Find three numbers in the ratio 2:3:5, the sum of whose squares is 608

1.	Find three numbers in the ratio 2:3:5, the sum of whose squares is 608
Answer» Step-by-step explanation: LET the three numbersa, b, c, be 2X, 3x, 5x. we know that (2x) ^2+ (3x) ^2 + (5x)^2= 608 4x^2 + 9x^2 + 25x^2 = 608 38x^2 = 608 x^2= 608/38 x^2=16 x= √16 x= 4 a= 2x = 24=8 b=3x= 34=12 c = 5x= 5*4=20 Therefore, the three numbers are 8,12,20

Answer»

Step-by-step explanation:

LET the three numbersa, b, c, be 2X, 3x, 5x.

we know that

(2x) ^2+ (3x) ^2 + (5x)^2= 608

4x^2 + 9x^2 + 25x^2 = 608

38x^2 = 608

x^2= 608/38

x^2=16

x= √16

x= 4

a= 2x = 2*4=8

b=3x= 3*4=12

c = 5x= 5*4=20

Therefore, the three numbers are 8,12,20

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