Question

Let $A,G$ and $H$ be the arithmetic mean, geometric mean and harmonic mean, respectively of two distinct positive real numbers. If $\alpha$ is the smallest of the two roots of the equation $A(G-H)x^2+G(H-A)x+H(A-G)=0$, then

• A. $-2<\alpha <-1$
• B. $0<\alpha <1$
• C. $-1<\alpha <0$
• D. $1<\alpha <2$

Answer 1

Answer: (B)

$0<\alpha <1$

Consider two positive real numbers $a$ and $b$.

Then, $A=\frac{a+b}{2}$, $G=\sqrt{ab}$ and $H=\frac{2ab}{a+b}$.

From that we can conclude that $AH=G^2$

Now consider the given equation,

One of the root is $1$.

Given the other roots is $\alpha$.

Product of roots $=$ $\alpha$ $=$ $\frac{H(A-G)}{A(G-H)}$.

Now substitute $H=\frac{G^2}{A}$ in the equation of $\alpha$.

$\Rightarrow$ $\alpha =\frac{G(A-G)}{A(A-G)}$

$\Rightarrow$ $\alpha =\frac{G}{A}$

We know that $A>G>H$ (Because the numbers are distinct and positive)

$\Rightarrow$ $0<\frac{G}{A}<1$

$\Rightarrow$ $0<\alpha <1$

Therefore, option B is correct.