SOLUTION: when a stone is dropped into a deep well, the number of seconds until the sound of a splash is heard is given by the formula{{{t=sqrt(x)/4+x/1100}}} where x is the depth of the wel

Algebra ->  Length-and-distance -> SOLUTION: when a stone is dropped into a deep well, the number of seconds until the sound of a splash is heard is given by the formula{{{t=sqrt(x)/4+x/1100}}} where x is the depth of the wel      Log On


   

Question 1040517: when a stone is dropped into a deep well, the number of seconds until the sound of a splash is heard is given by the formulat=sqrt%28x%29%2F4%2Bx%2F1100 where x is the depth of the well feet. For one particular well, the splash is heard 14 seconds after the stone is released.
1. How deep (to the nearest foot) is the well?
2. For a different well, the sound of the splash is heard 2 seconds after the stone is release. How deep (to the nearest foot) is the well?
Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
sqrt(x)/4+x/1100 = 14
(275*sqrt(x)+x)/1100 = 14
275*sqrt(x)+x = 15400
275*sqrt(x) = 15400-x
75625x = (15400-x)^2
75625x = x^2-30800x+237160000
-x^2+106425x-237160000 = 0
x^2-106425 x+237160000 = 0
x^2-106425 x = -237160000
x^2-106425 x+11326280625/4 = 10377640625/4
(x-106425/2)^2 = 10377640625/4
x-106425/2 = (1375*sqrt(5489))/2 or x-106425/2 = -(1375*sqrt(5489))/2
x = 106425/2+(1375*sqrt(5489))/2 or x-106425/2 = -(1375*sqrt(5489))/2
x = 106425/2+(1375*sqrt(5489))/2 or x = 106425/2-(1375*sqrt(5489))/2
sqrt(x)/4+x/1100 => 1/4 sqrt(106425/2-(1375*sqrt(5489))/2)+(106425/2-(1375 sqrt(5489))/2)/1100 = 1/4*sqrt(106425/2-(1375*sqrt(5489))/2)+(106425/2-(1375*sqrt(5489))/2)/1100 ~~ 14
So this solution is correct. You can do the other one and you will see it's incorrect. Therefore, the solution is:
x = 106425/2-(1375*sqrt(5489))/2 You don't need me to finish it off, use your calculator.