Maths problem

[Solar]

steady. My son is crazy about math. Your boy and my boy can be good friends [laugh]

 

Wow.. my nerdy boy loves math too.

[Vid]

Wow.. my nerdy boy loves math too.

[Angcheek]

Gave me son your answer but he says it's not all correct

 

He says cannot use brute force and there's another answer which is -9.99999 something something ...

During my time there was no brute force ... everything was gentle .

[Angcheek]

Wow.. my nerdy boy loves math too. ð

 

Can u imagine he was so disappointed when he wasn't qualified to join his school's math club.

 

Told him to continue his interests at home so that he's able to try again when it's open to everyone else after year 2

He said he picked it from his science class when they learn pressure..

Here's the formula

Yours have the answer also right? ð

I know i 12 yr old ... but i think that is definition of pressure . Force per area .

 

Wa wu tark cher one hor

[Vid]

During my time there was no brute force ... everything was gentle .

[Angcheek]

Radx always brute force you wor [laugh]

Hahahha i got to ask the new bride

[Solar]

Yes, can imagine the disappointment. My boy spent more than 5 hours on that qns [sweatdrop]

 

Which sec sch your boy got to eventually? Cat high?

RI. Maybe can arrange them meet Edited October 14, 2018 by Solar

[Lala81]

Yes, can imagine the disappointment. My boy spent more than 5 hours on that qns [sweatdrop]

 

Which sec sch your boy got to eventually? Cat high?

It doesn't matter if he got the answer. The fact that he can spend so many hours trying to work this problem speaks volumes of his determination/tenacity. I think I would have given up in ten mins lol.

 

Not gone UK yet?

RI. Maybe can arrange them meet ð¤£

And so ironic they didn't allow him to join the math club

Lppl he joined the physics club instead

I got tak chek also don't understand.

Almost failed mine and my parents had to enroll me to tuition classes ð¢

My maths is half past six also. Edited October 14, 2018 by Lala81

[Vid]

RI. Maybe can arrange them meet

[Vid]

It doesn't matter if he got the answer. The fact that he can spend so many hours trying to work this problem speaks volumes of his determination/tenacity. I think I would have given up in ten mins lol.

 

He works really hard until my wife and I have to tell him to stop all the time to rest [laugh]

[Solar]

Those at RI must be geniuses that even your boy cannot get into the math club [sweatdrop]

 

Do you buy any assessment books for him? Any recommendation? i find it hard to buy books that he finds challenging.

The questions vary. Some can be very easy

 

In terms of cheemilogy, he said first the Australian Math Olympiad questions his school sells, followed by those sold by Popular bookstore then Singapore math Olympiad.

 

Might be others out there but these are the ones he's been exposed to

Edited October 14, 2018 by Solar

[Vid]

The questions vary. Some can be very easy

 

In terms of cheemilogy, he said first the Australian Math Olympiad questions his school sells, followed by those sold by Popular bookstore then Singapore math Olympiad.

 

Might be others out there but these are the ones he's been exposed to

 

  Thank you very much. That is very helpful

[Angcheek]

RI. Maybe can arrange them meet ð¤£

And so ironic they didn't allow him to join the math club

Lppl he joined the physics club instead

I got tak chek also don't understand.

Almost failed mine and my parents had to enroll me to tuition classes ð¢

Ai yo .. if i know you 20 yrs back , i give u free tuition.

[Angcheek]

He works really hard until my wife and I have to tell him to stop all the time to rest [laugh]

U never tell him cannot use brute force

[mersaylee]

It doesn't matter if he got the answer. The fact that he can spend so many hours trying to work this problem speaks volumes of his determination/tenacity. I think I would have given up in ten mins lol.

 

Not gone UK yet?

My maths is half past six also.

10mins!? I’d spend 5s to read the question, 4.5s to look left and right to see if it’s possible to copy...0.5s to come to the decision to just give it up.

 

See? I tried...

[1fast1]

How to solve for x in this equation?

 

2x^4 + 4x^3 + 6x^2 + 4x + 2 = 0

 

I figured out that it can probably only be factorised into 2 quadratic equations, both of which do not have real roots?

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.

Edited October 14, 2018 by Turboflat4

[1fast1]

Math question time ...

 

image001.png

 

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).

[Enye]

wah nowadays sec 1 maths so tough?