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This is my best pair of simultaneous equations by far - I have spent the last $2$ - $3$ weeks finding a really extreme pair. I challenge anybody to get full marks for this. You have until $14^{th}$ November, $1:30$pm to get all $22$ marks.

I found a few complex solutions too (where $j = \sqrt{-1}$ ). 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.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewest@Vinayak Srivastava, @Frisk Dreemurr

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This is my best pair of simultaneous equations by far - I have spent the last $2$ - $3$ weeks finding a really extreme pair. I challenge anybody to get full marks for this. You have until $14^{th}$ November, $1:30$pm to get all $22$ marks.

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Ok!!!

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Me with the help of internet.

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

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That's not a solution.

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From eq 2,

$a = b+2$

Substituting in eq 1,

$(b+b+2)^{3}(b+b+2)^{2}=b^{4}+ 3b^{3} + b^{2} - b$

$(2b+2)^{3}(2b+2)^{2}=b^{4}+ 3b^{3} + b^{2} - b$

$(2b+2)^{5}=b^{4}+ 3b^{3} + b^{2} - b$

$32(b^{5} + 5b^{4} + 10b^{3} + 10b^{2} + 5b + 1 )=b^{4}+ 3b^{3} + b^{2} - b$

$32b^{5} + 160b^{4} + 320b^{3} + 320b^{2} + 160b + 32=b^{4}+ 3b^{3} + b^{2} - b$

$32b^{5} + 159b^{4} + 317b^{3} + 319b^{2} + 161b + 32 = 0$ (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 -

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

$a = 0.464537, b = -1.53546$

$a = 1.42518, b = -0.574825$

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I don't get when you change $(2b + 2)^5 = b^4 + 3b^3 + b^2 - b$ into $32(b^5 + 5b^4 + 10b^3 + 10b^2 + 5b + 1) = b^4 + 3b^3 + b^2 - b$ can you explain plz.

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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)

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I found a few complex solutions too (where $j = \sqrt{-1}$ ). 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:

$a_1 \approx 1.0708 + 0.5192 j \\ b_1 \approx -0.9292 + 0.5192 j$

Second complex solution:

$a_2 \approx 1.0708 - 0.5192 j \\ b_2 \approx -0.9292 - 0.5192 j$

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

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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.).

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Update - I have changed equation $1$ to compensate for my solution...

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

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

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bruh...

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