SOLUTION: (3y+2)^2+400=0

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Question 346505: (3y+2)^2+400=0
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%283y%2B2%29%5E2%2B400=0
If we subtract 400 from each side we get:
%283y%2B2%29%5E2+=+-400
At this point, if we're paying attention to what the equation is saying, we can figure out that there is no solution (at least within the set of Real numbers) to this equation. The equation is saying that something squared equals a negative number. Within the set of Real numbers squaring a number always results in either a positive number or zero. So squaring a number and getting -400 is not possible.

If you are working with the set of Complex numbers, then this equation is possible. So I will continue a solution with the understanding that the answer will be a complex number.

We can proceed by finding the square root(s) of each side:
sqrt%28%283y%2B2%29%5E2%29+=+sqrt%28-400%29
Simplifying this:
abs%283y%2B2%29+=+sqrt%28400%2A%28-1%29%29
3y%2B2+=+0+%2B-+sqrt%28400%29%2Asqrt%28-1%29
(Note: The "0" is used here because Algebra.com's formula software will not let me use "+-" without it.)
3y%2B2+=+0+%2B-+20%2Ai
Now we can solve for y. Add -2 to each side:
3y+=+-2+%2B-+20%2Ai
Multiply each side by 1/3:
y+=+-2%2F3+%2B-+%2820%2F3%29i
These are the complex number solutions to your equation.