The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 − 12x + 9. If the sum of the perimeters of both squares is 64, what is the value of x?	Pick:  A) 0  B) 2  C) 4.8  D) 4.67  E) 10
spotting the pattern of equations both are in form of (X+4.8)^2 so   A1= (x+5)^2A2= (2x-3)^2   L1= x+5L2= 2x-3   P1 = 4( x+5)P2=4(2x-3)   P1+P2=64   4( x+5) +4(2x-3)=64............&gt; X=4.8   Answer: 4.8
<result>C</result>
The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 − 12x + 9. If the sum of the perimeters of both squares is 64, what is the value of x?