Let the arithmetic mean of 1 a and 1 ~b be 5 16, a>2. If α is such that a, 4, α, ~b are in A.P., then the equation α x^2-a x+2(α-2 b)=0 has:

Given AM of $\frac{1}{a}$a1​ and $\frac{1}{b}$b1​ is $\frac{5}{16}$165​ with $a > 2$a>2. Since $a, 4, \alpha, b$a,4,α,b are in AP:

$$4 - a = \alpha - 4 = b - \alpha$$4−a=α−4=b−α

This gives $\alpha = 8 - a$α=8−a and $b = 12 - 2a$b=12−2a.

From AM condition: $\frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{5}{16}$2a1​+b1​​=165​ leads to $\frac{a+b}{ab} = \frac{5}{8}$aba+b​=85​

Substituting: $\frac{20-3a}{96-28a+2a^2} = \frac{5}{8}$96−28a+2a220−3a​=85​

$$160 - 24a = 480 - 140a + 10a^2$$160−24a=480−140a+10a2

$10a^2 - 116a + 320 = 0$10a2−116a+320=0 or $5a^2 - 58a + 160 = 0$5a2−58a+160=0

Using the quadratic formula gives $a = 4$a=4 or $a = 8$a=8.

For $a = 4$a=4: $\alpha = 4, b = 4$α=4,b=4.

The equation becomes $4x^2 - 4x - 8 = 0$4x2−4x−8=0 or $x^2 - x - 2 = 0$x2−x−2=0 with roots 2 and $-1$−1.

One root lies in $(1,4)$(1,4) and the other in $(-2,0)$(−2,0).