Question

Consider the quadratic equation $(1+m)x^2-2(1+3m)x+(1+8m)=0$, (where $m\in R-\{-1\})$, then the set of values of ${}^{\prime }{m}^{\prime }$ such that the given quadratic equation has both roots positive are,

• A. $(\infty ,-1)\cup (3,\infty )$
• B. $(-\infty ,-1]\cup [3,\infty )$
• C. $(-\infty ,-1)\cup [3,\infty )$
• D. None of the above

Answer 1

The correct option is C $(-\infty ,-1)\cup [3,\infty )$

Given quadratic equation,

$(1+m)x^2-2(1+3m)x+(1+8m)=0$

Let $\alpha$ and $\beta$ be the roots of the equation,
$\alpha +\beta =\frac{2(1+3m)}{1+m}$
and $\alpha \beta =\frac{1+8m}{1+m}$
Condition for both roots to be positive is
$\alpha +\beta >0$ , $\alpha \beta >0$ and $D\ge 0$
$\Rightarrow \frac{1+3m}{1+m}>0$ and $\frac{1+8m}{1+m}>0$ and $4m^2-12m\ge 0$

$(1+3m)\frac{1}{(1+m)}>0$ and $(1+8m)\frac{1}{(1+m)}>0$ and $4m(m-3)\ge 0$
$\therefore m\in \left\{\right(-\infty ,-1)\cup (-\frac{1}{3},\infty )\}\cap \left\{\right(-\infty ,-1)\cup (-\frac{1}{8},\infty )\}\cap \{m\in (-\infty ,0]\cup [3,\infty \left)\right\}$

$\Rightarrow m\in (-\infty ,-1)\cup [3,\infty )$