Question

For what values of $m$, the equation$(1+m)x^2-2(1+3m)x+(1+8m)=0$ has $(m\in R)$ at least one negative root.

• A. $m\in (-\infty ,0)\cup (3,\infty )$
• B. $m\in (-\infty ,-1)\cup (3,\infty )$
• C. $m\in (-1,\frac{1}{8})$
• D. $m\in (-1,-\frac{1}{8})$

Answer 1

The correct option is D $m\in (-1,-\frac{1}{8})$

$(1+m)x^2-2(1+3m)x+(1+8m)=0$
We need to find the values of $m$ for which the above equation has at-least one negative root.

Lets consider that the above equation has both non-negative roots, then the condition for both roots are non-negative is "Products of roots $\ge 0$" and "Sum of roots $\ge 0$"

(i) Product of roots $\ge 0$
$\Rightarrow \frac{1+8m}{1+m}\ge 0$
$\Rightarrow m\in (\infty ,-1)\cup [\frac{-1}{8},\infty )$
(ii) Sum of roots $\ge 0$
$\frac{2(1+3m)}{(1+m)}\ge 0$
$\Rightarrow m\in (\infty ,-1)\cup [\frac{-1}{3},\infty )$

Taking intersection of (i) and (ii), we get
$m\in (\infty ,-1)\cup [-\frac{1}{8},\infty )$

Now for the final values of $m$ for which the equation have at-least one negative root, is

$m\in R-\text{intersection of (i) and (ii)}$

$m\in R-(\infty ,-1)\cup [-\frac{1}{8},\infty )$

$\Rightarrow m\in [-1,-\frac{1}{8})$

Checking at the corner values, we get that $m=-1$ does not satisfies the equation conditions.

$\therefore m\in (-1,-\frac{1}{8})$

Hence, option D.