Respuesta :
Width: x
Length: x + 3
Area = lw
154 = (x+3)(x)
154= x^2 + 3x
0= x^2 + 3x -154
0= (x+14)(x-11)
Possible solutions: x+14=0 >> x= -14, x-11=0 >> x=11
The answer is x=11 (or in other words the width is 11m) because the width cannot be negative.
Length: x + 3
Area = lw
154 = (x+3)(x)
154= x^2 + 3x
0= x^2 + 3x -154
0= (x+14)(x-11)
Possible solutions: x+14=0 >> x= -14, x-11=0 >> x=11
The answer is x=11 (or in other words the width is 11m) because the width cannot be negative.
Let x = the width of the rectangle.
Let x+3 = the length of the rectangle.
Since area of a rectangle is A=l*w
154 = x(x + 3)
154 = x^2 + 3x
-154. -154
0 = x^2 + 3x - 154
0 = (x + 14)(x - 11)
Set x + 14 = 0 and x - 11 = 0
Then x = -14 and x = 11
Reject x= -14 since lengths cannot be negative.
Therefore, the width is 11 m and the length is 11+3 or 14 m.
Let x+3 = the length of the rectangle.
Since area of a rectangle is A=l*w
154 = x(x + 3)
154 = x^2 + 3x
-154. -154
0 = x^2 + 3x - 154
0 = (x + 14)(x - 11)
Set x + 14 = 0 and x - 11 = 0
Then x = -14 and x = 11
Reject x= -14 since lengths cannot be negative.
Therefore, the width is 11 m and the length is 11+3 or 14 m.