Author Topic: Recreational Problems (SM level) (Read 98477 times) Tweet Share

Ahmad · « **Reply #60 on:** January 11, 2008, 10:12:26 am »

You hit a correct approach, but you didn't see it through! Check your working out

Ahmad · « **Reply #61 on:** January 11, 2008, 10:22:47 am »

Correct! Well done.

dcc · « **Reply #62 on:** January 11, 2008, 08:43:40 pm »

Quote from: Ahmad on January 04, 2008, 07:39:34 pm

Right! I actually had another idea in mind, here's an extension:

17. Solve $6x^4 - 24x^3 + 12x^2 + 24x + 6 = 0$ .

$6x^4 - 24x^3 + 12x^2 + 24x + 6 = 0$

Dividing by 6 gives:

$x^4 - 4x^3 + 2x^2 + 4x + 1 = 0$

Let $u = x - 1, x = u +1$

Substituting x = u + 1 back into formula:

$(u+1)^4 - 4(u+1)^3 + 2(u+1)^2 + 4(u+1) + 1 = 0$

$u^4 + 4u^3 + 6u^2 + 4u + 1 -4u^3 - 12u^2 - 12u - 4 + 2u^2 + 4u + 2 + 4u + 4 + 1 = 0$

Simplifying:

$u^4 - 4u^2 + 4 = 0$

Let $q = u^2$

$q^2 - 4q + 4 = 0$

$(q-2)^2=0$

:. $q = 2 = u^2$

$u^2 = 2, u = \pm\sqrt{2}$

$u = \pm\sqrt{2} = x - 1$

$x = \pm\sqrt{2} +1$

Solutions for x are:
$x=1+\sqrt{2}\mbox{ and }x=1-\sqrt{2}$

Edit:
Just to show how I arrived at the subsitution x = u + 1, I have shown a little proofy thingy here:

Consider

$x^4 + bx^3 + cx^2 + dx + e = 0$

Now, we want to get rid of the 2nd term of this equation, so we say

$x = u + r$

Expanding this out, we get:

$u^4 + (4r + b)u^3 + (6r^2 + 3br + c)u^2 + (4r^3 + 3br^2 + 2cr + d)u + r^4 + br^3 + cr^2 + dr + e = 0$

Now, we want to get rid of the u^3 to make it easier to solve, so:

$4r + b = 0$

$4r = -b$

$r = -b / 4$

So, using the subsitution x = u - (b / 4), we can reduce the equation

$x^4 + bx^3 + cx^2 + dx + e = 0$ to the form:

$x^4 + Bx^2 + Cx + D = 0$ , which is alot easier to solve.

The reason that this question is particular easy to solve, is that the the coeffiecent C = 0, so we have
$x^4 + Bx^2 + D = 0\mbox{ which can be expressed as (for example): }u^2 + Bu + D = 0\mbox{ where u = x^2 }$

Ahmad · « **Reply #63 on:** January 11, 2008, 09:40:24 pm »

Yup that's one way to do it. Well done.

dcc · « **Reply #64 on:** January 12, 2008, 09:17:18 pm »

Quote from: Ahmad on January 04, 2008, 06:59:22 pm

14. Find the real roots of $\sqrt{x+3-4\sqrt{x-1}} + \sqrt{x+8-6\sqrt{x-1}} = 1$ .

Let $a = \sqrt{x - 1}$

$\sqrt{x+3-4a} + \sqrt{x+8-6a} = 1$

But since: $a = \sqrt{x - 1}$

then $a^2 = x - 1$

$x = a^2 + 1$

So the quation can be rewritten:

$\sqrt{(a^2 + 1) + 3 -4a} + \sqrt{(a^2 + 1) + 8 - 6a} = 1$

$\sqrt{a^2 - 4a + 4} + \sqrt{a^2 - 6a + 9} = 1$

These are both perfect squares!

$\sqrt{(a-2)^2} + \sqrt{(a-3)^2} = 1$

$a - 2 + a - 3 = 1$

$2a = 6, a = 3$

But remember $a = \sqrt{x - 1}$ , so:

$3 = \sqrt{x - 1}$

$9 = x - 1$

$x = 10$

Ahmad · « **Reply #65 on:** January 12, 2008, 09:48:07 pm »

Correct.

dcc · « **Reply #66 on:** January 13, 2008, 01:56:38 am »

ASKDASKDLA:Sdk

Ahmad · « **Reply #67 on:** January 13, 2008, 09:32:10 am »

Sorry, I should've made it more clear. There are n radical signs.

dcc · « **Reply #68 on:** January 17, 2008, 12:38:55 pm »

Easy one for someone to do:

Prove that the equation $c^n = a^n + b^n$

$\mbox{ a, b, c $\in$ $R^+$ where n $>$ 2}$ has no solutions in the non-zero integers.

Ahmad · « **Reply #69 on:** January 17, 2008, 12:50:18 pm »

For more information about the above problem, which should not be taken too seriously, click here.

dcc · « **Reply #70 on:** January 17, 2008, 12:57:29 pm »

I WAS BEING SERIOUS MAN
IF YOU DIDN'T TELL THEM IT TOOK 350 YEARS TO SOLVE THEN PEOPLE MIGHT NOT OF BEEN FLUSTERED BY ITS INPENETRABILITY!!!!

edit:

god work is boring

Mao · « **Reply #71 on:** January 17, 2008, 01:10:34 pm »

Quote from: Ahmad on January 17, 2008, 12:50:18 pm

For more information about the above problem, which should not be taken too seriously, click here.

ahmad that site told a story....

not the answer...

Collin Li · « **Reply #72 on:** January 17, 2008, 01:11:52 pm »

If you want to know the answer you will need to have to do a lot of study in seemingly disconnected specialisations of maths.

dcc · « **Reply #73 on:** January 17, 2008, 01:12:31 pm »

mao, to understand the answer you would need a copious amount of university education. Like, there was an interview with the guy who proved it. And he had made a mistake in one of the original proofs.

And the guy asked him like, 'What was the error?'

'Uh you wouldn't understand, you'd probably need to study that specific part of the manuscript for 3 months to understand the problem'

pure number theorists = crazy

Mao · « **Reply #74 on:** January 17, 2008, 01:20:05 pm »

Quote from: dcc on January 17, 2008, 01:12:31 pm

mao, to understand the answer you would need a copious amount of university education. Like, there was an interview with the guy who proved it. And he had made a mistake in one of the original proofs.

And the guy asked him like, 'What was the error?'

'Uh you wouldn't understand, you'd probably need to study that specific part of the manuscript for 3 months to understand the problem'

pure number theorists = crazy

i'll fkn die

LOL

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Author Topic: Recreational Problems (SM level) (Read 98477 times) Tweet Share

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