algebra and trigonometry

Question

if the arithmetic mean of two positive quantities a and b is three times their geometric mean, find the value of a/b in surd form.

Answer 1

say, the two quantities are. x & y

(x+y)/2=a

 

3*sqrt(x*y)= b

 

Solving the above two equations

(x^2)+(y^2)=4(a^2)-2*(b^2)/9

 

(x-y)^2=4*(a^2)-2*(b^2)/9

 

(x+y)/(x-y)=2a/sqrt((4*(a^2)-4*(b^2)/9)

 

x/y =[2a+sqrt((4*(a^2)-4*(b^2)/9)]/[2a-sqrt((4*(a^2)-4*     (b^2)/9)]

Well, I solved the question finding x/y We need to find a/b a/b = ((x+y)/2)/(3*sqrt(xy))

Answer 2

Set up the equation:

AM(a,b) = 3GM(a,b)

 

(a+b)/2 = 3√(ab)

 

(a+b)/√ab = 6

 

√(a/b) + √(b/a) = 6

 

Let x = √(a/b):

 

x+1/x = 6

 

x²-6x+1 = 0

 

x²-6x+9 = 8

 

(x-3)² = 8

 

x-3 = ±2√2

 

x = 3±2√2

 

a/b = x² = (3±2√2)² = 3²±2·3(2√2)+(2√2)² = 9±12√2+8 = 17±12√2

Answer 3

You are told that:

 

(a+b)/2=3sqrt(ab)

 

This equation can be rewritten as a quadratic equation in the variable x=(a/b) as follows.

 

(a+b)=6sqrt(ab)

 

a2+2ab+b2=36ab

 

a2-34ab+b2=0

 

Divide by b2

 

(a/b)2-34(a/b)+1=0

 

x2-34x+1=0

 

Use the quadratic solution formula to obtain:

 

x=(a/b)=17±sqrt(1152)/4

 

(a/b)=17±√288

 

Since √288=16.971, both roots are positive, hence can correspond to positive a and b.

 

(a/b)2