How do you solve p^ { 2} + 2p= - 4?

1 Answer
Jul 19, 2018

Below

Explanation:

p^2+2p=-4

p^2+2p+4=0

Using the quadratic formula,

p=(-b+-sqrt(b^2-4ac))/(2a)

p=(-2+-sqrt(4-4(1)(4)))/2

p=(-2+-sqrt(-12))/2

Therefore no solution as Delta <0


If the answer allows for complex numbers, then

p=(-2+-isqrt12)/2

p=(-2+-2isqrt3)/2

p=-1+-isqrt3

Therefore, p^2+2p+4=0 becomes
(p-(-1+isqrt3))(p-(-1-isqrt3))=0

(p+1-isqrt3)(p+1+isqrt3)=0