Question 1048957
What was the area of the original - did you mean to say, RECTANGLE?


{{{L=w+8}}}
{{{wL=A}}}, original area


Changed rectangle, {{{(L-9)(3w)=(3/2)A}}}, the 50% area increase.
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{{{3wL-27w=(3/2)A}}}
{{{3A-27w=(3/2)A}}}
{{{3A-(3/2)A=27w}}}
{{{A-(A/2)=27w}}}
{{{A/2=27w}}}
{{{highlight_green(A=54w)}}}-------steps might be not the best path, but this equation is still useful.


Re-substitute for A.
{{{wL=54w}}}
{{{(L+8)L=54(L-8)}}}------substituted for w;
{{{L^2+8L=54L-54*8}}}
{{{L^2+8L-54L+54*8=0}}}
{{{L^2-46L+8*54=0}}}
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Maybe the constant term is better factorable.
{{{8*54=2*2*2*27*2=2*2*2*2*3*3*3}}}-----or maybe this not too useful yet.

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{{{L=(46+- sqrt(46^2-4*8*54))/2}}}-----general solution of quadratic formula;
{{{L=(46+- sqrt(388))/2}}}
{{{L=L=46+- sqrt(4*97))/2}}}
{{{L=(46+- 2*sqrt(97))/2}}}
{{{L=23+- sqrt(97)}}}---------97 is a prime number.


Approx, {{{L=23+- 9.849}}}, and {{{w=23+- 9.849-8}}}



If taking the PLUS form of L,
{{{system(L=32.9849,w=24.849)}}}
Original area, 820


If taking the MINUS form of L,
{{{system(L=13.151,w=5.151)}}}
Original area, 68