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Question Number 131880 by Algoritm last updated on 09/Feb/21

	Answered by SEKRET last updated on 09/Feb/21

	$x = \frac{5}{2}$
	Commented by Algoritm last updated on 09/Feb/21

	$prove that$
	Commented by Raxreedoroid last updated on 09/Feb/21
	$4((5/2))^2 −5⌊2.5⌋+8{x}=19 x=⌊x⌋+{x}⇒{x}=x−⌊x⌋ 25−10+8(2.5−⌊2.5⌋)=19 15+8(0.5)=19 19=19 ∴ x=19$
	$4 {(\frac{5}{2})}^{2} - 5 ⌊ 2.5 ⌋ + 8 {x} = 19$ $x = ⌊ x ⌋ + {x} \Rightarrow {x} = x - ⌊ x ⌋$ $25 - 10 + 8 (2.5 - ⌊ 2.5 ⌋) = 19$ $15 + 8 (0.5) = 19$ $19 = 19$ $∴ x = 19$
	Answered by mr W last updated on 09/Feb/21
	$just for easy writing: let n=⌊x⌋, f={x} with 0≤f<1 x=n+f assume x>0. you can assume x<0 similarly, and you′ll see that it gives no solution. 4(n+f)^2 −5n+8f=19 4(n+1)^2 −5n>4(n+f)^2 −5n=19−8f>11 4(n+1)^2 −5n>11 4n^2 +3n−7>0 n_(1,2) =((−3±11)/8)=−(7/4), 1 n<−(7/4) (rejected) or n>1 ...(I) 4n^2 −5n<4(n+f)^2 −5n=19−8f≤19 4n^2 −5n<19 4n^2 −5n−19<0 n_(1,2) =((5±(√(329)))/8)≈−1.6, 2.9 ⇒−1.6<n<2.9 ...(II) from (I) and (II): 1<n<2.9 ⇒n=2 4(2+f)^2 −5×2+8f=19 4f^2 +24f−13=0 f=((−24±(√(24^2 +16×13)))/8)=(1/2), −((13)/2) (rejected) ⇒solution: n=2, f=(1/2) ⇒x=2+(1/2)=(5/2)$
	$j u s t f o r e a s y w r i t i n g :$ $l e t n = ⌊ x ⌋, f = {x} w i t h 0 ⩽ f < 1$ $x = n + f$ $a s s u m e x > 0 .$ $y o u c a n a s s u m e x < 0 s i m i l a r l y, a n d$ ${y o u}^{'} l l s e e t h a t i t g i v e s n o s o l u t i o n .$ $4 {(n + f)}^{2} - 5 n + 8 f = 19$ $4 {(n + 1)}^{2} - 5 n > 4 {(n + f)}^{2} - 5 n = 19 - 8 f > 11$ $4 {(n + 1)}^{2} - 5 n > 11$ $4 n^{2} + 3 n - 7 > 0$ $n_{1, 2} = \frac{- 3 \pm 11}{8} = - \frac{7}{4}, 1$ $n < - \frac{7}{4} (r e j e c t e d) o r n > 1 . . . (I)$ $4 n^{2} - 5 n < 4 {(n + f)}^{2} - 5 n = 19 - 8 f ⩽ 19$ $4 n^{2} - 5 n < 19$ $4 n^{2} - 5 n - 19 < 0$ $n_{1, 2} = \frac{5 \pm \sqrt{329}}{8} \approx - 1.6, 2.9$ $\Rightarrow - 1.6 < n < 2.9 . . . (I I)$ $f r o m (I) a n d (I I) :$ $1 < n < 2.9$ $\Rightarrow n = 2$ $4 {(2 + f)}^{2} - 5 \times 2 + 8 f = 19$ $4 f^{2} + 24 f - 13 = 0$ $f = \frac{- 24 \pm \sqrt{24^{2} + 16 \times 13}}{8} = \frac{1}{2}, - \frac{13}{2} (r e j e c t e d)$ $\Rightarrow s o l u t i o n : n = 2, f = \frac{1}{2}$ $\Rightarrow x = 2 + \frac{1}{2} = \frac{5}{2}$
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