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Question Number 156652 by Tawa11 last updated on 13/Oct/21

$Sirs, please give me the general solutions to a quadratic ineqality .$ $(1) {ax}^{2} + bx + c > 0$ $(2) {ax}^{2} + bx + c ⩾ 0$ $(3) {ax}^{2} + bx + c < 0$ $(4) {ax}^{2} + bx + c ⩽ 0$
	Answered by MJS_new last updated on 14/Oct/21
	$f(x)=ax^2 +bx+c 1. solve the equation ax^2 +bx+c=0 ⇒ x=((−b±(√(b^2 −4ac)))/(2a)) 2. how many real solutions? 2.1. b^2 −4ac>0 ⇒ 2 real solutions x_1 <x_2 2.2. b^2 −4ac=0 ⇒ 1 real solution x_1 2.3. b^2 −4ac<0 ⇒ no real solution 3. check a 3.1. a<0 ⇒ ♮hangingε parabola 3.2. a=0 ⇒ straight line 3.3. a>0 ⇒ ♮standingε parabola ⇒ all together a<0∧b^2 −4ac>0∧x_1 <x_2 ⇒ { ((f(x)<0; x<x_1 ∨x>x_2 )),((f(x)=0; x=x_1 ∨x=x_2 )),((f(x)>0; x_1 <x<x_2 )) :} a<0∧b^2 −4ac=0 ⇒ { ((f(x)<0; x≠x_1 )),((f(x)=0; x=x_1 )) :} a<0∧b^2 −4ac<0 ⇒ f(x)<0 a=0∧b<0 ⇒ { ((f(x)<0; x>x_1 )),((f(x)=0; x=x_1 )),((f(x)>0; x<x_1 )) :} a=0∧b=0 ⇒ { ((f(x)<0; c<0)),((f(x)=0; c=0)),((f(x)>0; c>0)) :} a=0∧b>0 ⇒ { ((f(x)<0; x<x_1 )),((f(x)=0; x=x_1 )),((f(x)>0; x>x_1 )) :} a>0∧b^2 −4ac>0∧x_1 <x_2 ⇒ { ((f(x)<0; x_1 <x<x_2 )),((f(x)=0; x=x_1 ∨x=x_2 )),((f(x)>0; x<x_1 ∨x>x_2 )) :} a>0∧b^2 −4ac=0 ⇒ { ((f(x)=0; x=x_1 )),((f(x)>0; x≠x_1 )) :} a>0∧b^2 −4ac<0 ⇒ f(x)>0$
	$f (x) = {a x}^{2} + b x + c$ $1 . solve the equation$ ${a x}^{2} + b x + c = 0$ $\Rightarrow x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $2 . how many real solutions ?$ $2.1 . b^{2} - 4 a c > 0 \Rightarrow 2 real solutions x_{1} < x_{2}$ $2.2 . b^{2} - 4 a c = 0 \Rightarrow 1 real solution x_{1}$ $2.3 . b^{2} - 4 a c < 0 \Rightarrow no real solution$ $3 . check a$ $3.1 . a < 0 \Rightarrow ♮ hanging ε parabola$ $3.2 . a = 0 \Rightarrow straight line$ $3.3 . a > 0 \Rightarrow ♮ standing ε parabola$ $\Rightarrow$ $all together$ $a < 0 \land b^{2} - 4 a c > 0 \land x_{1} < x_{2} \Rightarrow {\begin{cases} f (x) < 0; x < x_{1} \lor x > x_{2} \\ f (x) = 0; x = x_{1} \lor x = x_{2} \\ f (x) > 0; x_{1} < x < x_{2} \end{cases}$ $a < 0 \land b^{2} - 4 a c = 0 \Rightarrow {\begin{cases} f (x) < 0; x \neq x_{1} \\ f (x) = 0; x = x_{1} \end{cases}$ $a < 0 \land b^{2} - 4 a c < 0 \Rightarrow f (x) < 0$ $a = 0 \land b < 0 \Rightarrow {\begin{cases} f (x) < 0; x > x_{1} \\ f (x) = 0; x = x_{1} \\ f (x) > 0; x < x_{1} \end{cases}$ $a = 0 \land b = 0 \Rightarrow {\begin{cases} f (x) < 0; c < 0 \\ f (x) = 0; c = 0 \\ f (x) > 0; c > 0 \end{cases}$ $a = 0 \land b > 0 \Rightarrow {\begin{cases} f (x) < 0; x < x_{1} \\ f (x) = 0; x = x_{1} \\ f (x) > 0; x > x_{1} \end{cases}$ $a > 0 \land b^{2} - 4 a c > 0 \land x_{1} < x_{2} \Rightarrow {\begin{cases} f (x) < 0; x_{1} < x < x_{2} \\ f (x) = 0; x = x_{1} \lor x = x_{2} \\ f (x) > 0; x < x_{1} \lor x > x_{2} \end{cases}$ $a > 0 \land b^{2} - 4 a c = 0 \Rightarrow {\begin{cases} f (x) = 0; x = x_{1} \\ f (x) > 0; x \neq x_{1} \end{cases}$ $a > 0 \land b^{2} - 4 a c < 0 \Rightarrow f (x) > 0$
	Commented byTawa11 last updated on 14/Oct/21

	$Wow, I really appreciate your time sir . God bless you sir .$
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