Question

The equation $x^2+2(m-1)x+m+5=0$ has at least one +ive root. Determine the range for m.

Answer 1

Atleast one +ive root means :

(i) Both the roots are real ( There cannot be one complex root as they occur in conjugate pairs complex root as they occur in conjugate pairs

(ii) Both are not -ive.

Condition (i)

$\Rightarrow \Delta \ge 0$ or $4{(m-1)}^2-4(m+5)\ge 0$

or $m^2-2m+1-m-5\ge 0$

or $m^2-3m-4=+ive$ or $(m+1)(m-4)=+ive$ $\therefore m\le -1,m\ge 4\cdots \left(1\right)$

Both are -ive.$\therefore S=$ -ive

and $P=$ +ive$-2(m-1)=$-ive and$m+5=+ive$ $\therefore m>1orm>-5$ When $m>1,$ it is automatically greater than -5.

Hence $m>1$ is the condition for both the roots to be-ive . Hence when both the roots are not -ive i.e., at least one is +ive , we must have $m>1$ ...(2)

Plotting both (1) and (2) on real line and then taking their intersection i.e,. common region we get the answer : Condition (1) gives broken line region. Condition (2) gives dotted line region. Common region is given by $m\le -1.$ $\therefore mϵ(-\infty ,-1).$