Question

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

Answer 1

Answer: c

Explanation:

just did this

Answer 2

x = (-2 ±2sqrt(6))/5

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