# Quintic Simultaneous Equations

Solve the simultaneous equations: $(a + b)^3 (a + b)^2 = 32b^4 + 96b^3 + 32b^2 - 32b$ $a - b = 2$

($22$ marks)

Note by Yajat Shamji
1 month ago

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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)

$a = 0.464537, b = -1.53546$

$a = 1.42518, b = -0.574825$

- 1 month ago

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.

- 1 month ago

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)

- 1 month ago

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:

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

- 4 weeks, 1 day ago

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

- 4 weeks, 1 day ago

My solution (wasn't going to work without changing R.H.S of equation $1$):

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

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

Then divide both sides by $32$ and move the R.H.S to the L.H.S and you'll obtain $b^5 + 4b^4 + 5b^3 + 4b^2 + 4b$. Label this as equation $3$.

Divide both sides by $b$ to get $b (b^4 + 4b^3 + 5b^2 + 4b + 4)$. Label this equation $4$.

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

Perform algebraic long division to obtain $b^3 + 2b^2 + b + 2$. Label this equation $5$ and assign it to $f(x)$.

Perform factor theorem on equation $5$ - you'll find that $f(-2) = 0$.

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

$\therefore$ Factors of $b^5 + 4b^4 + 5b^3 + 4b^2 + 4b$ are:

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

Either:

$b = 0$

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

or:

$b^2 + 1 = 0, b^2 = - 1, b = \sqrt - 1 = \pm i$

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

When $b = 0$:

$a = 2 + 0 = 2$

When $b = - 2$:

$a = 2 + - 2 = 2 - 2 = 0$

When $b = i$:

$a = 2 + i$

When $b = - i$:

$a = 2 + - i = 2 - i$

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

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

- 2 weeks, 2 days ago

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

$(a_1, b_1) = (1,-1) \\ (a_2, b_2) \approx (1.7622,-0.2378) \\ (a_3, b_3) \approx (-0.1207,-2.1207) \\ (a_4, b_4) \approx (1.6793-1.3711 j, -0.3207-1.3711 j ) \\ (a_5, b_5) \approx (1.6793+1.3711 j, -0.3207+1.3711 j )$

- 2 weeks, 1 day ago

Algebraically, not graphically!

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

- 2 weeks ago

- 1 month ago

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.

- 1 month ago

Ok!!!

- 1 month ago

Me with the help of internet.

- 1 month ago

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

- 1 month ago

That's not a solution.

- 3 weeks, 5 days ago

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.).

- 3 weeks, 5 days ago

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

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

- 3 weeks, 3 days ago

bruh...

- 2 weeks ago

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