A tricky question from Oxford University Interview

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@higher_mathematics

#maths #math

[GaryMSanchez]

I did this in my head in two minutes…it had to be negative to begin…-5 was too big, tried -4, got it

[sameernoor]

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!!

[markharder]

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.

[niccolopaganinifranzliszt]

I solved it like that:
x²-x³=80
x²(1-x)=80
Now, since x² is always positive and 1-x is positive only when x<1, we know that x<1.
Also, x is a number where 80 is devisible by it's square.
There are only three squares that 80 is devisible by and are less than 1:
(-1)², (-2)² and (-4)². We can easily see it's not -1 or -2, since 1*2=2, 4*3=12. So, x=-4. And really, 16*5=80.
For the two other solutions, we will factor out x+4, and we will get:
(x+4)(-x²+5x-20)=-x³+x²-80
So,

[janibeg]

20 seconds of thinking/observation solves the problem

[TurkishKS]

Lol

"So, what you do is, you randomly break 80 into 16 and 64, and like, woah, look at that..."

[Jenjared]

This isn’t a solution. It’s just writing the solution down then demonstrating it with algebra the long way.

[GeorgeAlexanderOz]

The first step is some kind of constructed trick. Try x^2 - x^3 = 79.9

[andrewclifton]

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.

[KipIngram]

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.

[YAWTon]

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.

[momo.ru-chan]

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

[blueshoter]

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+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

[eduardojesus]

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.

[kpdywo]

80 is the sum of a square and a cube. When you know this everything becomes easy, but knowing it takes a lot of experience.

[garyinbellerose]

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.

[kanguru_]

x = -4 by inspection for one solution, then solve the remaining quadratic: x^2-5x+20 by the quadratic formula.

[rpocc]

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.

[ferro]

x^2-x^3=80
x^2*(1-x) = 2^4*5
x^2*(1-x) = 4^2*5
x^2*(1-x) = 4^2*(1-(-4))
x^2*(1-x) = (-4)^2*(1-(-4))

[NISHANTH-yf]

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.