x^4 −⌊5x^2 ⌋+4=0


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Question Number 117620 by Dwaipayan Shikari last updated on 12/Oct/20

$x^{4} - ⌊ 5 x^{2} ⌋ + 4 = 0$
	Answered by mr W last updated on 12/Oct/20

	$x^{4} = ⌊ 5 x^{2} ⌋ - 4 \in Z$ $\Rightarrow x \in Z$ $\Rightarrow 5 x^{2} \in Z$ $\Rightarrow x^{4} - 5 x^{2} + 4 = 0$ $\Rightarrow (x^{2} - 1) (x^{2} - 4) = 0$ $\Rightarrow x = \pm 1, \pm 2$
	Commented by floor(10²Eta[1]) last updated on 12/Oct/20

	$missing the solutions \pm^{4} \sqrt{14} and \pm^{4} \sqrt{15}$
	Commented by mr W last updated on 13/Oct/20

	$y e s, y o u a r e r i g h t!$
	Answered by AbduraufKodiriy last updated on 12/Oct/20
	$x^2 =t t^2 −⌊5t⌋+4=0 ⌊5t⌋=t^2 +4 ∈ Z {5t}=−(t^2 −5t+4) 0≤−(t^2 −5t+4)<1 ⇒ t∈[1; ((5−(√5))/2))∪(((5+(√5))/2); 4] Thus, (t^2 +4)∈[5; ((23−5(√5))/2))∪(((23+5(√5))/2); 20] approximately: (t^2 +4)∈[5; 5.9)∪(17.09; 20] So, 1) t^2 +4=5 ⇒ t=1 2) t^2 +4=18 ⇒ t=(√(14)); 3) t^2 +4=19 ⇒ t=(√(15)); 4) t^2 +4=20 ⇒ t=4 Answer: x_(1,2) =±1, x_(3,4) =±((14))^(1/4) , x_(5,6) =±((15))^(1/4) , x_(7,8) =±2$
	$x^{2} = t$ $t^{2} - ⌊ 5 t ⌋ + 4 = 0$ $⌊ 5 t ⌋ = t^{2} + 4 \in Z$ ${5 t} = - (t^{2} - 5 t + 4)$ $0 ⩽ - (t^{2} - 5 t + 4) < 1 \Rightarrow t \in [1; \frac{5 - \sqrt{5}}{2}) \cup (\frac{5 + \sqrt{5}}{2}; 4]$ $T h u s, (t^{2} + 4) \in [5; \frac{23 - 5 \sqrt{5}}{2}) \cup (\frac{23 + 5 \sqrt{5}}{2}; 20]$ $a p p r o x i m a t e l y : (t^{2} + 4) \in [5; 5.9) \cup (17.09; 20]$ $S o, 1) t^{2} + 4 = 5 \Rightarrow t = 1 2) t^{2} + 4 = 18 \Rightarrow t = \sqrt{14};$ $3) t^{2} + 4 = 19 \Rightarrow t = \sqrt{15}; 4) t^{2} + 4 = 20 \Rightarrow t = 4$ $A n s w e r : x_{1, 2} = \pm 1, x_{3, 4} = \pm \sqrt[4]{14}, x_{5, 6} = \pm \sqrt[4]{15}, x_{7, 8} = \pm 2$
	Answered by floor(10²Eta[1]) last updated on 12/Oct/20
	$x^4 −⌊5x^2 ⌋=−4∈Z but ⌊5x^2 ⌋∈Z⇒x^4 ∈N ⇒x∈Z or x=±^4 (√a), a∈N ⌊5x^2 ⌋=n⇒5x^2 =n+θ, n∈Z, 0≤θ<1 x^2 =((n+θ)/5) x^4 =((n^2 +2nθ+θ^2 )/(25)) ((n^2 +2nθ+θ^2 )/(25))−n+4=0 n^2 +(2θ−25)n+100+θ^2 =0 n=((25−2θ±(√(225−100θ)))/2)∈Z since 0≤θ<1 so: I case: 15≥(√(225−100θ))>11 ⇒40≥40−2θ≥25−2θ+(√(225−100θ))>36−2θ>34 ⇒20≥n>17∴n∈{18,19,20} II case: 8<10−2θ≤25−2θ−(√(225−100θ))<14−2θ≤14 4<n<7⇒n∈{5,6} ⇒n=5 x^2 =((5+θ)/5)=1+(θ/5)⇒1≤x^2 <(6/5)=1.2 x=±1 ⇒n=6(no sol.) ⇒n=18 x^2 =((18+θ)/5)⇒3.6≤x^2 <3.8⇒x∉N x^2 =(√a), a∈N⇒3.6^2 ≤a<3.8^2 ⇒a∈{13,14} ⇒x=±^4 (√(14)) ⇒n=19 x^2 =((19+θ)/5)⇒3.8≤x^2 <4⇒x∉N x^2 =(√a) ⇒3.8^2 ≤a<16⇒a=15 ⇒x=±^4 (√(15)) n=20 x^2 =((20+θ)/5)⇒4≤x^2 <5⇒x=±2 all the solutions are: x={±1, ±^4 (√(14)), ±^4 (√(15)), ±2}$
	$x^{4} - ⌊ {5 x}^{2} ⌋ = - 4 \in Z$ $but ⌊ {5 x}^{2} ⌋ \in Z \Rightarrow x^{4} \in N$ $\Rightarrow x \in Z or x = \pm^{4} \sqrt{a}, a \in N$ $⌊ {5 x}^{2} ⌋ = n \Rightarrow {5 x}^{2} = n + θ, n \in Z, 0 ⩽ θ < 1$ $x^{2} = \frac{n + θ}{5}$ $x^{4} = \frac{n^{2} + 2 n θ + θ^{2}}{25}$ $\frac{n^{2} + 2 n θ + θ^{2}}{25} - n + 4 = 0$ $n^{2} + (2 θ - 25) n + 100 + θ^{2} = 0$ $n = \frac{25 - 2 θ \pm \sqrt{225 - 100 θ}}{2} \in Z$ $since 0 ⩽ θ < 1 so :$ $I case :$ $15 ⩾ \sqrt{225 - 100 θ} > 11$ $\Rightarrow 40 ⩾ 40 - 2 θ ⩾ 25 - 2 θ + \sqrt{225 - 100 θ} > 36 - 2 θ > 34$ $\Rightarrow 20 ⩾ n > 17 ∴ n \in {18, 19, 20}$ $II case :$ $8 < 10 - 2 θ ⩽ 25 - 2 θ - \sqrt{225 - 100 θ} < 14 - 2 θ ⩽ 14$ $4 < n < 7 \Rightarrow n \in {5, 6}$ $\Rightarrow n = 5$ $x^{2} = \frac{5 + θ}{5} = 1 + \frac{θ}{5} \Rightarrow 1 ⩽ x^{2} < \frac{6}{5} = 1.2$ $x = \pm 1$ $\Rightarrow n = 6 (no sol .)$ $\Rightarrow n = 18$ $x^{2} = \frac{18 + θ}{5} \Rightarrow 3.6 ⩽ x^{2} < 3.8 \Rightarrow x \notin N$ $x^{2} = \sqrt{a}, a \in N \Rightarrow 3 . 6^{2} ⩽ a < 3 . 8^{2} \Rightarrow a \in {13, 14}$ $\Rightarrow x = \pm^{4} \sqrt{14}$ $\Rightarrow n = 19$ $x^{2} = \frac{19 + θ}{5} \Rightarrow 3.8 ⩽ x^{2} < 4 \Rightarrow x \notin N$ $x^{2} = \sqrt{a} \Rightarrow 3 . 8^{2} ⩽ a < 16 \Rightarrow a = 15$ $\Rightarrow x = \pm^{4} \sqrt{15}$ $n = 20$ $x^{2} = \frac{20 + θ}{5} \Rightarrow 4 ⩽ x^{2} < 5 \Rightarrow x = \pm 2$ $all the solutions are :$ $x = {\pm 1, \pm^{4} \sqrt{14}, \pm^{4} \sqrt{15}, \pm 2}$
	Commented by Dwaipayan Shikari last updated on 12/Oct/20

	$T h a n k i n g a l l o f y o u$
	Answered by MJS_new last updated on 12/Oct/20

	$let x = t^{1 / 4}$ $t - ⌊ 5 \sqrt{t} ⌋ + 4 = 0$ $n ⩽ 5 \sqrt{t} < n + 1 \Rightarrow \frac{n^{2}}{25} ⩽ t < \frac{{(n + 1)}^{2}}{25}$ $let t = \frac{{(n + α)}^{2}}{25} with 0 ⩽ α < 1 \Rightarrow ⌊ 5 \sqrt{t} ⌋ = n$ $\frac{{(n + α)}^{2}}{25} + - n + 4 = 0$ $n^{2} + (2 α - 25) n + α^{2} + 100 = 0$ $n = \frac{25}{2} - α \pm \frac{5}{2} \sqrt{9 - 4 α}$ $0 ⩽ α < 1 \Rightarrow 5 ⩽ n_{-} < \frac{23 - 5 \sqrt{5}}{2} \approx 5.9 \land \frac{23 + 5 \sqrt{5}}{2} \approx 17.09 < n_{+} ⩽ 20$ $n \in N \Rightarrow$ $α = 0 \land n_{-} = 5$ $α = 0 \land n_{+} = 20$ $α = - 19 + 5 \sqrt{15} \land n_{+} = 19$ $α = - 18 + 5 \sqrt{14} \land n_{+} = 18$ $\Rightarrow$ $t = 1 \Rightarrow x = \pm 1$ $t = 16 \Rightarrow x = \pm 2$ $t = 15 \Rightarrow x = \pm 15^{1 / 4}$ $t = 14 \Rightarrow x = \pm 14^{1 / 4}$
	Commented by MJS_new last updated on 12/Oct/20

	$tried to find a shorter path . . .$
	Answered by 1549442205PVT last updated on 13/Oct/20
	$x^4 −⌊5x^2 ⌋+4=0(1).Put x^4 +4=t(t≥4) ⇒x^2 =(√(t−4)).Then(1)⇔[5(√(t−4))]=t(∗) By the definition of function [x]we have 0≤5(√(t−4))−t<1 ⇔ { ((5(√(t−4))≥t(1))),((5(√(t−4))<t+1(2))) :}(∗∗) (1)⇔25(t−4)≥t^2 ⇔t^2 −25t+100≤0 ⇔(t−5)(t−20)≤0⇔5≤t≤20.Since t≥4,so by(∗)⇒t∈N⇒t∈{5,6,...,20}(3) (2)⇔25(t−4)<t^2 +2t+1 ⇔t^2 −23t+101>0⇔ t∈(−∞,((23−5(√5))/2))∪(((23+5(√5))/2),+∞) 5.9≈((23−5(√5))/2) and ((23+5(√5))/2)≈17.09 ⇒t∈{4,5}∪{18,19,20,21,...}(4) Combining (3)(4)we get t∈{5,18,19,20}are roots of the system of the inequalities (∗∗) i)t=5⇔x^4 +4=5⇔x^4 =1⇔x=±1 ii)t=18⇒x^4 +4=18⇔x=±^4 (√(14)) iii)t=19⇒x^4 +4=19⇔x=±^4 (√(15)) iv)t=20⇒x^4 +4=20⇔x=±2 Thus,roots of the system (∗∗) are x∈{±1;±2,±^4 (√(14)), ±^4 (√(15))}that is answer of our problem.$
	$x^{4} - ⌊ 5 x^{2} ⌋ + 4 = 0 (1) . Put x^{4} + 4 = t (t ⩾ 4)$ $\Rightarrow x^{2} = \sqrt{t - 4} . Then (1) \Leftrightarrow [5 \sqrt{t - 4}] = t ()$ $By the definition of function [x] we$ $have 0 ⩽ 5 \sqrt{t - 4} - t < 1$ $\Leftrightarrow {\begin{cases} 5 \sqrt{t - 4} ⩾ t (1) \\ 5 \sqrt{t - 4} < t + 1 (2) \end{cases} ( )$ $(1) \Leftrightarrow 25 (t - 4) ⩾ t^{2} \Leftrightarrow t^{2} - 25 t + 100 ⩽ 0$ $\Leftrightarrow (t - 5) (t - 20) ⩽ 0 \Leftrightarrow 5 ⩽ t ⩽ 20 . Since$ $t ⩾ 4, so by () \Rightarrow t \in N \Rightarrow t \in {5, 6, . . ., 20} (3)$ $(2) \Leftrightarrow 25 (t - 4) < t^{2} + 2 t + 1$ $\Leftrightarrow t^{2} - 23 t + 101 > 0 \Leftrightarrow$ $t \in (- \infty, \frac{23 - 5 \sqrt{5}}{2}) \cup (\frac{23 + 5 \sqrt{5}}{2}, + \infty)$ $5.9 \approx \frac{23 - 5 \sqrt{5}}{2} and \frac{23 + 5 \sqrt{5}}{2} \approx 17.09$ $\Rightarrow t \in {4, 5} \cup {18, 19, 20, 21, . . .} (4)$ $Combining (3) (4) we get$ $t \in {5, 18, 19, 20} are roots of the$ $system of the inequalities (* )$ $i) t = 5 \Leftrightarrow x^{4} + 4 = 5 \Leftrightarrow x^{4} = 1 \Leftrightarrow x = \pm 1$ $ii) t = 18 \Rightarrow x^{4} + 4 = 18 \Leftrightarrow x = \pm^{4} \sqrt{14}$ $iii) t = 19 \Rightarrow x^{4} + 4 = 19 \Leftrightarrow x = \pm^{4} \sqrt{15}$ $iv) t = 20 \Rightarrow x^{4} + 4 = 20 \Leftrightarrow x = \pm 2$ $Thus, roots of the system ( *) are$ $x \in {\pm 1; \pm 2, \pm^{4} \sqrt{14}, \pm^{4} \sqrt{15}} that is$ $answer of our problem .$
	Commented by floor(10²Eta[1]) last updated on 13/Oct/20

	$\pm^{4} \sqrt{14} and \pm^{4} \sqrt{15} are solutions too$
	Commented by 1549442205PVT last updated on 13/Oct/20

	$I mistaked and corrected . Thank sir$
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