Question Number 32382 by hizmzm1 last updated on 24/Mar/18
$$\mathrm{If}\:\mathrm{the}\:\mathrm{equation}\:{ax}^{\mathrm{2}} +\mathrm{2}{bx}−\mathrm{3}{c}=\mathrm{0}\:\mathrm{has} \\ $$$$\mathrm{no}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{and}\:\left(\frac{\mathrm{3}{c}}{\mathrm{4}}\right)<\:{a}+{b},\:\mathrm{then} \\ $$
Commented by MJS last updated on 24/Mar/18
$$\mathrm{I}\:\mathrm{think}\:\mathrm{that}'\mathrm{s}\:\mathrm{not}\:\mathrm{enough}\:\mathrm{information} \\ $$
Commented by MJS last updated on 24/Mar/18
![x^2 +((2b)/a)x−((3c)/a)=0 exactly one root: (x−r)^2 =x^2 +((2b)/a)x−((3c)/a) I. −2r=((2b)/a) ⇒ r^2 =(b^2 /a^2 ) II. r^2 =−((3c)/a) ⇒ −((3c)/a)=(b^2 /a^2 ) ⇒ c=−(b^2 /(3a)) but now we have 1 root. It disappears when we add a real number >0 ⇒ ⇒ c>−(b^2 /(3a)) ((3c)/4)<a+b ⇒ c<(4/3)(a+b) −(b^2 /(3a))<c<(4/3)(a+b) [⇒ −(b^2 /(3a))<(4/3)(a+b) case 1: a<0 −b^2 >4a(a+b) −4a^2 −4ab−b^2 >0 4a^2 +4ab+b^2 <0 (2a+b)^2 <0 ⇒ no solution case 2: a>0 leads to (2a+b)^2 >0 ⇒ a>0]](https://www.tinkutara.com/question/Q32410.png)
$${x}^{\mathrm{2}} +\frac{\mathrm{2}{b}}{{a}}{x}−\frac{\mathrm{3}{c}}{{a}}=\mathrm{0} \\ $$$$\mathrm{exactly}\:\mathrm{one}\:\mathrm{root}: \\ $$$$\left({x}−{r}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\frac{\mathrm{2}{b}}{{a}}{x}−\frac{\mathrm{3}{c}}{{a}} \\ $$$$\mathrm{I}.\:−\mathrm{2}{r}=\frac{\mathrm{2}{b}}{{a}}\:\Rightarrow\:{r}^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$\mathrm{II}.\:{r}^{\mathrm{2}} =−\frac{\mathrm{3}{c}}{{a}} \\ $$$$\Rightarrow\:−\frac{\mathrm{3}{c}}{{a}}=\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:\Rightarrow\:{c}=−\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}} \\ $$$$\mathrm{but}\:\mathrm{now}\:\mathrm{we}\:\mathrm{have}\:\mathrm{1}\:\mathrm{root}.\:\mathrm{It} \\ $$$$\mathrm{disappears}\:\mathrm{when}\:\mathrm{we}\:\mathrm{add}\:\mathrm{a} \\ $$$$\mathrm{real}\:\mathrm{number}\:>\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:{c}>−\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}} \\ $$$$ \\ $$$$\frac{\mathrm{3}{c}}{\mathrm{4}}<{a}+{b}\:\Rightarrow\:{c}<\frac{\mathrm{4}}{\mathrm{3}}\left({a}+{b}\right) \\ $$$$ \\ $$$$−\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}}<{c}<\frac{\mathrm{4}}{\mathrm{3}}\left({a}+{b}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\left[\Rightarrow\:−\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}}<\frac{\mathrm{4}}{\mathrm{3}}\left({a}+{b}\right)\right. \\ $$$$\:\:\:\:\:\mathrm{case}\:\mathrm{1}:\:{a}<\mathrm{0} \\ $$$$\:\:\:\:\:−{b}^{\mathrm{2}} >\mathrm{4}{a}\left({a}+{b}\right) \\ $$$$\:\:\:\:\:−\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{ab}−{b}^{\mathrm{2}} >\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{ab}+{b}^{\mathrm{2}} <\mathrm{0} \\ $$$$\:\:\:\:\:\left(\mathrm{2}{a}+{b}\right)^{\mathrm{2}} <\mathrm{0}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$$$\:\:\:\:\:\mathrm{case}\:\mathrm{2}:\:{a}>\mathrm{0}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\:\:\:\:\:\left(\mathrm{2}{a}+{b}\right)^{\mathrm{2}} >\mathrm{0} \\ $$$$\left.\:\:\:\:\:\Rightarrow\:{a}>\mathrm{0}\right] \\ $$
Answered by Joel578 last updated on 24/Mar/18
$$\mathrm{no}\:\mathrm{real}\:\mathrm{roots}\:\rightarrow\:{D}\:<\:\mathrm{0} \\ $$$$\left(\mathrm{2}{b}\right)^{\mathrm{2}} \:−\:\mathrm{4}{a}\left(−\mathrm{3}{c}\right)\:<\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}{b}^{\mathrm{2}} \:+\:\mathrm{12}{ac}\:<\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}^{\mathrm{2}} \:>\:\mathrm{3}{ac} \\ $$$$ \\ $$$$\:\:\:\mathrm{3}{c}\:<\:\mathrm{4}\left({a}\:+\:{b}\right) \\ $$$$\mathrm{3}{ac}\:<\:\mathrm{4}{a}^{\mathrm{2}} \:+\:\mathrm{4}{ab} \\ $$$$\mathrm{3}{ac}\:<\:{b}^{\mathrm{2}} \\ $$
Commented by MJS last updated on 24/Mar/18
$$\mathrm{3}{c}<\mathrm{4}\left({a}+{b}\right)\:\Rightarrow\:\mathrm{3}{ac}<\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{ab}\:\mathrm{with}\:{a}>\mathrm{0} \\ $$$$\mathrm{3}{c}>\mathrm{4}\left({a}+{b}\right)\:\Rightarrow\:\mathrm{3}{ac}>\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{ab}\:\mathrm{with}\:{a}<\mathrm{0} \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{show}\:{a}>\mathrm{0}? \\ $$
Answered by saru53424@gmail.com last updated on 24/Mar/18
$$ \\ $$$${thanks} \\ $$