PLEASE answer my question which is quite a challenging one?

Question

If the arithmetic mean between a and b is twice as great as the geometric mean, show that
a : b = 2 + /3 : 2 - /3
/ stands for square root

Answer 1

arithmetic mean = (a+b)/2
geometric mean = sqrt ab

as per given condition (using sqrt as short form of sqare root)

(a+b)/2 = 2 sqrt ab
rewriting above eqn

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

applying componendo and dividendo

a+b+2*sqrt (ab)/a+b -2*sqrt (ab) = 3/1

(sqrt a + sqrt b)^2/(sqrt a - sqrt b)^2 = 3/1

hence
(sqrt a + sqrt b)/(sqrt a - sqrt b) = sqrt 3

again doing componendo and dividendo
we get

sqrt a / sqrt b = sqrt 3 + 1/sqrt 3 - 1

now sqaring both sides

a/b= (3+1+2*sqrt 3)/(3+1-2*sqrt 3)
a/b = (2+sqrt 3) /(2-sqrt 3)

done!!!

Answer 2

If I want to double /3, it could get mistaken for 2/3. Therefore I shall use SQRT ("squirt") to stand for square root.

The arithmetic mean is (a+b)/2. The geometric mean is sqrt(ab). Double this, and the givens is that you get the arithmetic mean; thus, 2*sqrt(ab) = (a + b)/2. Square both sides to get 4ab = (a^2 + 2ab + b^2)/4. Multiply by 4, subtract the "ab" term on the left side, and get something of the form, 0 = ta^2 + uab + vb^2, or equivalently, ta^2 + uab + vb^2 = 0, with t, u, and v as constants. Solve for a, using the quadratic formula for Ax^2 + Bx + C = 0, where A = t, B = ub and C = vb^2. You should get something of the form a = something * b. Divide by b, and if the something isn't of the form (2 + sqrt 3)/(2 - sqrt 3), then show that they are the same, by multiplying the numerator & denominator by 2 - sqrt 3.

Answer 3

So you have

(a+b)/2 = 2sqrt(ab)
---> a + b = 4sqrt(ab)
---> (a+b)² = 16ab
---> a² + 2ab + b² = 16ab
---> a² - 14ab + b² = 0
---> a²/b² - 14a/b + 1 = 0 (dividing by b² throughout)

This is a quadratic equation for a/b which yields the solutions
a/b = [14 ± sqrt192] / 2
---> a/b = 7 ± 4sqrt3.
The two solutions just reflect the fact that either a or b could be the bigger number. Taking a as the bigger number, we'll use the solution that is bigger than 1, i.e.
a/b = 7 + 4sqrt3
= (7+4sqrt3)(2-sqrt3) / (2-sqrt3)
= (14 - 7sqrt3 + 8sqrt3 - 12) / (2-sqrt3)
= (2 + sqrt3) / (2-sqrt3).
QED.

Answer 4

(a+b)/2 = 2*sqrt ab===> a^2 +b^+2ab = 16ab
dividing both sides by b^2
(a/b)^2 +1 +2*a/b=16*(a/b) and calling a/b=x

x^2 -14x+1=0===> x= ( 14+-sqrt (192))/2 but sqrt 192 =8sqrt3

so x= 7+-4sqrt 3 let´s see (2+sqrt3)/(2-sqrt3) =(2+sqrt3)^2/(4-3)= 7+4sqrt 3 which is the value of x taking+ sign

Answer 5

Here is how:

The AM is (a+b)/2; the GM is sqrt ab

(a+b)/2 = 2 sqrt ab

a + b = 4 sqrt (ab)

(a + b)^2 = 16 ab

a^2 + 2ab + b^2 = 16ab

a^2 - 14ab = -b^2

Complete the square

(a - 7b)^2 = -b^2 + 49b^2

(a - 7b)^2 = 48b^2

a - 7b = b sqrt 48

a - 7b = 4b sqrt 3

a = 4b sqrt 3 + 7b

a = b(4 sqrt 3 + 7)

a/b = 4 sqrt 3 + 7


but (2 + sqrt 3)/(2 - sqrt 3) = 4 sqrt 3 + 7 rationalize the denominator)

so they are equal

Hope that helps

Answer 6

well,
let me tell u a simple one.
2+√3
-------
2-√3
now,rationalise.
the rationalising factor for 2-√3 is 2+√3 so,multiplying both denominator and numerator with this rationalisinf factor,u dont change anything in the prob so.

(2+√3)(2+√3)/ (2-√3)(2+√3)
= 4+3+4√3/ 4-3
(using (a+b)^2 formula and (a-b)(a+b)=a^2-b^2) formula)
=7+4√3

hope this helps u...

Answer 7

2+/3=1/2-/3.
if a/b=2+/3/2-/3=1/2-/3\2+/3=1/1=1
a+b/2#/a*/b
hence the desired proof is impossible if it is wrong please send the correct one to vaishu_meridian_princess

Answer 8

a=7
b=2/3
7+2/3
7+2/3=2+/3:2-/3

Answer 9

hold down key. type in 251. Release key. result: √

Answer 10

it is 7+2/3. Hope I helped. and to the rest of u answerers : DO NOT COPY FROM ME!