imaginary numbers in quadratic equations?

Question

Ok this is the third time i posted this question becasue I keep doing the wrong one. So. Lets try this again:

Find all real and imaginary solutions of the quadratic equations.
(y+12)^2 + 400 = 0
I know you get it to (y+12)^2 = - 400

Then we take the sqare root of -400 but then I dont get how you do that. I know that i^2 in -1 but i dont know how to put it to application here.

Answer 1

(y+12)^2=-400=-1*400=>i^2*400
y+12=plus or minus 20i and so
y=-12+20i or -12-20i

Answer 2

Listen to this carefully:
sqrt(-400)=sqrt{(400)*(-1)}
=sqrt(400)*sqrt(-1) [by laws of surds]
=(+20 or -20)*(+i or -i)=+20i or -20i;
Hence, now we have:
y+12=+20i or -20i;
y= -12+20i or -12-20i
Here, -12 is called the real part while +20i or -20i is called the imaginery part of the answer which, in turn, is called a complex number, i.e. a number having both a real as well as an imaginery part.

Answer 3

first of all, the square root of -1 = i

so...
(y+12)^2 + 400 = 0
y + 12 = sqrt(- 400)
y + 12 = sqrt(400) x sqrt(-1)
y + 12 = ±20i
y = ±20i - 12
y = ±5i - 3

the rest you can work out yourself

Answer 4

The sqare root of -400 is 20i.
Your answer will have "i" in it if you have the square root of a negative number.

y^2 + 12y= -544

Solve it yourself and you'll remember it this time.

Answer 5

take the squareroot of -400

(y+12)^2=+/- sqrt(-400)

sqrt(-400) is 20i (i being the imaginary number)

now you have y1=-12+20i and y2=-12-20i

Answer 6

The answer is that you can't take the square root of a negative number. The answer is does not exist.

Answer 7

DO YOUR OWN HOMEWORK, THEN YOU'LL HAVE LEARNED IT AND WON'T HAVE TO GO ON THE INTERNET ASKING OTHERS TO DO YOUR WORK FOR YOU.

Answer 8

listen to sundance, he has the correct answer