Question Number 72628 by Rio Michael last updated on 30/Oct/19
$${solve}\:{the}\:{inequality}\: \\ $$$$\:\:{log}_{\mathrm{3}} \left(\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{9}{x}\:+\:\mathrm{9}\right)\:<\:\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 30/Oct/19
![⇒((ln(2x^2 +9x +9))/(ln3))<0 ⇒ln(2x^2 +9x+9)<0 (3>1 ⇒ln(3)>0) we must have 2x^2 +9x +9>0 Δ=81−4.2.9 =81−72=9 ⇒x_1 =((−9+3)/4)=−(3/2) x_2 =((−9−3)/4)=−3 the equation is defined on ]−∞,−3[∪]−(3/2),+∞[ (e) ⇒2x^2 +9x +9<1 ⇒2x^2 +9x +8<0 Δ=81−4.2.8 =81−64 =17 ⇒t_1 =((−9+(√(17)))/4) and t_2 =((−9−(√(17)))/4) t_1 −(−3) =((−9+(√(27)))/4) +3 >0 ⇒t_1 >−3 t_1 −(−(3/2))=((−9+(√(27)))/4)+(3/2) =((−9+(√(27))+6 =(√(27))−3)/4)>0 ⇒t_1 >−(3/2) its clear thst t_2 <−3 x −∞ t_2 −3 −(3/2) t_1 +∞ x^2 +9x+8 + − − − + the set of so<ution is ]((−9−(√(17)))/4),−3[[∪]−(3/2),((−9+(√(17)))/4)[](https://www.tinkutara.com/question/Q72635.png)
$$\Rightarrow\frac{{ln}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{9}{x}\:+\mathrm{9}\right)}{{ln}\mathrm{3}}<\mathrm{0}\:\:\:\:\:\:\:\:\Rightarrow{ln}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{9}{x}+\mathrm{9}\right)<\mathrm{0}\:\:\:\left(\mathrm{3}>\mathrm{1}\:\Rightarrow{ln}\left(\mathrm{3}\right)>\mathrm{0}\right) \\ $$$${we}\:{must}\:{have}\:\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{9}{x}\:+\mathrm{9}>\mathrm{0}\: \\ $$$$\Delta=\mathrm{81}−\mathrm{4}.\mathrm{2}.\mathrm{9}\:=\mathrm{81}−\mathrm{72}=\mathrm{9}\:\Rightarrow{x}_{\mathrm{1}} =\frac{−\mathrm{9}+\mathrm{3}}{\mathrm{4}}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left.{x}_{\mathrm{2}} =\frac{−\mathrm{9}−\mathrm{3}}{\mathrm{4}}=−\mathrm{3}\:\:\:{the}\:{equation}\:{is}\:{defined}\:{on}\:\right]−\infty,−\mathrm{3}\left[\cup\right]−\frac{\mathrm{3}}{\mathrm{2}},+\infty\left[\right. \\ $$$$\left({e}\right)\:\Rightarrow\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{9}{x}\:+\mathrm{9}<\mathrm{1}\:\Rightarrow\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{9}{x}\:+\mathrm{8}<\mathrm{0} \\ $$$$\Delta=\mathrm{81}−\mathrm{4}.\mathrm{2}.\mathrm{8}\:=\mathrm{81}−\mathrm{64}\:=\mathrm{17}\:\Rightarrow{t}_{\mathrm{1}} =\frac{−\mathrm{9}+\sqrt{\mathrm{17}}}{\mathrm{4}}\:{and}\:{t}_{\mathrm{2}} =\frac{−\mathrm{9}−\sqrt{\mathrm{17}}}{\mathrm{4}} \\ $$$${t}_{\mathrm{1}} −\left(−\mathrm{3}\right)\:=\frac{−\mathrm{9}+\sqrt{\mathrm{27}}}{\mathrm{4}}\:+\mathrm{3}\:>\mathrm{0}\:\Rightarrow{t}_{\mathrm{1}} >−\mathrm{3} \\ $$$${t}_{\mathrm{1}} −\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{−\mathrm{9}+\sqrt{\mathrm{27}}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}\:=\frac{−\mathrm{9}+\sqrt{\mathrm{27}}+\mathrm{6}\:=\sqrt{\mathrm{27}}−\mathrm{3}}{\mathrm{4}}>\mathrm{0}\:\Rightarrow{t}_{\mathrm{1}} >−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${its}\:{clear}\:{thst}\:{t}_{\mathrm{2}} <−\mathrm{3} \\ $$$${x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:{t}_{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:{t}_{\mathrm{1}} \:\:\:\:\:\:\:\:\:\:+\infty \\ $$$${x}^{\mathrm{2}} +\mathrm{9}{x}+\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$$$ \\ $$$$\left.{the}\:{set}\:{of}\:{so}<{ution}\:{is}\:\:\:\right]\frac{−\mathrm{9}−\sqrt{\mathrm{17}}}{\mathrm{4}},−\mathrm{3}\left[\left[\cup\right]−\frac{\mathrm{3}}{\mathrm{2}},\frac{−\mathrm{9}+\sqrt{\mathrm{17}}}{\mathrm{4}}\left[\right.\right. \\ $$
Commented by Rio Michael last updated on 31/Oct/19
$${thanks}\:{bt}\:{you}\:{didnt}\:{solve}\:{for}\:\:\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{9}{x}\:+\:\mathrm{9}\:<\:\mathrm{3}^{\mathrm{2}} \\ $$