Quintic Simultaneous Equations

Solve the simultaneous equations: (a+b)3(a+b)2=32b4+96b3+32b232b(a + b)^3 (a + b)^2 = 32b^4 + 96b^3 + 32b^2 - 32b ab=2a - b = 2

(2222 marks)

Note by Yajat Shamji
1 month ago

No vote yet
1 vote

  Easy Math Editor

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.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

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

Percy Jackson - 1 month ago

Log in to reply

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 - 1 month ago

Log in to reply

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)

Percy Jackson - 1 month ago

Log in to reply

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 - 4 weeks, 1 day ago

Log in to reply

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

Percy Jackson - 4 weeks, 1 day ago

Log in to reply

My solution (wasn't going to work without changing R.H.S of equation 11):

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

Then simplify the L.H.S into (a+b)5 (a + b)^5 and expand it to 32b5+160b4+256b3+192b2+96b+3232b^5 + 160b^4 + 256b^3 + 192b^2 + 96b + 32.

Then divide both sides by 3232 and move the R.H.S to the L.H.S and you'll obtain b5+4b4+5b3+4b2+4bb^5 + 4b^4 + 5b^3 + 4b^2 + 4b. Label this as equation 33.

Divide both sides by bb to get b(b4+4b3+5b2+4b+4)b (b^4 + 4b^3 + 5b^2 + 4b + 4). Label this equation 44.

Then assign equation 44 to the function f(x) f(x) and perform factor theorem on it. You'll find that f(2)=0 f(-2) = 0. . Therefore (b+2)(b + 2) is a factor of b4+4b3+5b2+4b+4b^4 + 4b^3 + 5b^2 + 4b + 4.

Perform algebraic long division to obtain b3+2b2+b+2b^3 + 2b^2 + b + 2. Label this equation 55 and assign it to f(x) f(x).

Perform factor theorem on equation 55 - you'll find that f(2)=0 f(-2) = 0.

Then perform 'intelligent guessing' - you'll obtain (b+2)(b2+1) (b + 2) (b^2 + 1)

\therefore Factors of b5+4b4+5b3+4b2+4bb^5 + 4b^4 + 5b^3 + 4b^2 + 4b are:

b(b+2)2(b2+1)=0b(b + 2)^2(b^2 + 1) = 0

Either:

b=0b = 0

b+2=0,b=2b + 2 = 0, b = - 2

or:

b2+1=0,b2=1,b=1=±ib^2 + 1 = 0, b^2 = - 1, b = \sqrt - 1 = \pm i

Finally, substitute the values of bb into a=2+ba = 2 + b:

When b=0b = 0:

a=2+0=2a = 2 + 0 = 2

When b=2b = - 2:

a=2+2=22=0a = 2 + - 2 = 2 - 2 = 0

When b=ib = i:

a=2+ia = 2 + i

When b=ib = - i:

a=2+i=2ia = 2 + - i = 2 - i

Therefore the solutions to the simultaneous equations are (in the form (a,b)(a, b)):

(2,0)(2, 0) (0,2)(0, -2) (i,2+i)(i, 2 + i) (i,2i)(-i, 2 - i)

Yajat Shamji - 2 weeks, 2 days ago

Log in to reply

Here are the solutions I found for your revised equation (j=1 j = \sqrt{-1} ):

(a1,b1)=(1,1)(a2,b2)(1.7622,0.2378)(a3,b3)(0.1207,2.1207)(a4,b4)(1.67931.3711j,0.32071.3711j)(a5,b5)(1.6793+1.3711j,0.3207+1.3711j) (a_1, b_1) = (1,-1) \\ (a_2, b_2) \approx (1.7622,-0.2378) \\ (a_3, b_3) \approx (-0.1207,-2.1207) \\ (a_4, b_4) \approx (1.6793-1.3711 j, -0.3207-1.3711 j ) \\ (a_5, b_5) \approx (1.6793+1.3711 j, -0.3207+1.3711 j )

Steven Chase - 2 weeks, 1 day ago

Log in to reply

Algebraically, not graphically!

(Sorry if it appears I'm shouting - I'm actually just pressed for time!)

Yajat Shamji - 2 weeks ago

Log in to reply

Log in to reply

This is my best pair of simultaneous equations by far - I have spent the last 22 - 33 weeks finding a really extreme pair. I challenge anybody to get full marks for this. You have until 14th14^{th} November, 1:301:30pm to get all 2222 marks.

Yajat Shamji - 1 month ago

Log in to reply

Ok!!!

Ravi Singh - 1 month ago

Log in to reply

Me with the help of internet.

Half pass3 - 1 month ago

Log in to reply

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

Half pass3 - 1 month ago

Log in to reply

That's not a solution.

Yajat Shamji - 3 weeks, 5 days ago

Log in to reply

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 - 3 weeks, 5 days ago

Log in to reply

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 - 3 weeks, 3 days ago

Log in to reply

bruh...

Percy Jackson - 2 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...