Maths problem

[AndrewLee72]

Who can solve the ship.. Captain's age question?

[Atrecord]

Wah the maths problems all very cheem...

 

Came across this:

 

[Lala81]

I understand this as much as i would understand written sanksrit.   

 

 

Sorry for the late response, I was away.

The polynomial (this is a quartic, meaning power 4) on the left hand side (LHS) - let's call it P(x) can be represented as 2(x^2+x+1)^2.

So the solutions are simply x = -1/2 +/- i sqrt(3)/2. Those are repeated roots, and that accounts for all 4 roots of the quartic.

 

Those are actually the complex cube roots of one. And that's basically how I found the factorisation.

 

Firstly, by reducing to a monic (leading coefficient one), you have to solve x^4 + 2x^3 + 3x^2 + 2x + 1 = 0.

 

There is a useful theorem known as rational root theorem. This can immediately be employed to state that *if* a rational root where to exist, for this equation, it would have to divide one. Which means the root has to be either 1 or -1. Since neither of these is a solution, there are no rational roots.

 

That means that any roots will be real irrational or complex. Those are tough to solve in general.

 

There are methods to solve cubic and quartics exactly in terms of surds (roots of integers), but they are damn tedious.

 

But I spotted something that helped with this question. I started by testing the complex cube roots of unity (one). This sort of intuition usually comes with experience. The complex cube roots of unity are usually represented as omega (ω) and its conjugate, which is omega with a bar on top. The latter can also be expressed as omega squared (ω^2), which I'll use here. These are rather nice complex numbers that also satisfy the quadratic x^2 + x + 1 = 0.

 

If you substitute ω into P(x), you get P(ω) = 0. This is because ω^3 = 1 and ω^4 = ω^3.ω = 1.ω = ω. You can also use the rather nice identity ω^2 + ω + 1 = 0 (see quadratic above) to do further simplification. The upshot is that you've now established that ω is a root.

 

Complex roots of polynomials with real coefficients always come in conjugate pairs (the same real part but the imaginary part has its sign reversed). So the other root is ω^2, which is the conjugate of ω.

 

You can't show those are the only two roots without fully factorising the quartic. For that, I used polynomial division to divide P(x) by (x^2 + x + 1), to show that P(x) = 2(x^2 + x + 1)^2. That means that ω and ω^2 are repeated pairs of conjugate roots. There are no other solutions.

 

 

 

wah nowadays sec 1 maths so tough?

[Ender]

I understand this as much as i would understand written sanksrit.   

 

 

 

this one very high level lah. math club.

 

Yes, higher RI's math club level.. That's why I am not too concern when  my sec 3 son cannot solve. But he can learn from @turboflat4 's detail explanation.

[Lala81]

Yes, higher RI's math club level.. That's why I am not too concern when  my sec 3 son cannot solve. But he can learn from @turboflat4 's detail explanation.

 

Actually i think this is even beyond A level C maths from what i recall (F maths I'm not sure).

Definitely beyond A maths level at O level.

 

I mean what turboflat4 explained, i really don't even understand even 1%.

[SuPerBoRed]

Actually i think this is even beyond A level C maths from what i recall (F maths I'm not sure).

Definitely beyond A maths level at O level.

 

I mean what turboflat4 explained, i really don't even understand even 1%.

 

now i feel better... since the doctor dont even understand 1%... me is not even 0.1%... and im in finance... 

[Ender]

Wah the maths problems all very cheem...

 

Came across this:

 

Cat n table.jpeg

This one chinese or math question?   

I think it's 130cm..

[ALTK]

How come I totally cannot understand what u guys are talking about....

[mersaylee]

How come I totally cannot understand what u guys are talking about....

I know what they are talking about but like you I don’t understand what they are talking about.

[Enye]

brute force = trial and error?

 

even method name also sound so cheem

 

 

[1fast1]

I understand this as much as i would understand written sanksrit.   

 

 

 

this one very high level lah. math club.

 

 

Bro, it's not that tough. It really isn't. And not that high level.

 

Remember Remainder and Factor theorems? That was definitely covered in "just" A math at O-levels. Remainder Theorem states that a polynomial P(x), when divided by a linear factor (x-a) [where "a" is a constant] leaves a remainder of P(a). Factor Theorem is a trivial corollary (meaning a damn simple consequence, duh!) of that - P(a) is zero if and only if (x-a) is a factor of P(x) (i.e. the remainder is zero in this case, simply because (x-a) is a factor - that's the definition of a factor).

 

With me so far?

 

Now the only "extra" thing you need to basically take for granted is that everything I said there holds even if "a" is a complex number constant.

 

The thing about cube roots of unity (one) is quite simple also.

 

How do you find the cube root of a number?

 

Just set x^3 = that number lor...

 

So let's set x^3 = 1

 

Now we may be tempted to just take cube roots of both sides, giving the "obvious" answer of 1 (again, duh!).

 

But let's do it properly.

 

Rearrange to x^3 - 1 = 0

 

Let's call P(x) = x^3 - 1.

 

Now put x = 1. That gives P(1) = 0.

 

By what we just discussed (i.e. Factor Theorem) that means (x-1) is a factor of P(x).

 

If you divide P(x) by (x-1), you will get a remainder of zero (of course!) and a quotient of (x^2 + x + 1). You can divide the polynomial by old fashioned polynomial division or apply a slightly fancy technique called synthetic division (I'm pretty sure this was taught in A Math, at least in RI). That makes dividing by linear factors like (x-1) damn easy. Either way, you get:

 

P(x) = (x-1)(x^2+x+1)

 

So the original equation is (x-1)(x^2+x+1) = 0.

 

I'm sure you can see the solutions are x = 1 (that's the "obvious" cube root of 1) and whatever the solutions to the quadratic x^2 + x + 1 = 0 are.

 

The latter quadratic has complex roots which equal -1/2 + i*sqrt(3)/2 and -1/2 - i*sqrt(3)/2

 

The former I designate ω. By squaring this (you remember how to square complex numbers? It's as simple as remembering your basic identities like (a+b)^2 = a^2 + b^2 + 2ab)), you can verify that the latter is ω^2. In fact one number squared is the other, and it works both ways. Pretty cool, right?

 

So those are the other two numbers that satisfy x^3 = 1. Those complex numbers are called the complex cube roots of one (or unity).

 

I think you should be able to follow the rest now.

Edited October 15, 2018 by Turboflat4

[mersaylee]

Ahhh....Remainder and Factor theorems. I remember now, it’s one of the many topics taught by my math chers during my school times.

 

Been a poor student I had nothing to offer in order to show my gratification...I returned everything...kept only four topics - addition,subtraction,division and multiplication.

[Angcheek]

Wah the maths problems all very cheem...

 

Came across this:

 

Cat n table.jpeg

 

hahah my friend also posted this to me ...... took me awhile to figure out the answ . 

 

Must use primary school reasoning 

 

answ C 

Edited October 15, 2018 by Angcheek

[Cheesey74]

Bro, it's not that tough. It really isn't. And not that high level.

 

Remember Remainder and Factor theorems? That was definitely covered in "just" A math at O-levels. Remainder Theorem states that a polynomial P(x), when divided by a linear factor (x-a) [where "a" is a constant] leaves a remainder of P(a). Factor Theorem is a trivial corollary (meaning a damn simple consequence, duh!) of that - P(a) is zero if and only if (x-a) is a factor of P(x) (i.e. the remainder is zero in this case, simply because (x-a) is a factor - that's the definition of a factor).

 

With me so far?

 

Now the only "extra" thing you need to basically take for granted is that everything I said there holds even if "a" is a complex number constant.

 

The thing about cube roots of unity (one) is quite simple also.

 

How do you find the cube root of a number?

 

Just set x^3 = that number lor...

 

So let's set x^3 = 1

 

Now we may be tempted to just take cube roots of both sides, giving the "obvious" answer of 1 (again, duh!).

 

But let's do it properly.

 

Rearrange to x^3 - 1 = 0

 

Let's call P(x) = x^3 - 1.

 

Now put x = 1. That gives P(1) = 0.

 

By what we just discussed (i.e. Factor Theorem) that means (x-1) is a factor of P(x).

 

If you divide P(x) by (x-1), you will get a remainder of zero (of course!) and a quotient of (x^2 + x + 1). You can divide the polynomial by old fashioned polynomial division or apply a slightly fancy technique called synthetic division (I'm pretty sure this was taught in A Math, at least in RI). That makes dividing by linear factors like (x-1) damn easy. Either way, you get:

 

P(x) = (x-1)(x^2+x+1)

 

So the original equation is (x-1)(x^2+x+1) = 0.

 

I'm sure you can see the solutions are x = 1 (that's the "obvious" cube root of 1) and whatever the solutions to the quadratic x^2 + x + 1 = 0 are.

 

The latter quadratic has complex roots which equal -1/2 + i*sqrt(3)/2 and -1/2 - i*sqrt(3)/2

 

The former I designate ω. By squaring this (you remember how to square complex numbers? It's as simple as remembering your basic identities like (a+b)^2 = a^2 + b^2 + 2ab)), you can verify that the latter is ω^2. In fact one number squared is the other, and it works both ways. Pretty cool, right?

 

So those are the other two numbers that satisfy x^3 = 1. Those complex numbers are called the complex cube roots of one (or unity).

 

I think you should be able to follow the rest now.

 

Haiz.. now i understand why he turboflat4 while i'm inline 4 nia, really ashamed, considering i got A1 for A math...

[Lala81]

Haiz.. now i understand why he turboflat4 while i'm inline 4 nia, really ashamed, considering i got A1 for A math...

 

i feel like i'm those 50cc engine scooters that they use in tsukiji market   

[Lala81]

I didn't burn my notes after A levels but i might as well have set fire to all these knowledge.

I was a poor mathematics student at best. Just getting by with repeated ten year series in O levels and never really understood the higher concepts. Similar to my physics which i struggled at JC level.

 

Haha ok, i'll watch some online videos on what u mentioned earlier and try to refresh some brain cells 

 

Bro, it's not that tough. It really isn't. And not that high level.

 

Remember Remainder and Factor theorems? That was definitely covered in "just" A math at O-levels. Remainder Theorem states that a polynomial P(x), when divided by a linear factor (x-a) [where "a" is a constant] leaves a remainder of P(a). Factor Theorem is a trivial corollary (meaning a damn simple consequence, duh!) of that - P(a) is zero if and only if (x-a) is a factor of P(x) (i.e. the remainder is zero in this case, simply because (x-a) is a factor - that's the definition of a factor).

 

With me so far?

 

Now the only "extra" thing you need to basically take for granted is that everything I said there holds even if "a" is a complex number constant.

 

The thing about cube roots of unity (one) is quite simple also.

 

How do you find the cube root of a number?

 

Just set x^3 = that number lor...

 

So let's set x^3 = 1

 

Now we may be tempted to just take cube roots of both sides, giving the "obvious" answer of 1 (again, duh!).

 

But let's do it properly.

 

Rearrange to x^3 - 1 = 0

 

Let's call P(x) = x^3 - 1.

 

Now put x = 1. That gives P(1) = 0.

 

By what we just discussed (i.e. Factor Theorem) that means (x-1) is a factor of P(x).

 

If you divide P(x) by (x-1), you will get a remainder of zero (of course!) and a quotient of (x^2 + x + 1). You can divide the polynomial by old fashioned polynomial division or apply a slightly fancy technique called synthetic division (I'm pretty sure this was taught in A Math, at least in RI). That makes dividing by linear factors like (x-1) damn easy. Either way, you get:

 

P(x) = (x-1)(x^2+x+1)

 

So the original equation is (x-1)(x^2+x+1) = 0.

 

I'm sure you can see the solutions are x = 1 (that's the "obvious" cube root of 1) and whatever the solutions to the quadratic x^2 + x + 1 = 0 are.

 

The latter quadratic has complex roots which equal -1/2 + i*sqrt(3)/2 and -1/2 - i*sqrt(3)/2

 

The former I designate ω. By squaring this (you remember how to square complex numbers? It's as simple as remembering your basic identities like (a+b)^2 = a^2 + b^2 + 2ab)), you can verify that the latter is ω^2. In fact one number squared is the other, and it works both ways. Pretty cool, right?

 

So those are the other two numbers that satisfy x^3 = 1. Those complex numbers are called the complex cube roots of one (or unity).

 

I think you should be able to follow the rest now.

 

[Angcheek]

 

This is a nice question that illustrates some key learning points.

So we're asked to solve 2^x + x^2 = 100.

 

The first thing to note is that this is an example of a transcendental equation. Things with powers involving x added to polynomial functions usually fall into this category. Finding a "nice" solution is usually not possible in general, and numerical approximations are necessary. Even in special cases where "exact" solutions can be found, solutions can only be expressed in terms of special functions like the Lambert W function (which I'll mention soon, haha).

 

So this sort of question usually has at least one "obvious" and "nice" root - most often an integer. And it's usually not that hard to find.

 

It's quite easy to just "see" that x = 6 is a solution because 2^6 + 6^2 = 64 + 36 = 100.

 

But can that be made any more rigorous? Well, sort of.

 

Let's start with the assumption a natural number solution (i.e. a positive integer solution) exists. Clearly x cannot be 1 or 2.

 

Let's rearrange and factorise to give 2^x = (10+x)(10-x).

 

Note that both terms on the right hand side have the same even/odd parity. Since x > 2, the LHS has a factor of 4, so both terms on the RHS have to be even. Which means that x is even. Let's represent x = 2y.

 

So we get 2^(2y) = (10+2y)(10-2y)

 

Rearrange to 2^(2(y-1)) = (5+y)(5-y)

 

5^2 = y^2 + (2^(y-1))^2

 

From the primitive pythagorean triple (3,4,5), we should immediately be able to say that y = 3 or 4, and only y = 3 fits that equation.

 

So x = 6 is the solution to the original equation.

 

But realistically, simple inspection is enough for this part. This fancy stuff is only necessary when larger solutions are sought by hand and simple trial and error is not feasible.

 

The real trick is considering the full real domain rather than just natural numbers. Is there any other solution?

 

Answering that question requires studying the function f(x) = 2^x + x^2 and employing simple calculus. Note that f(x) > 0 for all real x. This function has the derivative (first differential) f'(x) = (ln 2)2^x + 2x, which is always positive for x> 0. Since the function is monotone increasing for positive x, x = 6 is the only real solution for this sub-domain.

 

However, if we consider the negative reals, there is exactly one more root. The minimum point (a global minimum) on the graph can be found by setting f'(x) = 0, and that can be solved in terms of the special Lambert W function that I mentioned a bit earlier. The minimum point is found at the x value x = -W(0.5(ln 2)^2)/(ln 2) which is approximately -0.2845. The global minimum is approx 0.902. I'm really not going to bother you with those details, but suffice it to say that when x decreases below that value (-0.2845), it starts increasing again and keeps increasing without bound, so it'll eventually hit and exceed 100 again. The x value where that occurs is a smidge above -10 (approx -9.99995 or thereabouts), as somebody's clever son has already figured out.

 

But what I did is essential to *proving* that those are the only 2 real roots (x = 6 and x approx -9.99995).

 

 

 

ai yo ...... made until so complicated .  

 

If you draw out the graph  2^x + x^2  ,   x^2 will give  2 answ . 2^x  will give a exponential effect to the result  

 

while x= 6 is a more absolute answ , x = -9.999995 is not exactly "real" as it will never hit 10.  just that  with x -> - infinity, 

the  2^x became extremely small or negligible 

 

    primary school theory 

[Enye]

Yes, higher RI's math club level.. That's why I am not too concern when my sec 3 son cannot solve. But he can learn from @turboflat4 's detail explanation.

my sec 3 son got the answer from calculator