5000 = (3000/(1+x)) + (4000/(1+x)^2) solve for x

[bm8643]

Hello,
I am a former math whiz who hasn't used math in any meaningful way in probably 12 years. A friend who is taking a finance class presented this problem to me at a party last night (he's been stuck on it), and now I am determined to recover some of my long lost math skills and solve this thing. It's kind of depressing to think that a few years ago, I would have looked at this problem and instantly known how to solve it, and now I am stuck and asking for online help! Anyway, I hope I am posting this in the proper forum, and I hope the '^' symbol represents exponents (my (1+x) at the end of the problem is squared).

So the first thing I did was subtract 3000/(1+x) from both sides to get this:
5000 - (3000/(1+x)) = 4000/(1+x)^2

Then I multiplied both sides by (1+x), giving me this:
5000(1+x) - 3000 = 4000/(1+x)

And this is where I am forgetting what to do next (presuming what I've done so far is correct). I tried multiplying both sides by (1+x) again, but that just spins me in a circle and rearranges the equation, but puts me no closer to a solution. I'm sure I'm forgetting something simple that's going to cause me to slap my forehead, but the light bulb won't turn on. Anybody have a hint they could offer me? Thanks!

Brian

 

[galactus]

If you don/t use it you lose it, huh?.

Here we go. You'll recollect soon enough.

\(\displaystyle \L\\\frac{3000}{1+x}+\frac{4000}{(1+x)^{2}}=5000\)

Make the denominators the same:

\(\displaystyle \L\\\frac{(1+x)}{(1+x)}\cdot\frac{3000}{1+x}+\frac{4000}{(1+x)^{2}}=5000\)

\(\displaystyle \L\\\frac{3000(1+x)+4000}{(1+x)^{2}}=5000\)

\(\displaystyle \L\\\frac{7000+3000x}{(1+x)^{2}}=5000\)

\(\displaystyle \L\\7000+3000x=5000(1+x)^{2}\)

I'll let you finish. Can you solve the resulting quadratic?.

 

[soroban]

Hello, Brian!

Solve: \(\displaystyle \L\: 5000 \:=\:\frac{3000}{1\,+\,x} \,+\,\frac{4000}{(1\,+\,x)^2}\)

First, divide by 1000: \(\displaystyle \L\; 5 \;= \;\frac{3}{1\,+\,x} \;+\;\frac{4}{(1\,+\,x)^{^2}}\;\;\) . . . sheesh!

Multiply by \(\displaystyle (1\,+\,x)^2:\L\;\;5(1\,+\,x)^2\;=\;3(1\,+\,x)\,+\,4\)

This simplifies to the quadratic: \(\displaystyle \L\:5x^2\,+\,7x\,-\,2\:=\:0\)

Apply the Quadratic Formula: \(\displaystyle \L\:\fbox{x \;= \;\frac{-7\,\pm\,\sqrt{89}}{10}}\)

 

[bm8643]

Thank you both for your help. I was able to whittle it down to this after the first reply:

5x^2 + 7x - 2 = 0

Question.....can I use the factoring method to solve for x or is this an instance where only the Quadratic Formula will work (and that x won't be a nice round number)?

I was trying to remember how to factor, but I can't seem to make it work. I had this:

(5x 2) (x 1)

....but using the FOIL method I was unable to make the + and - signs work. So that means that factoring does not always work, right? And my friend will have to input the quadratic formula for his answer for x?

Thanks again! Now I feel determined to start studying math again.

 

[galactus]

Quadratic is a good way to go. No, it doesn't factor nicely.

You could also complete the square and arrive at

\(\displaystyle 5(x+\frac{7}{10})^{2}-\frac{89}{20}\)

The quadratic formula is more efficient, unless you want to do the completing the square for kicks.