solve the inequality log_3 (2x^2 + 9x + 9)


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Question Number 72628 by Rio Michael last updated on 30/Oct/19

$s o l v e t h e i n e q u a l i t y$ ${l o g}_{3} (2 x^{2} + 9 x + 9) < 0$
Commented bymathmax by abdo last updated on 30/Oct/19

$\Rightarrow \frac{l n (2 x^{2} + 9 x + 9)}{l n 3} < 0 \Rightarrow l n (2 x^{2} + 9 x + 9) < 0 (3 > 1 \Rightarrow l n (3) > 0)$ $w e m u s t h a v e 2 x^{2} + 9 x + 9 > 0$ $Δ = 81 - 4.2.9 = 81 - 72 = 9 \Rightarrow x_{1} = \frac{- 9 + 3}{4} = - \frac{3}{2}$ $x_{2} = \frac{- 9 - 3}{4} = - 3 t h e e q u a t i o n i s d e f i n e d o n] - \infty, - 3 [\cup] - \frac{3}{2}, + \infty [$ $(e) \Rightarrow 2 x^{2} + 9 x + 9 < 1 \Rightarrow 2 x^{2} + 9 x + 8 < 0$ $Δ = 81 - 4.2.8 = 81 - 64 = 17 \Rightarrow t_{1} = \frac{- 9 + \sqrt{17}}{4} a n d t_{2} = \frac{- 9 - \sqrt{17}}{4}$ $t_{1} - (- 3) = \frac{- 9 + \sqrt{27}}{4} + 3 > 0 \Rightarrow t_{1} > - 3$ $t_{1} - (- \frac{3}{2}) = \frac{- 9 + \sqrt{27}}{4} + \frac{3}{2} = \frac{- 9 + \sqrt{27} + 6 = \sqrt{27} - 3}{4} > 0 \Rightarrow t_{1} > - \frac{3}{2}$ $i t s c l e a r t h s t t_{2} < - 3$ $x - \infty t_{2} - 3 - \frac{3}{2} t_{1} + \infty$ $x^{2} + 9 x + 8 + - - - +$ $t h e s e t o f s o < u t i o n i s] \frac{- 9 - \sqrt{17}}{4}, - 3 [[\cup] - \frac{3}{2}, \frac{- 9 + \sqrt{17}}{4} [$
Commented byRio Michael last updated on 31/Oct/19

$t h a n k s b t y o u d i d n t s o l v e f o r 2 x^{2} + 9 x + 9 < 3^{2}$
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