Question 1200265
<font color=black size=3>
9 hrs 36 min = 9.6 hrs
since 36 min = 36/60 = 0.6 hrs


x = speed of the boat in still water
where x > 7


When going upstream, the boat is slowed down to a speed of x-7 km/hr.
When going downstream, the boat is sped up to a speed of x+7 km/hr.


Upstream:
distance = rate*time
9 = (x-7)*t1
t1 = 9/(x-7)
It takes 9/(x-7) hours to row upstream.


Downstream:
distance = rate*time
9 = (x+7)*t2
t2 = 9/(x+7)
It takes 9/(x+7) hours to row downstream.


The two time values (t1 and t2) must add to the total 9.6 hours mentioned earlier.
{{{t[1]+t[2] = 9.6}}}


{{{9/(x-7)+9/(x+7) = 9.6}}}


{{{9(x+7)+9(x-7) = 9.6(x-7)(x+7)}}} Multiply both sides by the LCD (x-7)(x+7) to clear out the fractions.


{{{9(x+7)+9(x-7) = 9.6(x^2-49)}}}


{{{9x+63+9x-63 = 9.6x^2-470.4}}}


{{{18x = 9.6x^2-470.4}}}


{{{0 = 9.6x^2-470.4-18x}}}


{{{9.6x^2-18x-470.4 = 0}}}


{{{96x^2-180x-4704 = 0}}}


{{{12(8x^2-15x-392) = 0}}}


{{{8x^2-15x-392 = 0}}}
We have an equation in the form ax^2+bx+c = 0
a = 8
b = -15
c = -392
Plug those into the quadratic formula.
{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-(-15)+-sqrt((-15)^2-4(8)(-392)))/(2(8))}}}


{{{x = (15+-sqrt(12769))/(16)}}}


{{{x = (15+-  113)/(16)}}}


{{{x = (15+113)/(16)}}} or {{{x = (15-113)/(16)}}}


{{{x = (128)/(16)}}} or  {{{x = (-98)/(16)}}}


{{{x = 8}}} or  {{{x = -6.125}}}
A negative speed value doesn't make sense, so we ignore {{{x = -6.125}}}
The only practical answer is {{{x = 8}}}



Answer: The speed of the boat in still water is 8 km/hr.
</font>