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Question Number 29820 by Tinkutara last updated on 12/Feb/18

Commented by ajfour last updated on 12/Feb/18

$$\left(1\right)inG.P.$$

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

$${ax}^2+2bx+c=0]\times d$$$${dx}^2+2ex+f=0]\times a$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$2x(bd-ae)=af-cd$$$$x=\frac{af-cd}{2(bd-ae)}=\frac{ac(\frac{f}{c}-\frac{d}{a})}{2ab(\frac{d}{a}-\frac{e}{b})}$$$$As\frac{d}{a},\frac{e}{b},\frac{f}{c}areinA.P.with$$$$commondifferenceD\left(say\right)$$$$sox=\frac{ac\left(2D\right)}{2ab(-D)}=-\frac{c}{b}$$$$Andas{ax}^2+2bx+c=0$$$$\frac{{ac}^2}{b^2}+2b\left(\frac{-c}{b}\right)+c=0$$$$\Rightarrow {ac}^2-2b^{2}c+b^{2}c=0$$$$\Rightarrow ac=b^2,orc=0.$$

Commented by Tinkutara last updated on 13/Feb/18

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Commented by rahul 19 last updated on 24/Feb/18

$$\mathrm{wow}!$$

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