Question 163887
Area of original square:


{{{A=s^2}}} where A is the area and s is the side length



Since "the sides of a square are lengthened by 6cm the area becomes 169cm^2", this means that we replace A with 169 and s with {{{s+6}}} to form the area of the enlarged square. So we get:


{{{169=(s+6)^2}}}



{{{169=s^2+12s+36}}} FOIL



{{{0=s^2+12s+36-169}}} Subtract 169 from both sides



{{{0=s^2+12s-133}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{as^2+bs+c}}} where {{{a=1}}}, {{{b=12}}}, and {{{c=-133}}}



Let's use the quadratic formula to solve for s



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



{{{s = (-(12) +- sqrt( (12)^2-4(1)(-133) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=12}}}, and {{{c=-133}}}



{{{s = (-12 +- sqrt( 144-4(1)(-133) ))/(2(1))}}} Square {{{12}}} to get {{{144}}}. 



{{{s = (-12 +- sqrt( 144--532 ))/(2(1))}}} Multiply {{{4(1)(-133)}}} to get {{{-532}}}



{{{s = (-12 +- sqrt( 144+532 ))/(2(1))}}} Rewrite {{{sqrt(144--532)}}} as {{{sqrt(144+532)}}}



{{{s = (-12 +- sqrt( 676 ))/(2(1))}}} Add {{{144}}} to {{{532}}} to get {{{676}}}



{{{s = (-12 +- sqrt( 676 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{s = (-12 +- 26)/(2)}}} Take the square root of {{{676}}} to get {{{26}}}. 



{{{s = (-12 + 26)/(2)}}} or {{{s = (-12 - 26)/(2)}}} Break up the expression. 



{{{s = (14)/(2)}}} or {{{s =  (-38)/(2)}}} Combine like terms. 



{{{s = 7}}} or {{{s = -19}}} Simplify. 



So the possible answers are {{{s = 7}}} or {{{s = -19}}} 


  
However, since a negative length isn't possible, this means that the only answer is {{{s=7}}}