Quadratic EquationMedium
Question
Let the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{\text{ }b}$ be $\frac{5}{16},a > 2$. If $\alpha$ is such that $a,4,\alpha,\text{ }b$ are in A.P., then the equation $\alpha x^{2} - ax + 2(\alpha - 2b) = 0$ has:
Options
A.One root in $(1,4)$ and another in $( - 2,0)$
B.One root in $(0,2)$ and another in $( - 4, - 2)$
C.Complex roots of magnitude less than 2
D.Both roots in the interval $( - 2,0)$
Solution
$a = 4 - d,\alpha = 4 + d,b = 4 + 2\text{ }d$
$${\Rightarrow (4 + d)x^{2} - (4 - d)x + 2(4 + d - 8 - 4\text{ }d) = 0 }{\Rightarrow (4 + d)x^{2} - (4 - d)x + 2( - 4 - 3\text{ }d) = 0 }$$Also $\frac{\frac{1}{a} + \frac{1}{\text{ }b}}{2} = \frac{5}{16}$
$${\Rightarrow \frac{\frac{1}{4 - d} + \frac{1}{4 + 2\text{ }d}}{2} = \frac{5}{16} }{\Rightarrow d = 2 }$$Equation becomes
$${6x^{2} - 2x - 20 = 0 }{3x^{2} - x - 10 = 0 }{x = 2,\frac{- 5}{3}}$$
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