Which of the following correctly completes the square for the equation below?
x2 + 4x = 27
A. (x + 2)2 = 31
B. (x + 2)2 = 43
C. (x + 4)2 = 31
D. (x+4)2 = 43
SUBMI
PREVIOUS

A. (x + 2)^2 = 31.

Step-by-step explanation:

x^2 + 4x = 27

x^2 + 4x - 27 = 0

A. (x + 2)^2 = 31

x^2 + 4x + 4 - 31 = 0

x^2 + 4x - 27 = 0

This option corresponds to the given equation, so it is correct.

B. (x + 2)^2 = 43

x^2 + 4x + 4 - 43 = 0

x^2 + 4x - 39 = 0

This does not correspond to the equation.

C. (x + 4)^2 = 31

x^2 + 8x + 16 - 31 = 0

x^2 + 8x - 15 = 0

This does not correspond to the equation.

D. (x+4)^2 = 43

x^2 + 8x + 16 - 43 = 0

x^2 + 8x - 27 = 0

This does not correspond to the equation.

So, your answer is A. (x + 2)^2 = 31.

Make sure to SUBMIT and check your PREVIOUS answers to make sure they are right! XD Also, hope this helps!

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

a

step-by-step explanation: