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Question Number 204511 by Ghisom last updated on 20/Feb/24

$$\mathrm{solve}\mathrm{for}x$$$$x^2-10\lfloor x\rfloor +\frac{57}{4}=0$$

Answered by mr W last updated on 20/Feb/24

$$\lfloor x\rfloor =n\in {\mathbb{Z}}^{+}$$$$x=n+fwith0⩽f<1$$$${(n+f)}^2-10n+\frac{57}{4}=0$$$$10n-\frac{57}{4}={(n+f)}^2⩾n^2$$$$n^2-10n+\frac{57}{4}⩽0$$$$\Rightarrow n⩾5-\sqrt{25-\frac{57}{4}}\Rightarrow n⩾2$$$$\Rightarrow n⩽5+\sqrt{25-\frac{57}{4}}\Rightarrow n⩽8$$$$10n-\frac{57}{4}={(n+f)}^2<{(n+1)}^2$$$$n^2-8n+\frac{61}{4}>0$$$$\Rightarrow n<4-\sqrt{16-\frac{61}{4}}\Rightarrow n⩽3$$$$\Rightarrow n>4+\sqrt{16-\frac{61}{4}}\Rightarrow n⩾5$$$$\Rightarrow n=2,3,5,6,7,8$$$$\Rightarrow x=\sqrt{10n-\frac{57}{4}}$$$$n=2:x^2=20-\frac{57}{4}\Rightarrow x=\frac{\sqrt{23}}{2}✓$$$$n=3:x^2=30-\frac{57}{4}\Rightarrow x=\frac{\sqrt{63}}{2}✓$$$$n=5:x^2=50-\frac{57}{4}\Rightarrow x=\frac{\sqrt{143}}{2}✓$$$$n=6:x^2=60-\frac{57}{4}\Rightarrow x=\frac{\sqrt{183}}{2}✓$$$$n=7:x^2=70-\frac{57}{4}\Rightarrow x=\frac{\sqrt{223}}{2}✓$$$$n=8:x^2=80-\frac{57}{4}\Rightarrow x=\frac{\sqrt{263}}{2}✓$$

Commented by Ghisom last updated on 20/Feb/24

thank you

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