Quintic Simultaneous Equations

Solve the simultaneous equations:

(a+b)3(a+b)2=32b(8b2+6b+3)+b(b33b25b10)=8(a + b)^3 (a + b)^2 = 32b(8b^2 + 6b + 3) + b(b^3 - 3b^2 - 5b - 10) = - 8 ab=2a - b = 2

Give all answers in the form (a,b)(a, b).

(2222 marks)

Note by Yajat Shamji
8 months, 3 weeks ago

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New solution (hence the editing of the first equation):

Leave the =8= - 8 for later, we're focusing on the first two parts of the first equation and the second equation.

First, rearrange the second equation to obtain a=2+ba = 2 + b.

Then, substitute into (a+b)3(a+b)2(a + b)^3(a + b)^2.

Simplify to (2b+2)5(2b + 2)^5.

Expand and simplify to 32b(b4+5b3+10b2+10b+5+32b(b^4 + 5b^3 + 10b^2 + 10b + 5 +1b)\frac{1}{b}).

Divide all sides of the equation by bb.

Move the 32(8b2+6b+3)32(8b^2 + 6b + 3) to the LHS of the equation.

Divide the LHS by 3232.

Group like terms and you should be left with b4+5b3+2b2+6b+8+b^4 + 5b^3 + 2b^2 + 6b + 8 +1b=\frac{1}{b} = b33b25b10b^3 - 3b^2 - 5b - 10.

Move b33b25b10b^3 - 3b^2 - 5b - 10 to the LHS of the equation and group like terms - you should be left with b4+4b3+5b2+11b+18+1b=0b^4 + 4b^3 + 5b^2 + 11b + 18 + \frac{1}{b} = 0

Multiply by bb, resulting in b5+4b4+5b3+11b2+18b+1=0b^5 + 4b^4 + 5b^3 + 11b^2 + 18b + 1 = 0.

Set the equation equal to 8- 8 from the first simultaneous equation - you should obtain b5+4b4+5b3+11b2+18b+9=0b^5 + 4b^4 + 5b^3 + 11b^2 + 18b + 9 = 0

Set the equation as f(b)f(b).

Perform factor theorem - you'll find out that f(1)=0f(- 1) = 0, therefore (b+1)(b + 1) is a factor of b5+4b4+5b3+11b2+18b+9=0b^5 + 4b^4 + 5b^3 + 11b^2 + 18b + 9 = 0

Perform the long polynomial division b5+4b4+5b3+11b2+18b+9b+1\frac{b^5 + 4b^4 + 5b^3 + 11b^2 + 18b + 9}{b + 1}

You should get b4+3b3+2b2+9b+9b^4 + 3b^3 + 2b^2 + 9b + 9.

Set this equation as f(b)f(b).

Perform factor theorem - you'll find out that f(1)=0f(- 1) = 0 and f(3)=0f(- 3) = 0, therefore (b+1)(b + 1) and (b+3)(b + 3) are factors of b4+3b3+2b2+9b+9b^4 + 3b^3 + 2b^2 + 9b + 9

Perform the long polynomial division of b4+3b3+2b2+9b+9(b+1)(b+3)\frac{b^4 + 3b^3 + 2b^2 + 9b + 9}{(b + 1)(b + 3)}

You should get b2b+3b^2 - b + 3.

Now, using the discriminant of b24acb^2 - 4ac, we get 11- 11, indicating imaginary solutions and the fact that this cannot be factored further.

Therefore, the factors of b5+4b4+5b3+11b2+18b+9b^5 + 4b^4 + 5b^3 + 11b^2 + 18b + 9 are (b+1)2(b+3)(b2b+3)(b + 1)^2(b + 3)(b^2 - b + 3)

Now, using Brahmagupta's quadratic formula and the discriminant, we solve b2b+3=0b^2 - b + 3 = 0, leading to 1±112\frac{1 \pm \sqrt{- 11}}{2}

This simplifies to 12±11i2\frac{1}{2} \pm \frac{11i}{2}

Therefore, the solutions for b=1,3b = - 1, - 3 and 12±11i2\frac{1}{2} \pm \frac{11i}{2}.

Substitute all values into ab=2a - b = 2 and a=1,3a = 1, - 3 and 52±11i2\frac{5}{2} \pm \frac{11i}{2}.

Therefore the solutions to the simultaneous equations:

(a+b)3(a+b)2=32b(8b2+6b+3)+b(b33b25b10)=8(a + b)^3 (a + b)^2 = 32b(8b^2 + 6b + 3) + b(b^3 - 3b^2 - 5b - 10) = - 8

ab=2a - b = 2

are (in the form (a,b)(a, b), as requested):

(1,1)(1, - 1)

(3,3)(- 3, - 3)

(5211i2,12+11i2)(\frac{5}{2} - \frac{11i}{2}, \frac{1}{2} + \frac{11i}{2})

(52+11i2,1211i2)(\frac{5}{2} + \frac{11i}{2}, \frac{1}{2} - \frac{11i}{2})

Note: Only 44 solutions are here to due to the repeated root of (b+1)2(b + 1)^2.

Test.

Yajat Shamji - 2 months, 3 weeks ago

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Ok!!!

Ravi Singh - 8 months, 3 weeks ago

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Me with the help of internet.

Half pass3 - 8 months, 3 weeks ago

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a=b4+3b3+b2b5ba=\sqrt[5]{b^4+3b^3+b^2-b}-b

Half pass3 - 8 months, 3 weeks ago

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That's not a solution.

Yajat Shamji - 8 months, 3 weeks ago

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From eq 2,

a=b+2a = b+2

Substituting in eq 1,

(b+b+2)3(b+b+2)2=b4+3b3+b2b(b+b+2)^{3}(b+b+2)^{2}=b^{4}+ 3b^{3} + b^{2} - b

(2b+2)3(2b+2)2=b4+3b3+b2b(2b+2)^{3}(2b+2)^{2}=b^{4}+ 3b^{3} + b^{2} - b

(2b+2)5=b4+3b3+b2b(2b+2)^{5}=b^{4}+ 3b^{3} + b^{2} - b

32(b5+5b4+10b3+10b2+5b+1)=b4+3b3+b2b32(b^{5} + 5b^{4} + 10b^{3} + 10b^{2} + 5b + 1 )=b^{4}+ 3b^{3} + b^{2} - b

32b5+160b4+320b3+320b2+160b+32=b4+3b3+b2b32b^{5} + 160b^{4} + 320b^{3} + 320b^{2} + 160b + 32=b^{4}+ 3b^{3} + b^{2} - b

32b5+159b4+317b3+319b2+161b+32=032b^{5} + 159b^{4} + 317b^{3} + 319b^{2} + 161b + 32 = 0 (I don't know how to find roots of this mammoth...)


@Yajat Shamji - This next part took a long time to calculate, mostly because I brute-forced the roots of the 5th degree term after graphing it...

One solution is A = 1 and B = -1, and other real solutions can be found when graphed -

Real Solutions -

a=1,b=1a = 1, b = -1 (got it easily using Remainder Theorem)

a=0.464537,b=1.53546a = 0.464537, b = -1.53546

a=1.42518,b=0.574825a = 1.42518, b = -0.574825

A Former Brilliant Member - 8 months, 3 weeks ago

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I don't get when you change (2b+2)5=b4+3b3+b2b(2b + 2)^5 = b^4 + 3b^3 + b^2 - b into 32(b5+5b4+10b3+10b2+5b+1)=b4+3b3+b2b32(b^5 + 5b^4 + 10b^3 + 10b^2 + 5b + 1) = b^4 + 3b^3 + b^2 - b can you explain plz.

Half pass3 - 8 months, 3 weeks ago

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Sure.

(2b+2) = 2(b+1)

(2b+2)^5 = 2^5 x (b+1)^5 = 32 x (b+1)^5 = 32(b^5+5b^4+10b^3+10b^2+5b+1)

The last expansion is done by (b+1) x (b+1)x(b+1) x (b+1) x (b+1)

A Former Brilliant Member - 8 months, 3 weeks ago

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I found a few complex solutions too (where j=1 j = \sqrt{-1} ). I was also able to find the same real-valued solutions you did. This makes sense, since there should be five solutions for a quintic equation.

First complex solution:

a11.0708+0.5192jb10.9292+0.5192j a_1 \approx 1.0708 + 0.5192 j \\ b_1 \approx -0.9292 + 0.5192 j

Second complex solution:

a21.07080.5192jb20.92920.5192j a_2 \approx 1.0708 - 0.5192 j \\ b_2 \approx -0.9292 - 0.5192 j

Steven Chase - 8 months, 3 weeks ago

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Nice @Steven Chase! That's all five solutions of the quintic equation. @Yajat Shamji - Challenge complete!!!

A Former Brilliant Member - 8 months, 3 weeks ago

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Well, @Percy Jackson, @Steven Chase, you have completed the challenge, yes!

But keep it a secret for now - I want other people to solve it...

Also, I'll post my solution as well. It is done an entirely different way to yours.

And Percy, there was a simpler way than brute force (that's all I'm telling you.).

Yajat Shamji - 8 months, 3 weeks ago

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Update - I have changed equation 11 to compensate for my solution...

@Percy Jackson, @Steven Chase, @Half pass3, @Ravi Singh

You have until tomorrow to solve it... again!

Yajat Shamji - 8 months, 2 weeks ago

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bruh...

A Former Brilliant Member - 8 months, 1 week ago

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@Frisk Dreemurr

Yajat Shamji - 2 months, 3 weeks ago

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Uhh...

Hello Yaj

Anything you'd love to say?

Frisk Dreemurr - 2 months, 3 weeks ago

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I posted a new solution and edited it - it just took me 66 months to figure it out...

Yajat Shamji - 2 months, 3 weeks ago

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@Yajat Shamji wait aren't you the one who made the question?

Half pass3 - 2 months, 3 weeks ago

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@Half Pass3 Yes, but can't I post my solution as well?

Yajat Shamji - 2 months, 3 weeks ago

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@Yajat Shamji you can, but you said you took 6 months to figure out your own question.

Half pass3 - 2 months, 3 weeks ago

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@Half Pass3 I did say that. But did anybody notice? No.

Plus, I was busy with my studies. So, not a lot of time to actually attempt to correct in these 66 months.

I only managed to correct and figure it out a few days ago.

So, technically, it took me 4545 minutes in 66 months...

Anyway, let's not keep arguing. Since you're here, what do you think of the solution? (since @Frisk Dreemurr won't answer.)

Yajat Shamji - 2 months, 3 weeks ago

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@Yajat Shamji I'd say it's not for my level, it was very detailed indeed. ( I'm pretty dumb by the way)

Half pass3 - 2 months, 3 weeks ago

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@Half Pass3 Ok. I get detailed a lot of the time.

Yajat Shamji - 2 months, 3 weeks ago

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@Yajat Shamji Can you check on my new note, I really need some suggestions for my next part thx.

Half pass3 - 2 months, 3 weeks ago

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@Half Pass3 Ok. Give me the link.

Yajat Shamji - 2 months, 3 weeks ago

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@Yajat Shamji https://brilliant.org/discussions/thread/4th-dimension-non-euclidean-geometry-part-1/

Half pass3 - 2 months, 3 weeks ago

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@Half Pass3 Not my area of specialty (my area of specialties are number theory, primes and algebra), but I'll try my best!

Yajat Shamji - 2 months, 3 weeks ago

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@Yajat Shamji cool👍

Half pass3 - 2 months, 3 weeks ago

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@Half Pass3 And as for you being dumb, use this website freely.

Yajat Shamji - 2 months, 3 weeks ago

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@Half Pass3 It'll help immensely.

Yajat Shamji - 2 months, 3 weeks ago

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Second to top comment. What do you think?

Yajat Shamji - 2 months, 3 weeks ago

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Let me know soon!

:)

Yajat Shamji - 2 months, 3 weeks ago

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Asking cuz I can see that this post is approx. six months old...

Frisk Dreemurr - 2 months, 3 weeks ago

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