Q. If ,G, and H be the A.M;G.M; and H.M. respectively of two distinct positive integers,then the equation `Ax^2-|G|x - H=0` has:-

A) both roots as fractions

B) atleast one root as a negative fraction

C) exactly one positive root

D) atleast one root as integer

### 2 Answers | Add Yours

Given `Ax^2-|G|x-H=0` where A,G,H are the arithmetic mean, geometric mean, and the harmonic mean of two positive integers:

I. Note that `H=G^2/A` ; we can use the quadratic formula to look at the roots:

`Ax^2-|G|x-H=0 ==>`

`x=(G+-sqrt(G^2-4(A)(G^2/A)))/(2A)`

`=(G+-sqrt(5G^2))/(2A)`

`=(G(1+-sqrt(5)))/(2A)`

Now A,G>0. `1+sqrt(5)>0,1-sqrt(5)=0` so there is exactly 1 positive root, and 1 negative root.

**The answer is C.** (Typically fractions are defined as a ratio of integers; the roots here will never be rational so A,B, and D cannot be correct.)

II. An alternative: let a,b be positive integers. Then `A=(a+b)/2,G=sqrt(ab),H=(2ab)/(a+b)` so we have:

`(a+b)/2 x^2-sqrt(ab)x-(2ab)/(a+b)=0` Use the quadratic formula:

`x=(sqrt(ab)+-sqrt(ab-4((a+b)/2)((-2ab)/(a+b))))/(2((a+b)/2))`

`=(sqrt(ab)+-sqrt(5ab))/(a+b)`

`=(sqrt(ab)(1+-sqrt(5)))/(a+b)`

Again, `sqrt(ab)>0,a+b>0,1+sqrt(5)>0,1-sqrt(5)<0` so we have a positive and negative root, neither of which is rational.

An example: let a=6, b=8 then A=7, `G=4sqrt(3),H=48/7`

The roots are `x=(4sqrt(3)+-4sqrt(15))/14=(4sqrt(3)(1+-sqrt(5)))/14`

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