SOLUTION: A rectangle and a square have the same area. The width of the rectangle is 48 cm less than the side of the square and the length of the rectangle is 128 cm more than twice the widt

Question 988637: A rectangle and a square have the same area. The width of the rectangle is 48 cm less than the side of the square and the length of the rectangle is 128 cm more than twice the width of the square. What are the dimensions of the rectangle?
I am having trouble translating the application into a formula.
Found 2 solutions by Boreal, lwsshak3:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
L=rectangle length
W=width
side of a square is S
S-48=W
2S+128=L since square has width S.
Rectangle has area (2S+128)(s-48)=S^2
2S^2-96S+128S-6144=S^2
S^2+32S-6144=0
S=-(1/2){-32 +/- sqrt (32^2+4*6144)}, the sqrt term equals 160; [-32+160]/2 using quadratic formula
S=64
W=16
L=128*2=256
Area of square is 64^2=4096 cm^2
Area of rectangle is 256*16=4096 cm^2
Rectangle is 256 cm by 16 cm

Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
A rectangle and a square have the same area. The width of the rectangle is 48 cm less than the side of the square and the length of the rectangle is 128 cm more than twice the width of the square. What are the dimensions of the rectangle?
let x=side(width) of square
x-48=width of rectangle
2x+128=length of rectangle
(x-48)(2x+128)=x^2
2x^2+32x-6144=x^2
x^2+32x-6144=0
(x-64)(x+96)=0
x=64
side(width) of square=64
x-48=width of rectangle=16 cm
2x+128=length of rectangle=256 cm
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