The area of one square is x^2 + 6x + 9 and the area of another square is 4x^2 − 20x + 25. If the sum of the perimeters of both squares is 52, what is the value of x? Answers:	A) 3	B) 4	C) 5	D) 6	E) 7
The areas are (x+3)^2 and (2x-5)^2.   The lengths of the sides are x+3 and 2x-5.   If we add the two perimeters:   4(x+3) + 4(2x-5) = 52   12x = 60   x = 5   The answer is 5.
<result>C</result>
The area of one square is x^2 + 6x + 9 and the area of another square is 4x^2 − 20x + 25. If the sum of the perimeters of both squares is 52, what is the value of x?