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

Question

1328 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-19T02:42:12

Below

#p^2+2p=-4#

#p^2+2p+4=0#

#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#