```Question 155378
# 3

Let x = original speed (ie slower speed)

{{{200=(x)t}}} Plug in {{{d=200}}} and {{{r=x}}}

{{{200/x=t}}} Divide both sides by "x" to isolate "t"

The statement "if he had gone 10mph faster, the trip would have taken 1 hour less" tells us that the new speed is {{{x+10}}} and the new time is {{{t-1}}}

{{{d=rt}}} Go back to the distance-rate-time formula

{{{200=(x+10)(t-1)}}} Plug in {{{d=200}}}, {{{r=x+10}}}, and replace {{{t}}} with {{{t-1}}}

{{{200=xt-x+10t-10}}} FOIL

{{{200=x(200/x)-x+10(200/x)-10}}} Plug in {{{t=200/x}}}

{{{200=200-x+2000/x-10}}} Multiply

{{{200x=200x-x^2+2000-10x}}} Multiply <b>every</b> term by the LCD "x" to clear the denominator

{{{0=200x-x^2+2000-10x-200x}}} Subtract 200x from both sides

{{{0=-x^2-10x+2000}}} Combine like terms

Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=-1}}}, {{{b=-10}}}, and {{{c=2000}}}

Let's use the quadratic formula to solve for x

{{{x = (-(-10) +- sqrt( (-10)^2-4(-1)(2000) ))/(2(-1))}}} Plug in  {{{a=-1}}}, {{{b=-10}}}, and {{{c=2000}}}

{{{x = (10 +- sqrt( (-10)^2-4(-1)(2000) ))/(2(-1))}}} Negate {{{-10}}} to get {{{10}}}.

{{{x = (10 +- sqrt( 100-4(-1)(2000) ))/(2(-1))}}} Square {{{-10}}} to get {{{100}}}.

{{{x = (10 +- sqrt( 100--8000 ))/(2(-1))}}} Multiply {{{4(-1)(2000)}}} to get {{{-8000}}}

{{{x = (10 +- sqrt( 100+8000 ))/(2(-1))}}} Rewrite {{{sqrt(100--8000)}}} as {{{sqrt(100+8000)}}}

{{{x = (10 +- sqrt( 8100 ))/(2(-1))}}} Add {{{100}}} to {{{8000}}} to get {{{8100}}}

{{{x = (10 +- sqrt( 8100 ))/(-2)}}} Multiply {{{2}}} and {{{-1}}} to get {{{-2}}}.

{{{x = (10 +- 90)/(-2)}}} Take the square root of {{{8100}}} to get {{{90}}}.

{{{x = (10 + 90)/(-2)}}} or {{{x = (10 - 90)/(-2)}}} Break up the expression.

{{{x = (100)/(-2)}}} or {{{x =  (-80)/(-2)}}} Combine like terms.

{{{x = -50}}} or {{{x = 40}}} Simplify.

So the possible answers are {{{x = -50}}} or {{{x = 40}}}

Since a negative speed doesn't make sense, this means that the only solution is {{{x=40}}}

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