
Justification:
Given: 
Split the middle term;

Factor:



QUESTION 3a
The given expression for the area of the rectangle is
.
This is equal to the indicated area which is 
This implies that




This implies that, the dimensions of the rectangle are equal;
Using the square root method, we obtain





We discard the negative value.
The side length of this rectangle is

QUESTION 3b
The given expression for the area of the rectangle is
.
This is equal to the indicated area which is 
This implies that





This implies that, the dimensions of the rectangle are equal;
Using the square root method, we obtain




We discard the negative value.
The side length of this rectangle is

QUESTION 3c
The given expression for the area of the rectangle is
.
This is equal to the indicated area which is 
This implies that




Applying the laws of indices gives;

This implies that;



The side length of this rectangle is

Dont forget that the square is also a rectangle.