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Question Number 156652 by Tawa11 last updated on 13/Oct/21
Sirs, please give me the general solutions to a quadratic ineqality. (1) ax^2 + bx + c > 0 (2) ax^2 + bx + c ≥ 0 (3) ax^2 + bx + c < 0 (4) ax^2 + bx + c ≤ 0
$$\mathrm{Sirs},\:\mathrm{please}\:\mathrm{give}\:\mathrm{me}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solutions}\:\mathrm{to}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{ineqality}. \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:>\:\:\:\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:\geqslant\:\:\:\:\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:<\:\:\:\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:\leqslant\:\:\:\:\mathrm{0} \\ $$
Answered by MJS_new last updated on 14/Oct/21
f(x)=ax^2 +bx+c 1. solve the equation ax^2 +bx+c=0 ⇒ x=((−b±(√(b^2 −4ac)))/(2a)) 2. how many real solutions? 2.1. b^2 −4ac>0 ⇒ 2 real solutions x_1 <x_2 2.2. b^2 −4ac=0 ⇒ 1 real solution x_1 2.3. b^2 −4ac<0 ⇒ no real solution 3. check a 3.1. a<0 ⇒ “hanging” parabola 3.2. a=0 ⇒ straight line 3.3. a>0 ⇒ “standing” parabola ⇒ all together a<0∧b^2 −4ac>0∧x_1 <x_2 ⇒ { ((f(x)<0; x<x_1 ∨x>x_2 )),((f(x)=0; x=x_1 ∨x=x_2 )),((f(x)>0; x_1 <x<x_2 )) :} a<0∧b^2 −4ac=0 ⇒ { ((f(x)<0; x≠x_1 )),((f(x)=0; x=x_1 )) :} a<0∧b^2 −4ac<0 ⇒ f(x)<0 a=0∧b<0 ⇒ { ((f(x)<0; x>x_1 )),((f(x)=0; x=x_1 )),((f(x)>0; x<x_1 )) :} a=0∧b=0 ⇒ { ((f(x)<0; c<0)),((f(x)=0; c=0)),((f(x)>0; c>0)) :} a=0∧b>0 ⇒ { ((f(x)<0; x<x_1 )),((f(x)=0; x=x_1 )),((f(x)>0; x>x_1 )) :} a>0∧b^2 −4ac>0∧x_1 <x_2 ⇒ { ((f(x)<0; x_1 <x<x_2 )),((f(x)=0; x=x_1 ∨x=x_2 )),((f(x)>0; x<x_1 ∨x>x_2 )) :} a>0∧b^2 −4ac=0 ⇒ { ((f(x)=0; x=x_1 )),((f(x)>0; x≠x_1 )) :} a>0∧b^2 −4ac<0 ⇒ f(x)>0
$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\mathrm{1}.\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$$\mathrm{2}.\:\mathrm{how}\:\mathrm{many}\:\mathrm{real}\:\mathrm{solutions}? \\ $$$$\mathrm{2}.\mathrm{1}.\:{b}^{\mathrm{2}} −\mathrm{4}{ac}>\mathrm{0}\:\Rightarrow\:\mathrm{2}\:\mathrm{real}\:\mathrm{solutions}\:{x}_{\mathrm{1}} <{x}_{\mathrm{2}} \\ $$$$\mathrm{2}.\mathrm{2}.\:{b}^{\mathrm{2}} −\mathrm{4}{ac}=\mathrm{0}\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{solution}\:{x}_{\mathrm{1}} \\ $$$$\mathrm{2}.\mathrm{3}.\:{b}^{\mathrm{2}} −\mathrm{4}{ac}<\mathrm{0}\:\Rightarrow\:\mathrm{no}\:\mathrm{real}\:\mathrm{solution} \\ $$$$\mathrm{3}.\:\mathrm{check}\:{a} \\ $$$$\mathrm{3}.\mathrm{1}.\:{a}<\mathrm{0}\:\Rightarrow\:“\mathrm{hanging}''\:\mathrm{parabola} \\ $$$$\mathrm{3}.\mathrm{2}.\:{a}=\mathrm{0}\:\Rightarrow\:\mathrm{straight}\:\mathrm{line} \\ $$$$\mathrm{3}.\mathrm{3}.\:{a}>\mathrm{0}\:\Rightarrow\:“\mathrm{standing}''\:\mathrm{parabola} \\ $$$$\Rightarrow \\ $$$$\mathrm{all}\:\mathrm{together} \\ $$$${a}<\mathrm{0}\wedge{b}^{\mathrm{2}} −\mathrm{4}{ac}>\mathrm{0}\wedge{x}_{\mathrm{1}} <{x}_{\mathrm{2}} \:\Rightarrow\:\begin{cases}{{f}\left({x}\right)<\mathrm{0};\:{x}<{x}_{\mathrm{1}} \vee{x}>{x}_{\mathrm{2}} }\\{{f}\left({x}\right)=\mathrm{0};\:{x}={x}_{\mathrm{1}} \vee{x}={x}_{\mathrm{2}} }\\{{f}\left({x}\right)>\mathrm{0};\:{x}_{\mathrm{1}} <{x}<{x}_{\mathrm{2}} }\end{cases} \\ $$$${a}<\mathrm{0}\wedge{b}^{\mathrm{2}} −\mathrm{4}{ac}=\mathrm{0}\:\Rightarrow\:\begin{cases}{{f}\left({x}\right)<\mathrm{0};\:{x}\neq{x}_{\mathrm{1}} }\\{{f}\left({x}\right)=\mathrm{0};\:{x}={x}_{\mathrm{1}} }\end{cases} \\ $$$${a}<\mathrm{0}\wedge{b}^{\mathrm{2}} −\mathrm{4}{ac}<\mathrm{0}\:\Rightarrow\:{f}\left({x}\right)<\mathrm{0} \\ $$$${a}=\mathrm{0}\wedge{b}<\mathrm{0}\:\Rightarrow\:\begin{cases}{{f}\left({x}\right)<\mathrm{0};\:{x}>{x}_{\mathrm{1}} }\\{{f}\left({x}\right)=\mathrm{0};\:{x}={x}_{\mathrm{1}} }\\{{f}\left({x}\right)>\mathrm{0};\:{x}<{x}_{\mathrm{1}} }\end{cases} \\ $$$${a}=\mathrm{0}\wedge{b}=\mathrm{0}\:\Rightarrow\:\begin{cases}{{f}\left({x}\right)<\mathrm{0};\:{c}<\mathrm{0}}\\{{f}\left({x}\right)=\mathrm{0};\:{c}=\mathrm{0}}\\{{f}\left({x}\right)>\mathrm{0};\:{c}>\mathrm{0}}\end{cases} \\ $$$${a}=\mathrm{0}\wedge{b}>\mathrm{0}\:\Rightarrow\:\begin{cases}{{f}\left({x}\right)<\mathrm{0};\:{x}<{x}_{\mathrm{1}} }\\{{f}\left({x}\right)=\mathrm{0};\:{x}={x}_{\mathrm{1}} }\\{{f}\left({x}\right)>\mathrm{0};\:{x}>{x}_{\mathrm{1}} }\end{cases} \\ $$$${a}>\mathrm{0}\wedge{b}^{\mathrm{2}} −\mathrm{4}{ac}>\mathrm{0}\wedge{x}_{\mathrm{1}} <{x}_{\mathrm{2}} \:\Rightarrow\:\begin{cases}{{f}\left({x}\right)<\mathrm{0};\:{x}_{\mathrm{1}} <{x}<{x}_{\mathrm{2}} }\\{{f}\left({x}\right)=\mathrm{0};\:{x}={x}_{\mathrm{1}} \vee{x}={x}_{\mathrm{2}} }\\{{f}\left({x}\right)>\mathrm{0};\:{x}<{x}_{\mathrm{1}} \vee{x}>{x}_{\mathrm{2}} }\end{cases} \\ $$$${a}>\mathrm{0}\wedge{b}^{\mathrm{2}} −\mathrm{4}{ac}=\mathrm{0}\:\Rightarrow\:\begin{cases}{{f}\left({x}\right)=\mathrm{0};\:{x}={x}_{\mathrm{1}} }\\{{f}\left({x}\right)>\mathrm{0};\:{x}\neq{x}_{\mathrm{1}} }\end{cases} \\ $$$${a}>\mathrm{0}\wedge{b}^{\mathrm{2}} −\mathrm{4}{ac}<\mathrm{0}\:\Rightarrow\:{f}\left({x}\right)>\mathrm{0} \\ $$
Commented by Tawa11 last updated on 14/Oct/21
Wow, I really appreciate your time sir. God bless you sir.
$$\mathrm{Wow},\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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