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Question Number 93705 by i jagooll last updated on 14/May/20

$$\mathrm{find}\mathrm{real}\mathrm{q}\mathrm{so}\mathrm{that}{\mathrm{x}}^4-{40\mathrm{x}}^2+\mathrm{q}=0$$$$\mathrm{has}\mathrm{four}\mathrm{real}\mathrm{solution}\mathrm{forming}$$$$\mathrm{AP}.$$

Commented by PRITHWISH SEN 2 last updated on 14/May/20

$$\mathrm{let}\mathrm{the}\mathrm{roots}\mathrm{are}\mathbf{a}-3\mathbf{b},\mathbf{a}-\mathbf{b},\mathbf{a}+\mathbf{b},\mathbf{a}+3\mathbf{b}$$$$\therefore \mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{roots}$$$$4\mathbf{a}=0\Rightarrow \mathbf{a}=0$$$$\mathbf{and}\mathbf{the}\mathbf{product}\mathbf{of}\mathbf{two}\mathbf{roots}\mathbf{at}\mathbf{a}\mathbf{time}$$$$-10{\mathbf{b}}^2=40\Rightarrow \mathrm{b}=\pm 2$$$$\mathrm{the}\mathrm{roots}\mathrm{are}-6,-2,2,6$$$$\therefore \mathbf{q}=36\times 4=144.\mathbf{please}\mathbf{check}.$$

Commented by i jagooll last updated on 14/May/20

correct sir. but my way is different

Commented by PRITHWISH SEN 2 last updated on 14/May/20

$$\mathrm{then}\mathrm{post}\mathrm{it}.\mathrm{I}\mathrm{have}\mathrm{also}\mathrm{lots}\mathrm{to}\mathrm{learn}.$$

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