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 well

Algebra ->  Decimal-numbers -> 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 well       Log On


   



Question 1113670: 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 well in feet. For one particular well, the splash is heard 14 seconds after the stone is released. How deep (to the nearest foot) is the well?
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
14 = sqrt%28x%29%2F4 + x%2F1100


Let y = sqrt%28x%29 be new variable.  Then the equation takes the form


14 =  y%2F4 + y%5E2%2F1100.


Multiply both sides by 1100 to get 


y%5E2+%2B+275y+-+15400 = 0


y%5B1%2C2%5D = %28-275+%2B-+sqrt%28275%5E2+%2B+4%2A15400%29%29%2F2 = %28-250+%2B-+370.4%29%2F2.


Only positive root works  y = %28-250+%2B+370.4%29%2F2 = 47.7.


Then  x = 47.7%5E2 =  2275.3 ft.


Check.  sqrt%282275.3%29%2F4 + 2275.3%2F1100 = 14.


Answer.  The depth of the well is equal to  2275.3 ft.

Solved.