Question Number 29820 by Tinkutara last updated on 12/Feb/18
Commented by ajfour last updated on 12/Feb/18
$$\left(\mathrm{1}\right)\:{in}\:{G}.{P}. \\ $$
Commented by math solver last updated on 13/Feb/18
$$\mathrm{yes}\:!\:\mathrm{they}\:\mathrm{will}\:\mathrm{be}\:\:\mathrm{in}\:\mathrm{g}.\mathrm{p}. \\ $$
Commented by 803jaideep@gmail.com last updated on 13/Feb/18
$$\mathrm{yes}\:\mathrm{i}\:\mathrm{also}\:\mathrm{think}\:\mathrm{gp} \\ $$
Answered by ajfour last updated on 13/Feb/18
![ax^2 +2bx+c=0 ]×d dx^2 +2ex+f=0 ]×a _____________ 2x(bd−ae)= af−cd x=((af−cd)/(2(bd−ae))) = ((ac((f/c)−(d/a)) )/(2ab((d/a)−(e/b)))) As (d/a), (e/b), (f/c) are in A.P. with common difference D (say) so x = ((ac(2D))/(2ab(−D))) = −(c/b) And as ax^2 +2bx+c=0 ((ac^2 )/b^2 )+2b(((−c)/b))+c=0 ⇒ ac^2 −2b^2 c+b^2 c =0 ⇒ ac = b^2 , or c=0 .](https://www.tinkutara.com/question/Q29871.png)
$$\left.{ax}^{\mathrm{2}} +\mathrm{2}{bx}+{c}=\mathrm{0}\:\right]×{d} \\ $$$$\left.{dx}^{\mathrm{2}} +\mathrm{2}{ex}+{f}=\mathrm{0}\:\right]×{a} \\ $$$$\:\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\mathrm{2}{x}\left({bd}−{ae}\right)=\:{af}−{cd} \\ $$$${x}=\frac{{af}−{cd}}{\mathrm{2}\left({bd}−{ae}\right)}\:=\:\frac{{ac}\left(\frac{{f}}{{c}}−\frac{{d}}{{a}}\right)\:}{\mathrm{2}{ab}\left(\frac{{d}}{{a}}−\frac{{e}}{{b}}\right)}\: \\ $$$${As}\:\:\:\frac{{d}}{{a}},\:\frac{{e}}{{b}},\:\frac{{f}}{{c}}\:\:{are}\:{in}\:{A}.{P}.\:{with} \\ $$$${common}\:{difference}\:{D}\:\:\left({say}\right) \\ $$$$\:\:\:{so}\:\:\:{x}\:=\:\frac{{ac}\left(\mathrm{2}{D}\right)}{\mathrm{2}{ab}\left(−{D}\right)}\:=\:−\frac{{c}}{{b}} \\ $$$${And}\:{as}\:\:{ax}^{\mathrm{2}} +\mathrm{2}{bx}+{c}=\mathrm{0} \\ $$$$\:\:\:\:\:\frac{{ac}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\mathrm{2}{b}\left(\frac{−{c}}{{b}}\right)+{c}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\:{ac}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} {c}+{b}^{\mathrm{2}} {c}\:=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\:\:\:{ac}\:=\:{b}^{\mathrm{2}} \:\:\:,\:\:{or}\:\:{c}=\mathrm{0}\:\:. \\ $$
Commented by Tinkutara last updated on 13/Feb/18
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Commented by rahul 19 last updated on 24/Feb/18
$$\mathrm{wow}! \\ $$