Question 548378
Draw this.  It's a square inside a larger square.  Call the lengths of the sides of the inner square, the picture, x.  Call the lengths of the sides of the larger square (the square made by the frame) x+4 because the 2 inch frame adds 2 inches on each side to the smaller square.<P>
Area of a square is {{{s^2}}}
For the smaller square that's {{{x^2}}}
For the larger square that's {{{(x+4)^2 = x^2 + 8x +16}}}
The problem indicates that the area of the small square is 1/3 the area of the large square.  In other words, three times the area of the small square equals the area of the large square.<P>
{{{3x^2 = x^2 + 8x + 16}}}
Move what's on the right to the left of the = by subtracting it from both sides.<P>
{{{2x^2 - 8x - 16 = 0}}}
Divide both sides by 2.
{{{x^2 - 4x - 8 = 0}}}
Factor by using the quadratic formula {{{(-b +- sqrt(b^2 -4ac))/2a}}}
{{{(4 +- sqrt((-4)^2 -4*1*-8))/(2*1) = (4 +- sqrt(16+32))/2=(4 +- sqrt(48))/2=(4 +- sqrt(3*16))/2=(4 +- 4sqrt(3))/2=2 +- 2sqrt(3)}}}

That gives two possible answers:

{{{x=2-2sqrt(3) <= 0}}}  The length cannot be negative so discard that.
{{{x=2+2sqrt(3) = 5.46 = 5.5}}}
Round it to the nearest quarter inch, so .46 rounds up to .5.<P>
Check if {{{3*5.5^2 = 9.5^2}}}. 
{{{90.75 = 90.25}}} Yes, it's close enough given we had to round up to the nearest quarter inch.
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