Let the arithmetic mean of 1a and 1b be 516, with a2.

Question

Let the arithmetic mean of $\dfrac1a$ and $\dfrac1b$ be $\dfrac{5}{16}$, with $a>2$. If $\alpha$ is such that $a, 4, \alpha, b$ are in A.P., then the equation $\alpha x^2 - ax + 2(\alpha - 2b) = 0$ has:

Accepted Answer

AM condition: $\dfrac1a+\dfrac1b = \dfrac58 \Rightarrow \dfrac{a+b}{ab} = \dfrac58 \Rightarrow 8(a+b)=5ab. \quad (i)$

Since $a,4,\alpha,b$ are in A.P. with common difference $d$: $4=a+d\Rightarrow d=4-a$, so $\alpha = a+2d = 8-a$ and $b = a+3d = 12-2a$.

Substituting into (i): $8(12-a) = 5a(12-2a) \Rightarrow 10a^2-68a+96=0 \Rightarrow 5a^2-34a+48=0$, giving $a = 4.8$ or $a=2$ (rejected since $a>2$).

So $a=4.8$, $\alpha = 3.2$, $b=2.4$. The equation becomes
$$3.2x^2 - 4.8x + 2(3.2-4.8) = 0 \implies 2x^2-3x-2=0 \implies (2x+1)(x-2)=0,$$
giving roots $x=2 \in (1,4)$ and $x=-\tfrac12 \in (-2,0)$.

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