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Question Number 32382 by hizmzm1 last updated on 24/Mar/18

If the equation ax^2 +2bx−3c=0 has  no real roots and (((3c)/4))< a+b, then

$$\mathrm{If}\:\mathrm{the}\:\mathrm{equation}\:{ax}^{\mathrm{2}} +\mathrm{2}{bx}−\mathrm{3}{c}=\mathrm{0}\:\mathrm{has} \\ $$ $$\mathrm{no}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{and}\:\left(\frac{\mathrm{3}{c}}{\mathrm{4}}\right)<\:{a}+{b},\:\mathrm{then} \\ $$

Commented byMJS last updated on 24/Mar/18

I think that′s not enough information

$$\mathrm{I}\:\mathrm{think}\:\mathrm{that}'\mathrm{s}\:\mathrm{not}\:\mathrm{enough}\:\mathrm{information} \\ $$

Commented byMJS 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]

$${x}^{\mathrm{2}} +\frac{\mathrm{2}{b}}{{a}}{x}−\frac{\mathrm{3}{c}}{{a}}=\mathrm{0} \\ $$ $$\mathrm{exactly}\:\mathrm{one}\:\mathrm{root}: \\ $$ $$\left({x}−{r}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\frac{\mathrm{2}{b}}{{a}}{x}−\frac{\mathrm{3}{c}}{{a}} \\ $$ $$\mathrm{I}.\:−\mathrm{2}{r}=\frac{\mathrm{2}{b}}{{a}}\:\Rightarrow\:{r}^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$ $$\mathrm{II}.\:{r}^{\mathrm{2}} =−\frac{\mathrm{3}{c}}{{a}} \\ $$ $$\Rightarrow\:−\frac{\mathrm{3}{c}}{{a}}=\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:\Rightarrow\:{c}=−\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}} \\ $$ $$\mathrm{but}\:\mathrm{now}\:\mathrm{we}\:\mathrm{have}\:\mathrm{1}\:\mathrm{root}.\:\mathrm{It} \\ $$ $$\mathrm{disappears}\:\mathrm{when}\:\mathrm{we}\:\mathrm{add}\:\mathrm{a} \\ $$ $$\mathrm{real}\:\mathrm{number}\:>\mathrm{0}\:\Rightarrow \\ $$ $$\Rightarrow\:{c}>−\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}} \\ $$ $$ \\ $$ $$\frac{\mathrm{3}{c}}{\mathrm{4}}<{a}+{b}\:\Rightarrow\:{c}<\frac{\mathrm{4}}{\mathrm{3}}\left({a}+{b}\right) \\ $$ $$ \\ $$ $$−\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}}<{c}<\frac{\mathrm{4}}{\mathrm{3}}\left({a}+{b}\right) \\ $$ $$ \\ $$ $$\:\:\:\:\:\left[\Rightarrow\:−\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}}<\frac{\mathrm{4}}{\mathrm{3}}\left({a}+{b}\right)\right. \\ $$ $$\:\:\:\:\:\mathrm{case}\:\mathrm{1}:\:{a}<\mathrm{0} \\ $$ $$\:\:\:\:\:−{b}^{\mathrm{2}} >\mathrm{4}{a}\left({a}+{b}\right) \\ $$ $$\:\:\:\:\:−\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{ab}−{b}^{\mathrm{2}} >\mathrm{0} \\ $$ $$\:\:\:\:\:\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{ab}+{b}^{\mathrm{2}} <\mathrm{0} \\ $$ $$\:\:\:\:\:\left(\mathrm{2}{a}+{b}\right)^{\mathrm{2}} <\mathrm{0}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$ $$\:\:\:\:\:\mathrm{case}\:\mathrm{2}:\:{a}>\mathrm{0}\:\mathrm{leads}\:\mathrm{to} \\ $$ $$\:\:\:\:\:\left(\mathrm{2}{a}+{b}\right)^{\mathrm{2}} >\mathrm{0} \\ $$ $$\left.\:\:\:\:\:\Rightarrow\:{a}>\mathrm{0}\right] \\ $$

Answered by Joel578 last updated on 24/Mar/18

no real roots → D < 0  (2b)^2  − 4a(−3c) < 0               4b^2  + 12ac < 0                                  b^2  > 3ac       3c < 4(a + b)  3ac < 4a^2  + 4ab  3ac < b^2

$$\mathrm{no}\:\mathrm{real}\:\mathrm{roots}\:\rightarrow\:{D}\:<\:\mathrm{0} \\ $$ $$\left(\mathrm{2}{b}\right)^{\mathrm{2}} \:−\:\mathrm{4}{a}\left(−\mathrm{3}{c}\right)\:<\:\mathrm{0} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}{b}^{\mathrm{2}} \:+\:\mathrm{12}{ac}\:<\:\mathrm{0} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}^{\mathrm{2}} \:>\:\mathrm{3}{ac} \\ $$ $$ \\ $$ $$\:\:\:\mathrm{3}{c}\:<\:\mathrm{4}\left({a}\:+\:{b}\right) \\ $$ $$\mathrm{3}{ac}\:<\:\mathrm{4}{a}^{\mathrm{2}} \:+\:\mathrm{4}{ab} \\ $$ $$\mathrm{3}{ac}\:<\:{b}^{\mathrm{2}} \\ $$

Commented byMJS last updated on 24/Mar/18

3c<4(a+b) ⇒ 3ac<4a^2 +4ab with a>0  3c>4(a+b) ⇒ 3ac>4a^2 +4ab with a<0  can you show a>0?

$$\mathrm{3}{c}<\mathrm{4}\left({a}+{b}\right)\:\Rightarrow\:\mathrm{3}{ac}<\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{ab}\:\mathrm{with}\:{a}>\mathrm{0} \\ $$ $$\mathrm{3}{c}>\mathrm{4}\left({a}+{b}\right)\:\Rightarrow\:\mathrm{3}{ac}>\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{ab}\:\mathrm{with}\:{a}<\mathrm{0} \\ $$ $$\mathrm{can}\:\mathrm{you}\:\mathrm{show}\:{a}>\mathrm{0}? \\ $$

Answered by saru53424@gmail.com last updated on 24/Mar/18

  thanks

$$ \\ $$ $${thanks} \\ $$

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