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Question Number 23208 by Joel577 last updated on 27/Oct/17

$$\mathrm{Is}\mathrm{it}\mathrm{possible}\mathrm{to}\mathrm{find}\mathrm{how}\mathrm{many}\mathrm{real}\mathrm{roots}$$$$\mathrm{exist}\mathrm{in}\mathrm{the}\mathrm{equation}$$$$x^4+\mid x\mid =3$$$$\mathrm{without}\mathrm{find}\mathrm{all}\mathrm{the}\mathrm{value}\mathrm{of}x?$$

Answered by mrW1 last updated on 28/Oct/17

$${\mathrm{x}}^4+\mid \mathrm{x}\mid ⩾0$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4+\mid \mathrm{x}\mid \mathrm{is}\mathrm{symmetric},\mathrm{i}.\mathrm{e}.$$$$\mathrm{f}(-\mathrm{x})=\mathrm{f}\left(\mathrm{x}\right)$$$$\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{for}\mathrm{x}>0$$$$\mathrm{and}\mathrm{strictly}\mathrm{decreasing}\mathrm{for}\mathrm{x}<0$$$$\mathrm{therefore}$$$${\mathrm{x}}^4+\mid \mathrm{x}\mid =\mathrm{a}(\mathrm{a}<0)\mathrm{has}\mathrm{no}\mathrm{real}\mathrm{root}$$$${\mathrm{x}}^4+\mid \mathrm{x}\mid =0\mathrm{has}\mathrm{one}\mathrm{real}\mathrm{root},\mathrm{x}=0$$$${\mathrm{x}}^4+\mid \mathrm{x}\mid =\mathrm{a}(\mathrm{a}>0)\mathrm{has}\mathrm{two}\mathrm{real}\mathrm{roots}$$

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