If the set of all a R-1, for which the roots of the equation (1-a)x^2+2(a-3)x+9=0 are posi…

Question

If the set of all $a\in R-{1}$, for which the roots of the equation $(1-a){x}^{2}+2(a-3)x+9=0$ are positive is $(-\infty,-\alpha ]\cup [\beta ,\gamma )$, then $2\alpha +\beta +\gamma $ is equal to $\,$ .

Answer: 7

Accepted Answer

$\begin{gathered} f(x)=(1-a){x}^{2}+2(a-3)x+9,f(0)=9>0 \ D\ge 0\Rightarrow 4(a-3{)}^{2}\ge 4(1-a)\cdot 9 \end{gathered}$

$\Rightarrow a\in (-\infty,-3]\cup [0,\infty)\,$

$\begin{gathered} {x}_{1}+{x}_{2}=\frac{-2(a-3)}{1-a},{x}_{1}{x}_{2}=\frac{9}{1-a} \ {x}_{1}+{x}_{2}>0\Rightarrow \frac{a-3}{a-1}>0\Rightarrow a\in (-\infty,1)\cup (3,\infty)\ldots \end{gathered}$

${x}_{1}{x}_{2}>0\Rightarrow 1-a>0\Rightarrow a\in (-\infty,1)\ldots ($ iii $)$

$\begin{gathered} \Rightarrow \ a\in (-\infty,-3]\cup [0,1) \ \Rightarrow \, \alpha =3,\beta =0,\gamma =1\Rightarrow 2\alpha +\beta +\gamma =7 \end{gathered}$

Solutions · 1