Question 884734
The width of a rectangle is 1 less than twice its length. If the area of the rectangle is 88 cm^2, what is the length of the diagonal?
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let x=length of rectangle
2x-1=width of rectangle
..
length*width=area
x(2x-1)=88
2x^2-x=88
2x^2-x-88=0
solve for x by quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
a=2, b=-1, c=-88
ans: 
x=length≈6.89
width≈2*6.89-1≈12.78
diagonal≈√[(6.89)^2+(12.78)^2]≈√(47.47+163.33)≈√210.80≈14.52
length of the diagonal≈14.52 cm