Question 149494

The area of any rectangle is {{{A=W*L}}}. Since the "length 8 inches longer than its width", this tells us that {{{L=W+8}}}



{{{A=W*L}}} Start with the given area equation.



{{{84=W(W+8)}}} Plug in {{{A=84}}} (the given area) and {{{L=W+8}}}



{{{84=W^2+8W}}} Distribute


{{{0=W^2+8W-84}}} Subtract 84 from both sides.



Let's use the quadratic formula to solve for W



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



{{{W = (-(8) +- sqrt( (8)^2-4(1)(-84) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=8}}}, and {{{c=-84}}}



{{{W= (-8 +- sqrt( 64-4(1)(-84) ))/(2(1))}}} Square {{{8}}} to get {{{64}}}. 



{{{W = (-8 +- sqrt( 64--336 ))/(2(1))}}} Multiply {{{4(1)(-84)}}} to get {{{-336}}}



{{{W = (-8 +- sqrt( 64+336 ))/(2(1))}}} Rewrite {{{sqrt(64--336)}}} as {{{sqrt(64+336)}}}



{{{W = (-8 +- sqrt( 400 ))/(2(1))}}} Add {{{64}}} to {{{336}}} to get {{{400}}}



{{{w = (-8 +- sqrt( 400 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{W = (-8 +- 20)/(2)}}} Take the square root of {{{400}}} to get {{{20}}}. 



{{{W = (-8 + 20)/(2)}}} or {{{W = (-8 - 20)/(2)}}} Break up the expression. 



{{{W = (12)/(2)}}} or {{{W =  (-28)/(2)}}} Combine like terms. 



{{{W = 6}}} or {{{W = -14}}} Simplify. 



Since a negative width is not possible, this means that the only solution is {{{W = 6}}}


So the width is 6 inches


{{{L=W+8}}} Go back to the second equation.



{{{L=6+8}}} Plug in {{{W = 6}}}



{{{L=14}}} Add


So the length is 14 inches




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Answer:



So the canvas is 6 in by 14 in