By rewriting -80 as -16-64, you can see that you get -4^2-4^3. Since the equation results to 0, you need (x^2-x^3) = 80. You can already see that one solution is x=-4, as (-4)^2=16 and (-4)^3=-64, so 16-(-64) = 80. To find the complex solution you just use synthetic division as we know that if x=-4 is a solution, then (x+4) is a factor. From there you get both the real and complex solutions!!
Not sure, if THIS rewriting is obvious, but we know, if there is an integer solution, it divides 80 and we know x²(1-x) is approx x³, so I started guessing: 5 too big, 4 small enough, so minus 4? Works! Then just polynomial divide and solve quadratic
@@hakanerci4372 but there are literally infinity possibilites of writing a-b = constant, which in general, if you dont "see" it because it's some power of 2 stuff, there are more reliable ways. When there is no rational solution, you probably have to go the longest but most reliable route.
@@zyklos229 when I was high school. And exam for going to university Usually find rational solution. All question for rational solution. İrrational doesnt use. For example i root (- 1) doesnt use and doesnt real solution
While solving by inspection is not generally applicable, it does require some thought and insight into the structure of the equation. For example, when you see that x^2 is greater than x^3, you know that the root must be negative. You next observe that 80 is the sum of powers of 4, so you guess that maybe x=-4 is a solution. Then you look for a way to use the known solution to find the remaining 2. You know that (x+4) is a factor, so you divide the cubic by x+4 and obtain the quadratic and thereby its two complex roots.
My method was to search first for integer solutions. Since x^2(1-x)=80 we're searching for square factors of 80, which are 4 and 16. x=-4 for x^2=16 works.
A cubic equation with an obvious solution: -4. Divide by (x+4) and solve the resulting quadratic equation via the well known formula. Not tricky at all.
Well if you instantly spot -4 yeah it's easy but not everyone will, and it's an interview not a test so they probably intend for the student to solve it the hard way
@@Ben-vl5ew not really, if you're expecting a real valued solution, and the only coefficient is 80, then you would know to try factors of 80. So 2,4,5,8,10,16,20,40,80. The entire algebraic work he did is a bit much. You can use synthetic division to try these quickly. One you find that x+4 is a factor, then the discriminate will show the other two solutions are complex, and you apply the quadratic formula to find them.
It takes a few seconds to see that x = -4 is an answer. There will actually be three answers, of course, but the other two are probably complex. Anyway, (-4)^2 is 16, and (-4)^3 is -64. 16 - (-64) = 16+64 = 80. Q.E.D.
Without reading the comments or looking at the solution, obviously x^2 is greater than x^3. that can only happen if x is a fraction - less than 1 - or x is negative. Trying to guess the approximate value - in other words applying a heuristic - we can quickly see that -3 (9+27) is too small and -5 (25+125) is too large. -4 gives us 16 + 64 which = 80 so x = -4 is the answer. Takes much longer to type than to work it out.
@@David-iy1zt it's found by rigor, dude, I just wrote a simple formal answer of why it must be 4, analytically and not evidence-based - check it if interested!
No, it's testing a student's mathematical intuition. Trying to solve it through the direct application of algebra doesn't. Many problems have been solved by making some educated guess about the form of the solution - Dirac's wave equation is one such famous example.
x ^ 2 * (1 - x) = 80 Now, divide both sides by x2 (assuming x≠0): 1 - x = 80/(x ^ 2) Rearrange to get a single variable: x = 1 - 80/(x ^ 2) x^2 - x^3 = 80 x^2(1 - x) = 80 At this point, you could use numerical methods or graphing to find approximate solutions for. You can also try guessing integer values to check if they satisfy the original equation. x^2 - x^3 = 4^2 - 4^3 = 16 - 64 = -48 x^2 - x^3 = 5^2 - 5^3 = 25 - 125 = -100 x^2 - x^3 = (-4)^2 - (-4)^3 = 16 - (-64) = 16 + 64 = 80 Thus, x = -4 is a solution to the equation x ^ 2 - x ^ 3 = 80 Therefore, the equation holds true for x = -4
Look at mine: x^2 - x^3 = x^2*(1-x) = 80 = 2^4*5 => x^2 divides 2^2k and x is positive integer, and k is either 1 or 2 => 1-x = 5*2^s but then x = 1-5*2^s (since 1-x must divide 5*2^s, where s is 2 or 1 respectively with k) => 2 does not divide x which is a contradiction, then 1-x = 5 => x = -4 ). Yours is also great, I like that you used my approach at factoring the equation!
so I solved it by guessing and factoring, by testing intergers in range of -5 to 5, I managed to found out x=-4 is one solution now we can factor it by x^3-x^2+80=0 (x^3+4x^2)-(5x^2+20x)+(20x+80)=0 (x+4)(x^2-5x+20)=0 after plugging in quadratic formula for x^2-5x+20=0, we managed to have 2 complex solutions and 1 real solution being -4
Hear, hear. I watched the video hoping to see a more rigorous procedure than "Hey, let's do this weird thing that I've reverse-engineered" and was disappointed.
@@mikehood6745 rigor: x^2 - x^3 = x^2*(1-x) = 80 = 2^4*5 => x^2 divides 2^2k and x is positive integer, and k is either 1 or 2 => 1-x = 5*2^s but then x = 1-5*2^s (since 1-x must divide 5*2^s, where s is 2 or 1 respectively with k) => 2 does not divide x which is a contradiction, then 1-x = 5 => x = -4 .
An integer coefficient polynomial will have an integer root if it divides it's zero-power term. By some trial and error, one could try to find such number, if it exists. After I found x = -4, it was a matter of factoring out (x+4) and solving the quadratic.
The 80 on the right hints you that the real solution might be an integer. Plotting the graphs of x^2 and x^3 in your mind, you realize that the real solution has to be negative. Try -5? Nah, too big. Try -4? You got it.
At step two of the solution he actually gives away the answer when he introduces the 16- 64 substitution for 80. The solution is actually arrived at when you realize that there is no positive value for x that works, it must be negative. There's no need for all these hieroglyphics.
From looking at the result and assuming simple answer, one can solve it in mind, testing negative integers from -8 up to 0 and find -4 as the obvious answer, but I wouldn’t be so sure that it’s the only root of the equation. It may have two other roots on a complex plane.
Take “x square”as common and u will have x^2(1-x)=80 . And then u just need to think which makes sense and juz like that I get (-4) . Which I can make 16(positive) due to square and 1-(-4)=5. Then (-4)square =16. Finally 16(5)=80.
If x=e, e=-80 and x=-80 so x=e. Judging by irrational number being 50:50 ratio in all numerical values apart from rational e^2+e^3 shows as 27 which is technically the difference between e^2-e^3. We now just have to add 27 or 30 to premise of e which is 50 and hence making the result as 80
Since x^2 - x^3 is a positive number (80), x has to be a negative number. since x is negative, we can rewrite the equation as x^2 +x^3=80. Now there are not many possibilities left, we find x=4 swiftly.
Here's how to arrive at the answer mentally in under 10 seconds: X^2 - x^3 = x^2 * (1 - x) = 80 > 0 implies that x < 1 since x^2>0 always. Trying x=1 and x=0 don't work, and x is almost certainly an integer,, so we can safely assume that x
You've got a solid approach there! The mental shortcut is brilliant-by transforming the problem and focusing on \( w = -x \), it simplifies the guessing process significantly. Since \( w^2(w + 1) = 80 \) becomes more manageable, testing the small integers (2, 3, and 4) quickly leads you to the right answer. It’s impressive how you narrowed the possibilities with logical steps like assuming \( x < 0 \) based on \( x^2(1 - x) = 80 \). I also agree with your point on the substitution. Choosing the values like \(-64 - 16\) seems arbitrary unless you already know \( x = -4 \). Your alternative, \(-72 - 8\), would indeed work if you tried fitting in different values. The key is recognizing the form of the equation and finding the right factorization pattern. It probably wouldn’t be an Oxford-level math exam question unless it’s about thinking quickly with patterns, but certainly a good mental challenge over a pint at the pub!
It’s easy to get one of the answer is -4 from the first sight, and therefore we could divide the original equation by x+4 to get the other two roots… this is a faster way to solve this problem I think.
Why does this video about a supposed Oxford interview question have an image of perhaps the must famous Cambridge applied mathematics professor since Newton?
x^2 - x^3 = x^2*(1-x) = 80 = 2^4*5 => x^2 divides 2^2k and x is positive integer, and k is either 1 or 2 => 1-x = 5*2^s but then x = 1-5*2^s (since 1-x must divide 5*2^s, where s is 2 or 1 respectively with k) => 2 does not divide x which is a contradiction, then 1-x = 5 => x = -4. Fairly ez, did it on my browser search bar cuz I was too lazy to take the notebook. And BTW understanding enough number theory will take you far - look at my logic, it's a lot simpler(once you understand it) than solving equations and invoking school formulas.
At a quick glance, X^3 will be larger than x^2 hence x is likely to be negative to give a positive sum of x^2 and x^3. 64 + 16 = 80 hence x= -4 is a clear solution. -4^2 = 16 and -4^3 = -64 then x^2 - x^3 =80.
@@sherlockpoirot9946 Yes though as the comments below show, to find the complex solution you just use synthetic division as we know that if x=-4 is a solution, then (x+4) is a factor. From there you get both the real and complex solutions.
I just figured: If (x^2 -x^3) > 0, and (x^2 -x^3) > 1, x has to be negative, and |x| has to be greater than 1. If x is positive and greater than 1, the result will always be negative since the cube is greater than the square. So now I have to find an x such that |x|^2 + |x|^3 = 80 And due to too much binary, the answer was pretty obvious I dunno if this made sense, it's 4:30 am lol
I don't think you have explained the solution to those people who will have difficulty with quickly seeing the solution. You should try a general approach on how to solve this equation.
Solved in 3 seconds. Obvious that it had to be a negative number. Larger than 80 when cubed won't work so -5 and lower is out immediately. The complex solution? Couldn't be arsed.
The answer is: -4 Rewrite the equation as (x+4)(-x²+5x-20)=0 The you need to solve. One answer is -4 and there is no other cause delta is less than zero
I was like... If it's a tricky question it probably has an integer solution and we're not just trying to use the standard methods for cubic equations. So if x^2(1-x)=80 has an integer solution then we're trying to factor 80 in a square factor and something else. We have 4×20 and 16×5. 16×5 works with x=-4. After dividing -x^3+x^2-80 by x+4 we find out it's the only solution.
I just wish my interview questions had been like that :) My humdinger was (the old chestnut) "how many zeroes are on the end of 365 factorial?" Not hard in retrospect, but a little more so when you're young and unprepped.
When I was young, I would have started working on the answer directly. Now that I'm older, my first reply probably would have been: why would you _want_ to know the number of trailing zeroes in 365 factorial in the first place? (Of course I understand the goal is simply to test the resourcefulness of the applicant, but the interviewer's response might be revealing and help me to inform my decision as to whether I'd want to work for them or not. An irritable 'this is just a test question we use to see if applicants have what it takes' might be a write off from my side. If they come up with a more original reaction, that may be a plus.
Guys it is a tricky one By simply observing we will see that cube of any number is always greater then square so in this case we will get negative number but Here given that resultant is +ve so definitely number will be -ve Then do hit and trail 16+ 64 = 80 Hence answer will be 4
i find this so frustrating; how would you solve this problem if it was 77 instead of 80 (and didn’t break so cleanly into the sum of a cube and a square)
Mental math and the answer is -4. Haven’t read the comments nor watched the video. In fact as I write this the YT ads are still running. There is NO way this is a Harvard question. Edit: I omitted the complex solutions, but that’s also easy considering we have two passive solutions in the form of x² and (1-x) so we must have 2 complex conjugates which simplifies the quadratic formula and gives us -22.5 +-rt(55)j/2
This problem has what is called the 100-million-dollar solution. A man when confronted about winning the lottery and dealing with the boredom from it, he confirmed the solution was to go take a nap on the couch and when he woke up, his problem was solved. Believe it or not, this problem is solved with this very same solution.
It took 2 seconds to solve this. Due to our education system which is unnecessary hard, any 2nd grade highschool student could solve this if that they are planning to go any engineering college. Because there is harder than this question must be solved within 30 seconds. The solution for me is if the x^2-x^3= 80 > x must be negative so: x^2+x^3 > x^2(x+1) which is equal to 80 from this equation x must be 4. so first I sait x must be negative so x = -4
This video demonstrates to every person (be they smart or not so smart) why studying mathematics sucks😂 Studying it at university isn't completely useless but it very clearly sucks
If "delta" is less than zero, the equation has no solution in the set of real numbers... I see no point in including answers from the range of imaginary numbers...
