cs101200
14.04.2020 •
Mathematics
Solve for x if 2(5x + 2)2 = 48. x = StartFraction negative 2 + 2 StartRoot 24 EndRoot Over 5 EndFraction and x = StartFraction negative 2 minus 2 StartRoot 24 EndRoot Over 5 EndFraction x = StartFraction negative 2 + 2 StartRoot 24 EndRoot Over 2 EndFraction and x = StartFraction negative 2 minus 2 StartRoot 24 EndRoot Over 2 EndFraction x = StartFraction negative 2 + 2 StartRoot 6 EndRoot Over 5 EndFraction and x = StartFraction negative 2 minus 2 StartRoot 6 EndRoot Over 5 EndFraction x = StartFraction negative 2 + 2 StartRoot 6 EndRoot Over 2 EndFraction and x = StartFraction negative 2 minus 2 StartRoot 6 EndRoot Over 2 EndFraction
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Ответ:
x = (-2 ±2sqrt(6))/5
Step-by-step explanation:
2(5x + 2)^2 = 48.
Divide each side by 2
2/2(5x + 2)^2 = 48/2
(5x + 2)^2 = 24
Take the square root of each side
sqrt((5x + 2)^2) = ±sqrt(24)
(5x + 2) = ±sqrt(6*4)
5x+2 = ±sqrt(6)sqrt(4)
5x+2 = ±2sqrt(6)
Subtract 2 from each side
5x+2-2 =-2 ±2sqrt(6)
5x =-2 ±2sqrt(6)
Divide each side by 5
5x/5 =(-2 ±2sqrt(6))/5
x = (-2 ±2sqrt(6))/5
Ответ: