Question 210931
Let {{{e}}}= distance travelled by the 
car going east when the car going north
has gone {{{24}}} mi
given:
Distance between the cars is
{{{3e + 4}}}
Using the pythagorean theorem, I can say
{{{24^2 + e^2 = (3e + 4)^2}}} 
{{{576 + e^2 = 9e^2 + 24e + 16}}}
{{{8e^2 + 24e - 560 = 0}}}
{{{e^2 + 3e - 70 = 0}}}
Using the quadratic formula:
{{{e = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = 1}}}
{{{b = 3}}}
{{{c = -70}}}
{{{e = (-3 +- sqrt( 3^2-4*1*(-70) ))/(2*1) }}}
{{{e = (-3 +- sqrt( 9 + 280 ))/2 }}}
{{{e = (-3 +- sqrt( 289 ))/2 }}}
{{{e = (-3 +- 17)/2 }}}
{{{e = (-3 - 17)/2}}} (Can't use this answer)
{{{e = (-3 + 17)/2}}}
{{{e = 14/2}}}
{{{e = 7}}}
and, since distance between the cars is
{{{3e + 4}}}
{{{3*7 + 4 = 25}}} mi
The distance between the cars is 25 mi
check:
{{{24^2 + 7^2 = 25^2}}}
{{{576 + 49 = 625}}}
{{{625 = 625}}}
OK