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Hey All..
I would like to know how would you find IRR when:
the solution to this answer is IRR = 28.08%
Thank you
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).
Why did the vector cross the road?
It wanted to be normal.
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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
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.
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I think it is much easier to solve the quadratic for t=1/(1+IRR) directly, then after that solve for IRR.
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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
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..
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.
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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
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
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
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.
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