For what values of m, the equation (1+m)x^2 −2(1+3m)x+(1+8m)=0 ; m ∈ R , has both roots positive ?


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Question Number 17359 by ajfour last updated on 04/Jul/17

$For what values of m, the equation$ $(1 + m) x^{2} - 2 (1 + 3 m) x + (1 + 8 m) = 0;$ $m \in R, has both roots positive ?$
Commented by ajfour last updated on 04/Jul/17

${book}^{'} s answer :$ $m \in (- \infty, - 1) \cup [3, \infty)$ $my answer differed . .$
Commented by ajfour last updated on 04/Jul/17

$thanks Sir . very precise .$
Commented by prakash jain last updated on 04/Jul/17
$roots are ((−b)/(2a))±((√(b^2 −4ac))/(2a)) −(b/(2a)) lies in between roots condition for +ve roots>0 −(b/(2a))≥0,αβ>0,D≥0 (((1+3m))/((1+m)))≥0⇒m∈(−∞,−1)∪[−(1/3),∞) (((1+8m))/((1+m)))>0⇒m∈(−∞,−1)∪(−(1/8),∞) D=4{(1+3m)^2 −(1+m)(1+8m)}≥0 ⇒m(m−3)≥0⇒m∈(−∞,0]∪[3,∞) answer m∈(−∞,−1)∪(−(1/8),0]∪[3,∞)$
$roots are \frac{- b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a}$ $- \frac{b}{2 a} lies in between roots$ $condition for + ve roots > 0$ $- \frac{b}{2 a} ⩾ 0, α β > 0, D ⩾ 0$ $\frac{(1 + 3 m)}{(1 + m)} ⩾ 0 \Rightarrow m \in (- \infty, - 1) \cup [- \frac{1}{3}, \infty)$ $\frac{(1 + 8 m)}{(1 + m)} > 0 \Rightarrow m \in (- \infty, - 1) \cup (- \frac{1}{8}, \infty)$ $D = 4 {{(1 + 3 m)}^{2} - (1 + m) (1 + 8 m)} ⩾ 0$ $\Rightarrow m (m - 3) ⩾ 0 \Rightarrow m \in (- \infty, 0] \cup [3, \infty)$ $a n s w e r$ $m \in (- \infty, - 1) \cup (- \frac{1}{8}, 0] \cup [3, \infty)$
Commented by Tinkutara last updated on 05/Jul/17

$ajfour Sir, which book is that ? Can you$ $please tell its description so that I can$ $prepare well from it ?$
Commented by ajfour last updated on 05/Jul/17

$S . K . {Goyal}^{'} s Algebra, Arihant$ $Prakashan .$
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