The area of a rectangle is 2 square feet. The perimeter of the rectangle is 9 ft. Find the length, l, and width, w, of the rectangle. What is l^2+w^2

Area of rectangle(A) = Length(l) *Width(w)
ie, 2=l*w
l=2/w this will be equation no 1.
Perimeter of rectangle(P) = 2*width(w) + 2* Length(l)
ie, 9=2l+2w this will be equation no 2.
By substituting equation 1 in equation 2 we will get
2*(2/w)+2w=9
4/w+2w=9
(4+2w^2)/w=9
4+2w^2=9w
2w^2-9w+4=0
w^2-(9/2)w+2=0
by using quadratic formula we will get,
w=1/2,4
let w=1/2 then l=2/(1/2)=4 (answer for the first question)
w^2=1/4, l^2=16
l^2+w^2=16+1/4 =65/4 (or) 16.25 (answer for the second question)

If the area of a rectangle is 2 feet, and its perimeter is 9 feet, we want to use the perimeter and area equations, as follows:
Area(a)= length(l) * width(w)
Perimeter(p) = 2*length(l)+2*width(w)

Because we have two variables (l and w) and two equations, we can use one equation to solve for l in terms of w, and then plug in that equation to the second equation, in order that we end up with one equation and one variable, as follows:

A=l*w =2
P=2l +2w = 9; l=(9-2w)/2
Thus, l*w=((9-2w)/2)*w = 2
We can re-write this equation in standard form quadratic:
-2w^2+9w-4 =0. This yields solutions (using the "quadratic formula," explained at the link below), w =.5, and w = 4. Whichever answer we select for the width (w) w will determine that the other answer is the length (l). Thus, if width (w) = .5, then length (l) = 4. It follows that w^2 = 1/4, and l^2=16.
https://www.purplemath.com/modules/quadform.htm