only REAL answer is 5.
Example: How many asterisks? *****
You can walk 5 yards. But can you walk 5i yards? Can you walk 5i steps, draw 5i trees?
Put another way, i is not a number, it is a tool for calculation. If it remains in the solution then it is not a solution at all.
It is still a number just not label as a "real" number.
Yes you can but the steps are not physical and you can draw five trees in you head.
That's not maths, it's just guesswork. Five is a more or less obvious solution so you factor out (X-5) and you get a second degree equation. Nothing interesting here.
if i am your teacher i will you 0 marks because . it's only one k=5 , but you have three values , your answer is completely wrong
You fail as a teacher.
Sorry I don't know if you are a teacher or a student. First, please repeat x.x.x - x.x =100 ==> x.x.x= 100+x.x (from here we know for sure k>0 and this is the solution condition of the equation). When solving the equation, we only have one solution, which is 5. There is no addiction that satisfies K>0. So if you are a student, you should study again, and if you are a teacher, you should try to gain more knowledge and not be too stupid to get views or likes.
x=5
Какой смысл получать ответы в мнимых числах и выдавать их за правильные ответы, ведь проверить правильность такого ответа подстановкой не получится.
It can be checked. It's not a big deal.
K= 100
5
K =5
K=5
K=5