Math Is Fun Forum

dadon11

Guest

rearranging to find IRR

Hey All..

I would like to know how would you find IRR when:

the solution to this answer is IRR = 28.08%

Thank you

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: rearranging to find IRR

Multiply both sides by (1+IRR)², then you can expand the brackets and be left with a quadratic in IRR.
This solves to give the answer you want (as well as a negative one which is presumably irrelevant in your context).

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Why did the vector cross the road?
It wanted to be normal.

dadon11

Guest

Re: rearranging to find IRR

Hey

Thanks for your reply..

This is what I get

6000IRR + 4000IRR^2 - 2000 = 0

Is that right?

How would I solve from here?

Thank you

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: rearranging to find IRR

That equation looks right, although you can divide it by 2000 if you want.
To solve it, use the quadratic equation:

---

Why did the vector cross the road?
It wanted to be normal.

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: rearranging to find IRR

I think it is much easier to solve the quadratic for t=1/(1+IRR) directly, then after that solve for IRR.

dadon11

Guest

Re: rearranging to find IRR

That equation looks right, although you can divide it by 2000 if you want.
To solve it, use the quadratic equation:

Hey..

Ok I divided by 2000 so:

3IRR + 2IRR^2 - 1 = 0

Then I used formula to get: x = 2/3 and x = -2

Have I done something wrong? as the answer should be 28.08 %

Thank you

dadon11

Guest

Re: rearranging to find IRR

I think it is much easier to solve the quadratic for t=1/(1+IRR) directly, then after that solve for IRR.

Hey

Thanks for your reply..

How does this method work? not quite sure how to apply it..

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: rearranging to find IRR

If you think of 1/(1+IRR) as a variable, then the original equation is already in quadratic form:
-4000 + 2000t + 4000t² = 0.

Solving this gives two values for t, and then you use t = 1/(1+IRR) --> IRR = (1/t) - 1 to get the answer.

---

Why did the vector cross the road?
It wanted to be normal.

dadon11

Guest

Re: rearranging to find IRR

If you think of 1/(1+IRR) as a variable, then the original equation is already in quadratic form:
-4000 + 2000t + 4000t² = 0.

Solving this gives two values for t, and then you use t = 1/(1+IRR) --> IRR = (1/t) - 1 to get the answer.

thank you

that seem a much more neat way..

also I found out where I was going wrong..
I plugged in the 'a' and 'b' values the oppsite way in the quadratic formula..
so yeah I do get the answer I was looking for...

what if the equation went up to higher terms... as what I am working out is the internal rate of return (IRR) and sumtimes the question can go up too powers of 5.

I get the quadratic stage which is probably the simplest.. and won't come in the exam.. how do I solve for higher powers??
are there more formula

Thank you

dadon11

Guest

Re: rearranging to find IRR

If you think of 1/(1+IRR) as a variable, then the original equation is already in quadratic form:
-4000 + 2000t + 4000t² = 0.

Solving this gives two values for t, and then you use t = 1/(1+IRR) --> IRR = (1/t) - 1 to get the answer.

thank you

that seem a much more neat way..

also I found out where I was going wrong..
I plugged in the 'a' and 'b' values the oppsite way in the quadratic formula..
so yeah I do get the answer I was looking for...

what if the equation went up to higher terms... as what I am working out is the internal rate of return (IRR) and sumtimes the question can go up too powers of 5.

I get the quadratic stage which is probably the simplest.. and won't come in the exam.. how do I solve for higher powers??
are there more formula

Thank you

dadon11

Guest

Re: rearranging to find IRR

If you think of 1/(1+IRR) as a variable, then the original equation is already in quadratic form:
-4000 + 2000t + 4000t² = 0.

Solving this gives two values for t, and then you use t = 1/(1+IRR) --> IRR = (1/t) - 1 to get the answer.

thank you

that seem a much more neat way..

also I found out where I was going wrong..
I plugged in the 'a' and 'b' values the oppsite way in the quadratic formula..
so yeah I do get the answer I was looking for...

what if the equation went up to higher terms... as what I am working out is the internal rate of return (IRR) and sumtimes the question can go up too powers of 5.

I get the quadratic stage which is probably the simplest.. and won't come in the exam.. how do I solve for higher powers??
are there more formula

Thank you

dadon11

Guest

Re: rearranging to find IRR

hi

sorry I didn't mean to double post.. there way an error so I went hit the 'back' button and now there's the same post.