A bridge on a river is modeled by the equation h = -0.2d^2 + 2.25d, where h is the height and d is the horizontal distance. For
cleaning and maintenance purposes a worker wants to tie a taut rope on two ends of the bridge so that he can slide on the rope. The rope is at an angle defined by the equation -d + 6h = 21.77. If the rope is attached to the bridge at points A and B, such that point B is at a higher level than point A, at what distance from the ground level is point A?
Since you are interested in values of h, it is convenient to eliminate d from the equations. The second equation tells you d = 6h - 21.77
Substituting that into the first equation gives h = (6h -21.77)(-0.2(6h -21.77) +2.25) = (6h -21.77)(-1.2h +6.604) 7.2h² -64.748h +143.76908 = 0 . . . . rearranging to standard form
The quadratic formula tells you the solution to ax²+bx+c=0 is ... x = (-b±√(b²-4ac))/(2a)
Using this formula on our quadratic, we have h = (64.748 ±√((-64.748)² -4(7.2)(143.76908)))/(2·7.2) h = (64.748 ±√51.754)/14.4 h ≈ 3.997 or 4.996
