How do you solve #(x+2)^2=-7#?

2 Answers
Mar 25, 2018

Answer:

No solution

Explanation:

#(x+2)^2=-7#
#x^2+4x+4=-7#
#x^2+4x+11=0#

Now I use the discriminant to determine if there is a solution

#b^2-4ac#

#4^2-(4.7)#
#16-28#

Already, discriminant is less than zero there there are no real solutions for x.

Mar 25, 2018

Answer:

#x=-2+-isqrt(7)#

Explanation:

There are many ways to tackle this; the easiest and quickest way here would be the square root method: take the square root of both sides:

#sqrt((x+2)^2)=+-sqrt(-7)#

#x+2=+-sqrt(-7)#

Since you cannot take the square root of a negative and get a real number, there is no real solution to this equation. However, we can simplify to get a complex solution:

#x+2=+-sqrt(7)*sqrt(-1)#

#x+2=+-isqrt(7)#

#x=-2+-isqrt(7)#