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

solve for x x^2 −10⌊x⌋+((57)/4)=0

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

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

⌊x⌋=n ∈ Z^+ x=n+f with 0≤f<1 (n+f)^2 −10n+((57)/4)=0 10n−((57)/4)=(n+f)^2 ≥n^2 n^2 −10n+((57)/4)≤0 ⇒n≥5−(√(25−((57)/4))) ⇒n≥2 ⇒n≤5+(√(25−((57)/4))) ⇒n≤8 10n−((57)/4)=(n+f)^2 <(n+1)^2 n^2 −8n+((61)/4)>0 ⇒n<4−(√(16−((61)/4))) ⇒n≤3 ⇒n>4+(√(16−((61)/4))) ⇒n≥5 ⇒n=2,3,5,6,7,8 ⇒x=(√(10n−((57)/4))) n=2: x^2 =20−((57)/4) ⇒x=((√(23))/2) ✓ n=3: x^2 =30−((57)/4) ⇒x=((√(63))/2) ✓ n=5: x^2 =50−((57)/4) ⇒x=((√(143))/2) ✓ n=6: x^2 =60−((57)/4) ⇒x=((√(183))/2) ✓ n=7: x^2 =70−((57)/4) ⇒x=((√(223))/2) ✓ n=8: x^2 =80−((57)/4) ⇒x=((√(263))/2) ✓

$$\lfloor{x}\rfloor={n}\:\in\:\mathbb{Z}^{+} \\ $$$${x}={n}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$$\left({n}+{f}\right)^{\mathrm{2}} −\mathrm{10}{n}+\frac{\mathrm{57}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{10}{n}−\frac{\mathrm{57}}{\mathrm{4}}=\left({n}+{f}\right)^{\mathrm{2}} \geqslant{n}^{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} −\mathrm{10}{n}+\frac{\mathrm{57}}{\mathrm{4}}\leqslant\mathrm{0} \\ $$$$\Rightarrow{n}\geqslant\mathrm{5}−\sqrt{\mathrm{25}−\frac{\mathrm{57}}{\mathrm{4}}}\:\Rightarrow{n}\geqslant\mathrm{2} \\ $$$$\Rightarrow{n}\leqslant\mathrm{5}+\sqrt{\mathrm{25}−\frac{\mathrm{57}}{\mathrm{4}}}\:\Rightarrow{n}\leqslant\mathrm{8} \\ $$$$\mathrm{10}{n}−\frac{\mathrm{57}}{\mathrm{4}}=\left({n}+{f}\right)^{\mathrm{2}} <\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} −\mathrm{8}{n}+\frac{\mathrm{61}}{\mathrm{4}}>\mathrm{0} \\ $$$$\Rightarrow{n}<\mathrm{4}−\sqrt{\mathrm{16}−\frac{\mathrm{61}}{\mathrm{4}}}\:\Rightarrow{n}\leqslant\mathrm{3} \\ $$$$\Rightarrow{n}>\mathrm{4}+\sqrt{\mathrm{16}−\frac{\mathrm{61}}{\mathrm{4}}}\:\Rightarrow{n}\geqslant\mathrm{5} \\ $$$$\Rightarrow{n}=\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{10}{n}−\frac{\mathrm{57}}{\mathrm{4}}} \\ $$$${n}=\mathrm{2}:\:{x}^{\mathrm{2}} =\mathrm{20}−\frac{\mathrm{57}}{\mathrm{4}}\:\Rightarrow{x}=\frac{\sqrt{\mathrm{23}}}{\mathrm{2}}\:\checkmark \\ $$$${n}=\mathrm{3}:\:{x}^{\mathrm{2}} =\mathrm{30}−\frac{\mathrm{57}}{\mathrm{4}}\:\Rightarrow{x}=\frac{\sqrt{\mathrm{63}}}{\mathrm{2}}\:\checkmark \\ $$$${n}=\mathrm{5}:\:{x}^{\mathrm{2}} =\mathrm{50}−\frac{\mathrm{57}}{\mathrm{4}}\:\Rightarrow{x}=\frac{\sqrt{\mathrm{143}}}{\mathrm{2}}\:\checkmark \\ $$$${n}=\mathrm{6}:\:{x}^{\mathrm{2}} =\mathrm{60}−\frac{\mathrm{57}}{\mathrm{4}}\:\Rightarrow{x}=\frac{\sqrt{\mathrm{183}}}{\mathrm{2}}\:\checkmark \\ $$$${n}=\mathrm{7}:\:{x}^{\mathrm{2}} =\mathrm{70}−\frac{\mathrm{57}}{\mathrm{4}}\:\Rightarrow{x}=\frac{\sqrt{\mathrm{223}}}{\mathrm{2}}\:\checkmark \\ $$$${n}=\mathrm{8}:\:{x}^{\mathrm{2}} =\mathrm{80}−\frac{\mathrm{57}}{\mathrm{4}}\:\Rightarrow{x}=\frac{\sqrt{\mathrm{263}}}{\mathrm{2}}\:\checkmark \\ $$

Commented by Ghisom last updated on 20/Feb/24

thank you

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