Question 547553
If the length of a side (in inches) is {{{x}}},
the perimeter of the square (in inches) is {{{4x}}},
and the area (in square inches) is {{{x^2}}}.
The wording says that {{{4x}}} is 3 more than {{{x^2}}}.
That means {{{4x=x^2+3}}} <--> {{{x^2-4x+3=0}}} <--> {{{(x-1)(x-3)=0}}}, which gives you two solutions: {{{x=1}}} and {{{x=3}}}.
A square of side 1 inch has an area of 1 square inch and a perimeter of 4 inches, with 4 being 3 more than 1.
A square of side 3 inches has an area of 9 square inch and a perimeter of 12 inches, with 12 being 3 more than 9.
If you graph {{{perimeter=4x}}} and {{{area+3=x^2+3}}}, you find that they cross at two points.
{{{graph(200,300,-1,4,-3,15,4x,x^2+3)}}}