Area is 560 the lenght is 8 yards greater than the width. What is the width

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May 4th, 2015

Area=Lenght . Width

Let's call x the width. As the lenght is 8 yards greater than the width, we can express lenght as x+8

So we have Area = (x+8).x

We just have to solve 560 = (x+8)*x 

x^2+8x-560=0

Then we calculate the discriminant D 

D=8^2-4*1*(-560)=2304

(In the general case of the equation a.x^2+b.x+c=0, D=b^2-4.a.c)

Solutions are x1=(-8+square root(2304))/(2*1) or  x1=(-8-square root(2304))/(2*1)

x1=(-8+48)/2 or x2=(-8-48)/2

x1=40/2=20 or x2=-28

x2 is a negative solution so we have to remove it. The only good solution is x1 and the width we are seeking is 20 yards. 

May 4th, 2015

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