SOLUTION: Here's another question from my teacher. Solve this equation by the means of your choice: p^2 + 100 = 0 My initial thought is that P would = -10, but that would would work

Algebra ->  Algebra  -> Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Here's another question from my teacher. Solve this equation by the means of your choice: p^2 + 100 = 0 My initial thought is that P would = -10, but that would would work       Log On

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Question 133465: Here's another question from my teacher.
Solve this equation by the means of your choice: p^2 + 100 = 0
My initial thought is that P would = -10, but that would would work out as:
-10^2+100=200. Somehow I have to get the equation to work out as -100+100=0, but I just can't figure it out.

Answer by solver91311(12126) About Me  (Show Source):
You can put this solution on YOUR website!
The problem that you are having is that there is no real number x such that x%5E2=-1, hence you are doomed to eternal frustration trying to find a real number solution to your problem. In order to deal with this problem, we created a number that is not a real number and called it an imaginary number represented by i and defined by the relationship: i%5E2=-1.

Using i, you can solve your problem.

p%5E2+%2B+100+=+0

p%5E2=-100

p%5E2=%28-1%29%28100%29

p=sqrt%28-1%29%2Asqrt%28100%29 or p=-sqrt%28-1%29%2Asqrt%28100%29 (remember, you must consider both the positive and negative square root because if x%5E2=4, then x=2 or x=-2.

Since i%5E2=-1, i=sqrt%28-1%29, so:

p=i%2Asqrt%28100%29 or p=-i%2Asqrt%28100%29
p=10i or p=-10i

Let's check the answer:
%2810i%29%5E2=100i%5E2, but i%5E2=-1, so 100i%5E2=-100 and

%28-10i%29%5E2=100i%5E2, but i%5E2=-1, so 100i%5E2=-100

Answer checks.

In general, anytime the value under a radical is less than zero, factor out a -1 and the square root of that is i, then take the positive square root of the positive number remaining under the radical. The square root is then the square root of the positive value times i.