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a and b are real numbers and satisfy ab^2 = 27/5 and a^2 b = 1080. Compute a+5b.

 Apr 2, 2022
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ab^2 = 27/5....................(1)  

a^2 b = 1080..................(2)

 

From (2), we have: a ==sqrt [1080 / b]

Sub this into (1) above and solve for b:

 

Solve for b:
(6 sqrt(30))/(1/b)^(3/2) = 27/5

Take the reciprocal of both sides:
(1/b)^(3/2)/(6 sqrt(30)) = 5/27

Multiply both sides by 6 sqrt(30):
(1/b)^(3/2) = (10 sqrt(10/3))/3

Raise both sides to the power of two:
1/b^3 = 1000/27

Take the reciprocal of both sides:
b^3 = 27/1000

Take the cube root of both sides:

 

b ==3 / 10  and   a==60

a + 5b ==60 + 15/10 ==123/2 ==61.50

 Apr 2, 2022

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