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

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Answer 1 · 2018-04-14T04:01:17

#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 " "#

Answer 2 · 2018-04-14T04:08:03

#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 :)

Answer 3 · 2018-04-14T04:11:43

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