Verified answer

2x^2 + 9x + 4 = 0

(2x + 1)(x + 4) = 0

2x + 1 = 0 or x + 4 = 0

2x = -1............x = -4

x = -1/2

A

2x²+4x+4= -5x

2x^2 + 4x + 5x + 4

2x^2 + 9x + 4

multiply the coefficient of x^2 by the constant------- 2 x 4 = 8

now find two numbers that multiplies to give 8 and add to give 9----- the numbers are 8 and 1

replace the 9x by the new values

2x^2 + 8x + x + 4

group terms

(2x^2 + 8x) + (x + 4)

factorize

2x ( x + 4) + (x +4)

factor the ( x + 4)

(2x + 1) ( x + 4)

equate both to zero to solve for x

2x + 1 = 0

x = -1/2

x + 4 = 0

x = -4

x can be -4 or -1/2 but from the choices only -4 is given

The answers is A

2x² + 4x + 4 = -5x

2x² + 9x + 4 = 0

(2x + 1)(x + 4) = 0

x = -½ or x = -4 ∙ ∙ ∙ ∙ ∙ so A and F