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Question Number 199654 by hardmath last updated on 06/Nov/23

$${\mathrm{x}}^4+{4\mathrm{x}}^3-8\mathrm{x}+1-\mathrm{m}=0$$$$\mathrm{has}4\mathrm{real}\mathrm{roots}$$$$\mathrm{find}\mathrm{m}=?$$

Answered by Frix last updated on 07/Nov/23

$$\mathrm{This}\mathrm{can}\mathrm{be}\mathrm{factorized}\mathrm{and}\mathrm{thus}\mathrm{solved}$$$$x=-1\pm \sqrt{3\pm \sqrt{m+3}}\left[4\mathrm{combinations}\mathrm{of}\mathrm{signs}\right]$$$$\Rightarrow$$$$m+3>0\wedge 3-\sqrt{m+3}>0$$$$\Rightarrow$$$$-3<m<6$$

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