Sylvia, let x be the side of the original square and y be the area. We start with two equations based on what we are given:

x^{2} = y

(x+3)^{2} = 4y

Now we substitute x^{2 }in for y in the second equation and get into standard quadratic form:

(x+3)^{2 }= 4x^{2}

(x+3)(x+3) = 4x^{2}

x^{2 }+ 6x + 9 = 4x^{2}

3x^{2 }- 6x - 9 = 0

We factor and solve the quadratic:

(3x + 3)(x - 3) = 0

x= -1,3

Since we can't have a negative length, only one answer makes sense

x = 3 in

Hope this helps!