Question

HELP ALGEBRA 2!

One side of a square is increased by 2 inches, while adjacent side is decreased by 2 inches. The area of the resulting rectangle is 32 square inches. What were the dimensions of the original square?

Answer 1

(x+2)(x-2)=32

(x+2)(x-2)=32x^2+2x-2x-4=32

(x+2)(x-2)=32x^2+2x-2x-4=32x^2-4-32=0

(x+2)(x-2)=32x^2+2x-2x-4=32x^2-4-32=0x^2-36=0

(x+2)(x-2)=32x^2+2x-2x-4=32x^2-4-32=0x^2-36=0(x+6)(x-6)=0

(x+2)(x-2)=32x^2+2x-2x-4=32x^2-4-32=0x^2-36=0(x+6)(x-6)=0x-6=0

x=6 the length of the original side of the square.

(6+2)(6-2)=32

(6+2)(6-2)=328*4=32

(6+2)(6-2)=328*4=3232=32