# Thread: Internal Rate of Return, Simplifying Equation Help

1. ## Internal Rate of Return, Simplifying Equation Help

Hi, I have an Internal Rate of Return question I need help with please.

I have been given the answer for r, which is 0.208, however I cannot understand how the following equation:

-300 + -300/(1+r) + 800/(1+r)^2 = 0

when multiplied by (1+r)^2, becomes:
300r^2 + 900r - 200 = 0

Could someone please explain step by step how this is achieved?

Thanks

Charlie

2. ## Re: Internal Rate of Return, Simplifying Equation Help

Have you done as advised? Multiplying both sides by (1 + r)^2 ? After you've done that, expand...

3. ## Re: Internal Rate of Return, Simplifying Equation Help

That's what I am struggling to do, how would you multiply this by (1+r)^2..... I can't get to 300r^2 + 900r - 200 = 0

4. ## Re: Internal Rate of Return, Simplifying Equation Help

So what did you get?

(If the problem is, as you stated, -300 + -300/(1+r) + 800/(1+r)^2 = 0 multiplied by (1+ r)^2, then the "given" answer, 300r^2 + 900r - 200 = 0, is NOT correct.)

5. ## Re: Internal Rate of Return, Simplifying Equation Help

That's what I thought, it's from a past exam paper at my university.

I would have made 1/(1+r) = x and then multiplied it out.

6. ## Re: Internal Rate of Return, Simplifying Equation Help

-300 + -300/(1+r) + 800/(1+r)^2 = 0
divide every term by $100$ ...

$-3 - \dfrac{3}{1+r} + \dfrac{8}{(1+r)^2} = 0$

$\dfrac{1}{1+r} = \dfrac{3 + \sqrt{9 +96}}{16} \implies r = \dfrac{16}{3+\sqrt{105}} - 1 \approx 0.208$

7. ## Re: Internal Rate of Return, Simplifying Equation Help

Originally Posted by charlie3
I would have made 1/(1+r) = x and then multiplied it out.
Easier this way: let x = 1+r

Then divide by 100 (as per Skeeter) to get:
3x^2 + 3x - 8 = 0

Solve using quadratic to get x = 1.2078...

1 + r = 1.2078...
r = .2078 which is ~20.8%

Yokay?

8. ## Re: Internal Rate of Return, Simplifying Equation Help

I love you all so much.

Thank you this really helped me!!!!