Question 269691: we are going to fence in a rectangular field and we know that for some reason we want the field to have an enclosed area of 75 feet squared. we also know that we want the width of the field to be 3 feet longer than the length. what are the dimensions of the field? round yound your answer to the nearest hundreth.
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! we are going to fence in a rectangular field and we know that for some reason we want the field to have an enclosed area of 75 feet squared. we also know that we want the width of the field to be 3 feet longer than the length. what are the dimensions of the field? round yound your answer to the nearest hundreth.
let the length be x
the width will be x+3
Area of rectangle = length * width
x*(x+3) = 75
x^2 +3x =75
x^2+3x-75=0
this is similar to the equation ax^2 + bx+c
So a= 1, b= 3 an c c=-75
The roots of the quadratic equation is given by
x1,x2 = - b +/ - (sqrt(b^2-4ac)) /2a
sqrt (b^2-4ac)= sqrt ( 9 +300)
= sqrt(309)= 17.58
x1= (-3 +17.58) / 2
x1 = 14.58/2
7.29
x2= (-3-17.58)/2
x2=-20.58 /2
x2= 10.29
The dimensions are 7.29 feet and 10.29 feet.
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