Can this equation be solved algebraically?

Question

8.712r + 1= (1+r)^10 Solve for r.

Of course I can just use guess and check, but where's the elegance in that?

Maddie can you put the Maple results in scientific notation?

The formula I came up with is the end result of a financial formula analyzing my IRA.

I started in April and have put in $339 (including employer contributions) a month since. I now have $3,292. That's a net gain of $245 or so. Now most of that growth was in the 4th Quarter according to the statement, and obviously the monthly return grows as the principle does, so I'm trying to figure out what the actual annual rate of return is compounding monthly.

That's what I came up with, having 12r as the Annual Rate. Logically, r must be a bit higher then .01, but not much higher.

Answer 1

r + 1 = (1+r)^10 can be solved algebraically, but not 8.712 r +1 = (1+r)^10.

Sorry, pal.

Here's one root for the first equation: r = 1/2(-3 - i Sqrt(3)). Forget about the other one.

Addendum: If you had asked for an approximate real root, the answer is

r = -0.0310601...

Which is the only real root, besides 0, and 8 other complex roots.

Note that this is NOT an "algebraic solution", and for everyone else's information on this page, there EXISTS NO algebraic solution for this equation, because of the factor 8. 712. Read up on polynomial functions and Abelian groups

Addendum 2: You made 9 payments of $339, or 3,051 total. The equation to be solved is

339((1+r)^9 -1)/((1+r)-1) = 3292, or

(1+r)^9 - (3292/339)r -1 = 0

This has a real root of r = 0.01889, which, compounded over 12 months, yields a 25.178% annual rate of return.

Unfortunately, the exact calculation depends when you reached $3,292. Was it right after you made your 9th payment, or a month later, after interest? This calculation assumes the former.

If you made 10 payments, then you lost money.

Answer 2

It has 10 complex roots but if you talk about real roots...
You notice that it must have other root besides 0. For this proceed like Sheen did and see that, after eliminate r, it boils down to a polynomial of odd degree. But any polynomial of odd degree has one real root which implies that the original equation has at least 2 real roots.

Moreover the second root cannot be positive
for this use Bernoulli inequality which says that
(1+r)^n > 1+r*n, for every real number r, r >=-1, and any integer n >=2.
Now, if r>0 then (1+r)^10 > 1 + 10r > 1+ 8.712 r

Indeed, as Maddie find out using Maple, the equation has other nonzero real root which is negative.

Moreover the second root is bigger than -1/8.712 because otherwise the left hand will be negative and the right hand positive. So they must be in a small interval.

Answer 3

I used algebra to find one solution (sort of):
The binomial expansion of (1+r)^10 is r^10 + 10r^9 + 45r^8 + 120r^7 + 210r^6 + 252r^5 + 210r^4 + 120r^3 + 45r^2 + 10r + 1. If you put that into your original equation and do a little algebra, you can make the equation say:
r^10 + 10r^9 + 45r^8 + 120r^7 + 210r^6 + 252r^5 + 210r^4 + 120r^3 + 45r^2 +(1.288)r + (0) = 0.
~I included + (0) on the left side only for emphasis.~
By the rational roots theorem, if there are any rational solutions, each of them must be expressible as a fraction consisting of a factor of the left-side constant term (0), over a factor of the leading coefficient (1). Thus, 0 is the only possible rational solution (though there may well be numerous real and complex solutions). Here, unfortunately, one must still *check* whether 0 is in fact a solution (how inelegant!). And, as others have pointed out, it is.

Addendum: Now that you've revealed that the problem arose from a real-world situation, I feel silly about having answered in the way I did. Good luck figuring it out!

Answer 4

Sorry about your math instructor! So what you do on the first one is....you're both using eliminating or substitution i'm guessing by technique of the concern. for that reason you've 2 variables that are ones so substitution might want to be a lot less complicated. So on the first (or 2d. although you choose) positioned y on one fringe of the = signal and the different stuff on the different facet. So ... y= -3x +16. I in basic terms moved the 3x over. then you definately change -3x+16 into the different equation for y. 2x + (-3x+16) = 11. Now clean up in many cases. So... (2x + -3x is -x by technique of ways) -x +16 = 11. Subtract and also you get -x= -5. change that to x=5. that is an same concern. Now all you want is the y. So plug the 5 into between both equations and clean up back. i'll apply the 2d one. 2(5) + y = 11. So y= a million. good success and that i desire I helped!

Answer 5

8.712r + 1= (1+r)^10

8.71r + 1 = 1 + 10r + 10 x 9 / (1x2) r^2 + 10 x 9 x 8 / (1x2x3)r^3 + ... .... ...... ... + 10r^9 + r^10
Transferring and simplifying and factorizing

r(r^9 + 10r^8 + ... .... ... ... + 1.3) = 0
will have 10 roots / values of r
r = 0 is one
Remaining 9 roots will be found by factorizing or trial/error method or by means of programmed calculator/ computer software

Answer 6

There are 10 roots by the fundamental theorem of algebra.

According to Maple, they are:
0., -2.269091916-.4408658322*I, -2.269091916+.4408658322*I, -1.713274699-1.119656568*I, -1.713274699+1.119656568*I, -.8656090231-1.285047693*I, -.8656090231+1.285047693*I, -.1364943246-.8678359722*I, -.1364943246+.8678359722*I, -0.3106007597e-1

If you have Maple, you can study the source code and see what techniques they use to find the roots; there are tons of them and they are not trivial.

Answer 7

r = 0 is the only solution

to do this, u can plug it into your calculator. just graph it
y = 8.712r + 1 - (1+r)^10
and see where the graph touches the x axis

or, you can always expand the binomial or use logarithms

Answer 8

R = 0 is one solution. It is almost obvious without using algebra.

Answer 9

8.712r + 1 = (1 + r) ^10

8.712r + 1 = 1^10 + r^10
8.712r+1 = 1 + r^10
8.712r = r^10
8.712 = r^9
ninth root of 8.712 = r

i dont have a calculator to do this but that is how it is solved...
Good luck!

:)

Answer 10

You could use log :

log(8.712r)+log1=10log(1+r)
log8.712+logr+0=10log(1+r)
log8.712=log[(1+r)^10/r]
(1+r)^10/r=8.712 (taking anti-log)

and then i got lost.if you get a solution please msg me.