Question-131880

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 {(\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

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 \dots (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 \dots (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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