SOLUTION: Dear Sir/Madam, I am confronted with the following problem: "Solve for x. Find all solutions. x^3 + 2x^2 + 5x = 0" I did the following: 1) x(x^2 + 2x + 5 = 0) 2) x = 0

Algebra ->  Algebra  -> Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Dear Sir/Madam, I am confronted with the following problem: "Solve for x. Find all solutions. x^3 + 2x^2 + 5x = 0" I did the following: 1) x(x^2 + 2x + 5 = 0) 2) x = 0      Log On

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Question 6598: Dear Sir/Madam,
I am confronted with the following problem:
"Solve for x. Find all solutions. x^3 + 2x^2 + 5x = 0"
I did the following:
1) x(x^2 + 2x + 5 = 0)
2) x = 0
3) x^2 + 2x + 5 = 0
4) x+=+%28-2+%2B-+sqrt%28+4-4%281%29%285%29+%29%29%2F%282%29+
5) x+=+%28-1+%2B-+sqrt%28+-16+%29%29+
So I assumed that the only answer is x = 0, however the answer is x = 0, -1+-2i. What is meant by 2i? Does i stand for imaginary number? And how do you get that answer?
Thanks in advance.
Regards,
-Mike
Found 2 solutions by ichudov, z3311638:
Answer by ichudov(499) About Me  (Show Source):
You can put this solution on YOUR website!
Great job, both in solving the problem as well as in using my formula system.
your step 5 is wrong, the answer is x=-1+%2B-+sqrt%28+-16+%29%2F2+, (you forgot /2)or
x=-1%2B-+2sqrt%28-1%29.
You should re-read the complex numbers chapter, where they explain what i is (i can be thought of as sqrt(-1).

Answer by z3311638(1) About Me  (Show Source):
You can put this solution on YOUR website!
1) x(x^2 + 2x + 5 = 0)
2) x = 0
3) x^2 + 2x + 5 = 0
4) x+=+%28-2+%2B-+sqrt%28+4-4%281%29%285%29+%29%29%2F%282%29+
5) x+=+%28-2+%2B-+sqrt%28+-16+%29%29%2F2+, i hope u aware of i = sqrt(-1),so
6) x=+%28-2+%2B-+sqrt%2816%2A-1%29%29%2F2+
7) x=+%28-2+%2B-+4%2Asqrt%28-1%29%29%2F2+
8) x=+%28-2+%2B-+4%2Ai%29%2F2+
9) +x+=+-1+%2B-+2i