# Did I solve this compound formula correctly?

Asked by chelle21689 (6507) November 20th, 2010

Here is the question.

“Use the formula A=P(1+r)^2 to find the rate R at which \$1500.00 increases to \$1749.60 in 2 years. (This formula is a compound interest formula for determining A the compound amount or future value of money based on the P the principle or present value, rate the annual rate of interest, and t time in years).

This is how I set it up

1749.60= 1500.00(1+r)^2

1749.60=1500.00+r^2

249.6= r^2

Square root both sides

Am I correct?

Observing members: 0 Composing members: 0

(a+c)^2 = a^2 +2ac +c^2, and you must distribute the product over the sum: a(b+c) = ab+ac.

ratboy (15142)

But 1 squared is still 1…which would still be 1500.00, no?

chelle21689 (6507)

@chelle21689 Your problem, though, is that you distributed the “squared” over the parenthesis. You can’t do that. (a+b)^2 doesn’t equal a^2+b^2, it equals a^2+2ab+b^2.

But also, in this situation you can avoid expanding your squared term if you want. Divide both sides by 1500, take the square root of both sides, then subtract one. Voila!

Mariah (23101)

A=P(1+r)^2
>A/P=(1+r)^2
>1+r=srt(A/P)
>r=srt(A/P) – 1

you should algebraically solve your equation prior to substituting the variables – later on in any math (and subset), this will be an important thing to do.

don’t forget that

P(1+r)^2

is basically

(P+PR)^2

BUT thats complicated, and complicated means too much work, and one should try to do as least work as possible in life. however, this doesn’t mean that you shouldn’t strive to be efficient! efficiency is key, young padawan!

Assuming that you put the correct numbers into the variables (A = \$1748.60, R = \$1500.00, and ?n = 2?) [which makes me kind of wonder if the basic formula is A=P(1+r)^n]
[sorry, i havent done annuities, or whatever this is, for a looong time…]

r = 0.08%

pennybooks (144)

Thanks, I’ll try again. Math is my worst subject, I’m lucky I have a C lol.

chelle21689 (6507)

@pennybooks you were doing well until you got to:
P(1+r)^2 = (P + Pr)^2

The order of execution is that the term raised to an exponent has to be executed prior to simple multiplication. So:
P(1+r)^2 = P( 1 + 2r + r^2) = P + 2Pr + Pr^2

This would certainly be different from the term you arrived at:
(P + Pr)^2 = P^2 + 2Pr + (Pr)^2

CyanoticWasp (20041)

@CyanoticWasp
FUUUUUUUUU———-

i totally knew that looked funky

pennybooks (144)

There’s an easier way to do this….

Divide everything by 1500. Then, take the square root of everything.

roundsquare (5512)

No, @roundsquare, @pennybooks had the right idea when she isolated r on the left side of the equation. It was after that that her solution went off the tracks.

CyanoticWasp (20041)

@CyanoticWasp
i’m pretty sure my answer is right though.

i just had a brain fart with that P(1-r)^2 thing

p.s. i have a dick

pennybooks (144)

I got 2.08% as the answer. is this right?

1749.60 = 1500.00 (1+r)^2

square root 729/625=square root (1+r) I do this to get rid of (1+r)^2

and you know the rest.

chelle21689 (6507)

nvm its wrong!

chelle21689 (6507)

BUT I did some how get a second answer to be -.08 I’ll try again.

chelle21689 (6507)

@chelle21689 Did you add one at the end instead of subtracting? Because I got .08. Note that this is the percent in decimal form, so the answer is 8%, not .08%.

Mariah (23101)

I got it :) THANKS EVERY ONE!!

chelle21689 (6507)

@Mariah
FUUUUUUUUU———-

yea its 8%

pennybooks (144)

or