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Answer 1 · 2017-03-21T17:36:14

Length 11 miles and width 6 miles

If the width is taken to be x miles, its length would be x+6 and area would be x(x+6). Thus,

x(x+6)=55.

This can be solved for x by factorising the quadratic expression $x^{2} + 6 x - 55 = 0$ $x^2+6x-55=0$

(x+11)(x-5)=0. this gives x=5 or -11. Since x cannot have a negative value, it would be x=5

Width is 5 miles hence length would be 6more that is 11 miles.

Answer 2 · 2017-03-21T17:53:04

Length $= 11$ $=11$ miles,width $= 5$ $=5$ miles

Let the length be $x$ $x$

then the width will be $x - 6$ $x-6$

$x (x - 6) = 55$ $x(x-6)=55$

$x^{2} - 6 x = 55$ $x^2-6x=55$

$x^{2} - 6 x - 55 = 0$ $x^2-6x-55=0$

$(x + 5) (x - 11) = 0$ $(x+5)(x-11)=0$

$x = - 5$ $x=-5$ or $x = 11$ $x=11$

substitute $x = 11$ $x=11$

$11 - 6 = 5$ $11-6=5$ miles $= w i d t h$ $=width$

Length $= 11$ $=11$ miles