Question 213295
Write 2 equations:
1 for A-B and another for B-C
A-B
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{{{d = r*t}}}
(1) {{{120 = r*t}}}
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B-C
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{{{d = r*t}}}
(2) {{{150 = (r + 10)(t + .1)}}} (note that 6 min = .1 hrs)
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{{{150 = rt + 10t + .1r + 1}}}
And substituting from (1)
{{{150 = 120 + 10t + .1r + 1}}}
{{{30 - 1 = 10t + .1r}}}
{{{.1r + 10t = 29}}}
Substituting from (1) again:
{{{t = 120/r}}}
{{{.1r + 10*120/r = 29}}}
{{{.1r^2 + 1200 = 29r}}}
{{{10r^2 + 12000 = 290r}}}
{{{10r^2 - 290r + 12000 = 0}}}
{{{r^2 - 29r + 1200 = 0}}}
Solve by completing the square:
{{{r^2 - 29r + (29/2)^2 = (29/2)^2 - 1200}}}
{{{r^2 - 29r + 210.25 = 210.25 - 1200}}}
{{{r - 14.5)^2 = -989.75}}}
I'm stuck- If I take the square root
of both sides, I get an imaginary on the right.
Can you see where I goofed?