Question 755384
a rectangle has its length of 2 ft greater than its width.if the length is increased by 3 ft and the width by 1 ft, the area of the new rectangle will be twice the area of the old one. what is the length and width of the original rectangle
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let x=width of original rectangle
x+2=length of original rectangle
area of original rectangle=x(x+2)=x^2+2x
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x+1=width of new rectangle
x+2+3=length of new rectangle
area of new rectangle=(x+1)(x+5)=x^2+6x+5
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x^2+6x+5=2(x^2+2x)
x^2+6x+5=2x^2+4x
x^2-2x-5=0
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solve for x using following quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=1, b=-2, c=-5
x=-1.55 (reject)
x=3.45
x+2=5.45
width of original rectangle=3.45 ft
length of original rectangle=5.45 ft