Question Number 228019 by mr W last updated on 10/Mar/26
$${t}\:{is}\:{the}\:{fractional}\:{part}\:{of}\:{a},\:{and} \\ $$$${a}^{\mathrm{2}} +{t}^{\mathrm{2}} =\mathrm{18}.\:{find}\:{t}=? \\ $$
Answered by Ghisom_ last updated on 10/Mar/26
![∀r∈R: fpart (r) =r−ipart (r) fpart (π) =π−3=.14... fpart (−π)=−π−(−3)=3−π=−.14... a^2 =18−t^2 ⇒ ∣a∣≤(√(18)) a=z+t∧∣z∣∈{0, 1, 2, 3, 4} (z+t)^2 +t^2 =18 t=−(z/2)±((√(36−z^2 ))/2) [sign t =sign z ∧∣t∣<1] ⇒ z_1 =−4∧t=2−(√5)∧a=−2−(√5) z_2 =4∧t=−2+(√5)∧a=2+(√5) t=±((√5)−2)](https://www.tinkutara.com/question/Q228020.png)
$$\forall{r}\in\mathbb{R}:\:\mathrm{fpart}\:\left({r}\right)\:={r}−\mathrm{ipart}\:\left({r}\right) \\ $$$$\mathrm{fpart}\:\left(\pi\right)\:=\pi−\mathrm{3}=.\mathrm{14}… \\ $$$$\mathrm{fpart}\:\left(−\pi\right)=−\pi−\left(−\mathrm{3}\right)=\mathrm{3}−\pi=−.\mathrm{14}… \\ $$$$ \\ $$$${a}^{\mathrm{2}} =\mathrm{18}−{t}^{\mathrm{2}} \:\Rightarrow\:\mid{a}\mid\leqslant\sqrt{\mathrm{18}} \\ $$$${a}={z}+{t}\wedge\mid{z}\mid\in\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\right\} \\ $$$$\left({z}+{t}\right)^{\mathrm{2}} +{t}^{\mathrm{2}} =\mathrm{18} \\ $$$${t}=−\frac{{z}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{36}−{z}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\left[\mathrm{sign}\:{t}\:=\mathrm{sign}\:{z}\:\wedge\mid{t}\mid<\mathrm{1}\right] \\ $$$$\Rightarrow\:{z}_{\mathrm{1}} =−\mathrm{4}\wedge{t}=\mathrm{2}−\sqrt{\mathrm{5}}\wedge{a}=−\mathrm{2}−\sqrt{\mathrm{5}} \\ $$$$\:\:\:\:\:\:{z}_{\mathrm{2}} =\mathrm{4}\wedge{t}=−\mathrm{2}+\sqrt{\mathrm{5}}\wedge{a}=\mathrm{2}+\sqrt{\mathrm{5}} \\ $$$${t}=\pm\left(\sqrt{\mathrm{5}}−\mathrm{2}\right) \\ $$
Answered by TonyCWX last updated on 11/Mar/26
Commented by Ghisom_ last updated on 11/Mar/26
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{the}\:\mathrm{fractional}\:\mathrm{part}\:\mathrm{of}\:\left(\mathrm{for}\right. \\ $$$$\left.\mathrm{example}\right)\:−\mathrm{2}.\mathrm{37}\:\mathrm{is}\:\mathrm{0}.\mathrm{63}\:\mathrm{because}\:\mathrm{also}\:\mathrm{the} \\ $$$$\mathrm{integer}\:\mathrm{part}\:\mathrm{of}\:{x}\:\mathrm{is}\:\mathrm{not}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\lfloor{x}\rfloor. \\ $$$$\mathrm{maybe}\:\mathrm{you}\:\mathrm{learned}\:\mathrm{a}\:\mathrm{different}\:\mathrm{definitions} \\ $$$$\mathrm{than}\:\mathrm{I}\:\mathrm{did}… \\ $$
Commented by TonyCWX last updated on 11/Mar/26
$$\mathrm{I}\:\mathrm{believe}\:\mathrm{I}\:\mathrm{made}\:\mathrm{a}\:\mathrm{mistake}. \\ $$$$ \\ $$$$\mathrm{I}'\mathrm{ll}\:\mathrm{try}\:\mathrm{again}. \\ $$