Question 120423
Let x=side length of original square


Remember the area of any square with side x is: {{{A=x*x=x^2}}}


Now if the side length is increased by 3, then the new side becomes: {{{x+3}}}



So the new area becomes: {{{A=(x+3)^2}}}



{{{A=(x+3)^2}}} Start with the given equation



{{{64=(x+3)^2}}} Plug in A=64 (this is the area of the new square)



{{{64=x^2+6x+9}}} Foil



{{{0=x^2+6x+9-64}}} Subtract 64 from both sides



{{{0=x^2+6x-55}}} Combine like terms




{{{(x+11)(x-5)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{x+11=0}}} or  {{{x-5=0}}} 


{{{x=-11}}} or  {{{x=5}}}    Now solve for x in each case



So our answer is 

 {{{x=-11}}} or  {{{x=5}}}  


However, since a negative length doesn't make sense, our only solution is {{{x=5}}}



So the original side length is 5 inches