Question 195713
First, we'll use the area of a rectangle formula {{{A=LW}}}. Also, since "its length is 6 meters greater than its width", this means that {{{L=W+6}}}



{{{A=LW}}} Start with the given equation.



{{{55=(W+6)W}}} Plug in {{{L=W+6}}}



{{{55=W(W+6)}}} Rearrange the terms.



{{{55=W^2+6W}}} Distribute



{{{0=W^2+6W-55}}} Subtract 55 from both sides.



Notice we have a quadratic equation in the form of {{{0=aW^2+bW+c}}} where {{{a=1}}}, {{{b=6}}}, and {{{c=-55}}}



Let's use the quadratic formula to solve for W



{{{W = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{W = (-(6) +- sqrt( (6)^2-4(1)(-55) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=6}}}, and {{{c=-55}}}



{{{W = (-6 +- sqrt( 36-4(1)(-55) ))/(2(1))}}} Square {{{6}}} to get {{{36}}}. 



{{{W = (-6 +- sqrt( 36--220 ))/(2(1))}}} Multiply {{{4(1)(-55)}}} to get {{{-220}}}



{{{W = (-6 +- sqrt( 36+220 ))/(2(1))}}} Rewrite {{{sqrt(36--220)}}} as {{{sqrt(36+220)}}}



{{{W = (-6 +- sqrt( 256 ))/(2(1))}}} Add {{{36}}} to {{{220}}} to get {{{256}}}



{{{W = (-6 +- sqrt( 256 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{W = (-6 +- 16)/(2)}}} Take the square root of {{{256}}} to get {{{16}}}. 



{{{W = (-6 + 16)/(2)}}} or {{{W = (-6 - 16)/(2)}}} Break up the expression. 



{{{W = (10)/(2)}}} or {{{W =  (-22)/(2)}}} Combine like terms. 



{{{W = 5}}} or {{{W = -11}}} Simplify. 



So the <i>possible</i> answers are {{{W = 5}}} or {{{W = -11}}}



However, a negative width doesn't make much sense. So this means we'll ignore {{{W = -11}}}



So the solution is {{{W = 5}}} which makes the width 5 meters.



{{{L=W+6}}} Go back to the second equation

  
  
{{{L=5+6}}} Plug in {{{z = 5}}}



{{{L=11}}} Add



So the length is 11 meters.