What are the solutions to x^4+9 = 0 ?Date: 03/31/2003 at 20:21:07 From: Christina Subject: What are the solutions to x^4+9=0 ? What are the solutions to x^4+9 = 0 ? My confusion has to do with how to work with complex numbers when the root is higher than two. Date: 04/01/2003 at 09:21:56 From: Doctor Ian Subject: Re: What are the solutions to x^4+9 = 0 ? Hi Christina, When I see 0 = x^4 + 9 I see 0 = (x^2)^2 - (3i)^2 which is a difference of squares, which means I can do this: 0 = (x^2 + 3i)(x^2 - 3i) The second factor is another difference of squares: 0 = (x^2 + 3i)(x + sqrt(3i))(x - sqrt(3i)) And if you look at it the right way, the first factor is also a difference of squares: 0 = (x^2 - (-3i))(x + sqrt(3i))(x - sqrt(3i)) = (x + sqrt(-3i))(x - sqrt(-3i))(x + sqrt(3i))(x - sqrt(3i)) So the four 4th roots of -9 would be sqrt(3i) sqrt(-3i) -sqrt(-3i) -sqrt(-3i) Let's check to see if those work: [sqrt(3i)]^4 = (3i)^2 = -9 [sqrt(-3i)]^4 = (-3i)^2 = -9 And I think you can see that the other two will work as well. So, what is sqrt(3i)? Obviously, sqrt(3i) = sqrt(3) * sqrt(i) So what is sqrt(i)? To find out, take a look at this answer from the Dr. Math archives: The Fourth Root of -1 http://mathforum.org/library/drmath/view/53846.html Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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