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.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

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.

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:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Leave the $= - 8$ 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 $a = 2 + b$.

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

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

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

Divide all sides of the equation by $b$.

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

Divide the LHS by $32$.

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

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

Multiply by $b$, resulting in $b^5 + 4b^4 + 5b^3 + 11b^2 + 18b + 1 = 0$.

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

Set the equation as $f(b)$.

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

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

You should get $b^4 + 3b^3 + 2b^2 + 9b + 9$.

Set this equation as $f(b)$.

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

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

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

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

Therefore, the factors of $b^5 + 4b^4 + 5b^3 + 11b^2 + 18b + 9$ are $(b + 1)^2(b + 3)(b^2 - b + 3)$

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

This simplifies to $\frac{1}{2} \pm \frac{11i}{2}$

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

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

Therefore the solutions to the simultaneous equations:

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

$a - b = 2$

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

$(1, - 1)$

$(- 3, - 3)$

$(\frac{5}{2} - \frac{11i}{2}, \frac{1}{2} + \frac{11i}{2})$

$(\frac{5}{2} + \frac{11i}{2}, \frac{1}{2} - \frac{11i}{2})$

Note: Only $4$ solutions are here to due to the repeated root of $(b + 1)^2$.Log in to reply

@Vinayak Srivastava, @Frisk Dreemurr

Log in to reply

Ok!!!

Log in to reply

Me with the help of internet.

Log in to reply

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

Log in to reply

That's not a solution.

Log in to reply

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)

$a = 0.464537, b = -1.53546$

$a = 1.42518, b = -0.574825$

Log in to reply

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.

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)

Log in to reply

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:

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

Log in to reply

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

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.).

Log in to reply

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!

Log in to reply

bruh...

Log in to reply

@Frisk Dreemurr

Log in to reply

Uhh...

Hello Yaj

Anything you'd love to say?

Log in to reply

I posted a new solution and edited it - it just took me $6$ months to figure it out...

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Plus, I was busy with my studies. So, not a lot of time to actually attempt to correct in these $6$ months.

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

So, technically, it took me $45$ minutes in $6$ months...

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Second to top comment. What do you think?

Log in to reply

Let me know soon!

:)

Log in to reply

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

Log in to reply