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 UA-cam the answer is almost always an integer, and usually a relatively small one. Not too many different things to try.
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.
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.
To really understand it I used numbers and accidentally chose the right one. It took less than 10 seconds. But I didn't look for complex number solutions.
Factor 100 to 2²*5² and then see it's 5*5*4 or 5*5*(5-1) , which means k = 5 is a solution. Rearrange the equation with all the terms on the left, and factor out (k - 5) to yield a quadratic equation that's easily solved.
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).
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.
@@aldillion9 .. Not a simple inspection. This is the “arithmetic method” that is often known before learning the “algebraic method”. After all, a logical thinking 12-year-old child will find a solution faster than a Harvard student.
It took me 10 seconds. k^3-k^2 = k^2(k-1) = 100 You have 100 only with 100*1, 50*2, 25*4, 10*10 and between 100, 50, 25 and 10 the only squared is 25 so K=5 ... 10 seconds.
@@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
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.
Worst explanation ever. Sorry to say. At 1:12 you basically skipped to one of the solutions by mentioning index form and disregard the rest Then you go on and on to tear the whole thing apart, even though you already have one of the possible (and in this case the only) solution(s). This has to be math ragebait 😀
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!
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 UA-cam the answer is almost always an integer, and usually a relatively small one. Not too many different things to try.
That's not true! The second solution is the Bible or TORA for now.😊
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。
No need for all the work. Use synthetic division to get the quadratic!!
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?
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 😉
# Python
for k in range(100):
if k*k*k-k*k == 100:
print(f"k * k * K - k * k = 100 | k = {k}")
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.
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.
Exactly!
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.
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.
I just look at it and PING (5.5.5)-(5.5)=100 125-25=100
To really understand it I used numbers and accidentally chose the right one. It took less than 10 seconds. But I didn't look for complex number solutions.
The Order is: Defend, Be Strong, Be Proud, Be Pure.
k = 5 by inspection. QED. LOL.
You would better state at the beginning that you expect \(k\) to be complex, not an integer, for example.
Since 100=5^2(5-1), hence k=5 to demonstrate the identity of expression.
k^3 - k^2 = 100
k^3 - k^2 - 100 = 0
(k - 5)(k^2 + 4k + 20) = 0
k = 5, -2 +/- 4i
perfect
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.
Factorise. k^2(k-1). Check with 2 or 3 naturals. 5 is solution: 25*4.
Factor 100 to 2²*5² and then see it's 5*5*4 or 5*5*(5-1) , which means k = 5 is a solution. Rearrange the equation with all the terms on the left, and factor out (k - 5) to yield a quadratic equation that's easily solved.
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).
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. 🙂
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
k³ - k² - 100 = 0
We can see k = 5, so (k - 5) is a factor:
(k - 5)*(k² + 4*k + 20) = 0
k = 5, -2 ± i*4
Since 2:30
K^3 - K^2=5^3-5^2
K = 5
k^3-k^2=k*k(k-1)=5*5*4 multipliers of 100 so k=5
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.
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.
Easy One...
Are you serious? This thing has a 14 minutes video 🤷🤷🤷..
5*5*5-5*5= 125-25=100
✅✅✅
When you decided that 100 was 125-25 (5^3 - 5^2), you gave the answer! K=5
thx
K=5 só de olhar
Coloquem valores que no sean evidentes
7th grade students will do that: k*k*k - k*k = 100 >>> k^2 (k-1) = 100 >>> (k-1) =100 / k^2
👉 (k-1) is an integer >>> 100 divisible by (k^2) >>> (k^2) = 2 4 5 10 20 25 50 😁
👉 (k^2) is the square number >>> k^2 = 25 >>> k = 5 😁 Very simply 👍
BUT, that's a solution by inspection...not "Algebraically".
@@aldillion9 .. Not a simple inspection. This is the “arithmetic method” that is often known before learning the “algebraic method”. After all, a logical thinking 12-year-old child will find a solution faster than a Harvard student.
Where did you get the numbers from. 4:57
Pulled it out of his ass
It took me 10 seconds.
k^3-k^2 = k^2(k-1) = 100
You have 100 only with 100*1, 50*2, 25*4, 10*10 and between 100, 50, 25 and 10 the only squared is 25 so K=5 ... 10 seconds.
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
Solution by inspection 5
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
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.
5 by inspection
K = 5, 5×5×5 - 5x5 = 100
My child solved this problem in 2 minutes
Did he ge the imaginary numbers.
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.
He should apply a formula just as we apply the formula for a quadratic equation. This solution is unsatisfactory and inadequate
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 👌
per tentativi orientati ( > oppure
and apply Ruffini's rule?
I answer 5 , less than 5 sec.
Just from Possibilties, probability
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//
As an assembly language programmer, I say k=10.
K = 5 a simple inspección.
Pour un niveau de l olympiade il faut aller plus vite
can we say
k.k.k-1=5×5×4,
hence, k=5.
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
Long-winded. I changed it to k^2(k+1)=100 and the only factor of 100 that fits this is 5.
K=5. 125-25
Nice
Thanks
Crazy
3k-2k=100
K= 100
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
Ur guess of 100 = 125 - 25 implies u knew k = 5 already. BOO.
My chat did it in 3 sec!
😂😂😂😂
Х^3-Х2=113
???
need 5 sec, no pen and paper
De hood of math.
K=100
125-25...
К =5
k=5
Geometric solution...K*k=1, k*k*k=6, 6-1=5,k=5
5
Ugly problem, ugly solution. There's a kind of beauty here.
K 1= 10 also K 2= 99
5 OK And the second element in the Bible or the Torah?
5
So algebra gives one real like 5 and two no real results. Keep that under mind that algebra, as and math, can describe not real things!
Its too cumbersome solution. We wont get that much time in exam. Its just a guess work.
Harvard? Realy? 5 sec.
K=5. แค่คิดในใจ555
This should be short. Unnecessary made it long.
That's not a good solution. How do you get to 125-25 out of the blue
By inspection
I just use Matlab symbolic…
K=5 sem mais nada de perda de tempo
😂😂😂
I did it in my head in 10 seconds, tops. This video is nonsensical.
75 % I was focused....
the last 25 % I just can't understand!⚠️
I think you complicated all unusefully!😢
Are the students at Hravard that stupid? In Romania, this is the 6th left problem at most of very low difficulty
Is this video a joke?!
k^2 (k-1) = 100. Looking at it & thinking for 10 seconds... Solution: k = 5
100 = 5^2×4 = 5^2(5-1)
k^3-k^2=k^2(k-1)
therefore
k=5
easy math
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
Worst explanation ever. Sorry to say. At 1:12 you basically skipped to one of the solutions by mentioning index form and disregard the rest
Then you go on and on to tear the whole thing apart, even though you already have one of the possible (and in this case the only) solution(s).
This has to be math ragebait 😀
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 .