How do you solve #x^2+ 4 = 0#?

Question

Hazel Anglin · Accepted Answer

No real solution

#x^2+4=0# Start by subtracting #4# from both sides #x^2 + 4 - 4 = 0 - 4# #x^2 = -4# Now we can take square root #x = +-sqrt(-4)# Thus, No real solutions

Abigail Allen · Answer

#x=+-2i#

For this, you need a concept of imaginary numbers

#x^2+4=0#

#rArr x^2+4-4=0-4#

(Subtracting #4# from both sides)

#rArr x^2=-4#

#rArr x=sqrt(-4)#

#rArr x=sqrt(-1xx4)#

#rArr x=+-2sqrt(-1)#

(The value of #sqrt(-1)# is #i# )

#rArr x=+-2i#

Hope this helps :)

Savannah Anderson · Answer

#x=+-2i " "# #x^2+4=0 " "# subtract 4 on both sides to get #x^2 = -4 " "# Take the square root of both sides to get #x=+-2i " "#

Abigail Allen · Answer

undefined To solve the equation $x^2 + 4 = 0$, you can start by subtracting 4 from both sides to isolate the term $x^2$. Then, you can take the square root of both sides to solve for $x$. However, there's a problem here because the square root of a real number is always non-negative, and the square root of a negative number is not a real number. Therefore, this equation has no real solutions. In the complex number system, you can express the solution as $x = \pm 2i$, where $i$ is the imaginary unit ($i^2 = -1$).

Ethan Adkins · Answer

undefined To solve $ x^2 + 4 = 0 $, follow these steps:

1. Subtract 4 from both sides of the equation:
   $ x^2 = -4 $.

2. Take the square root of both sides of the equation:
   $ x = \pm \sqrt{-4} $.

3. Since the square root of -4 is a complex number, it can be expressed as:
   $ x = \pm 2i $.

Therefore, the solutions to the equation $ x^2 + 4 = 0 $ are $ x = 2i $ and $ x = -2i $, where $ i $ is the imaginary unit ($ i = \sqrt{-1} $).