Find-all-solutions-of-x-real-numbers-such-that-2x-2-7x-6-15-1-x-x-

Question Number 64533 by naka3546 last updated on 19/Jul/19

F i n d a l l s o l u t i o n s o f x r e a l n u m b e r s s u c h t h a t

2 x^{2} - 7 x + 6 = 15 ⌊ \frac{1}{x} ⌋ ⌊ x ⌋

Answered by MJS last updated on 19/Jul/19

x = 0 \Rightarrow ⌊ \frac{1}{x} ⌋ not defined

x = 1 \Rightarrow no solution

(0 < x < 1 \Rightarrow ⌊ x ⌋ = 0) \lor (x > 1 \Rightarrow ⌊ \frac{1}{x} ⌋ = 0)

2 x^{2} - 7 x + 6 = 0 \Rightarrow x = \frac{3}{2} \lor x = 2

x < 0

⌊ \frac{1}{x} ⌋ ⌊ x ⌋ = f (x)

- 1 < x < 0

n \in N \land n ⩾ 2 : f (- \frac{1}{n - 1} < x ⩽ - \frac{1}{n}) = n

2 x^{2} - 7 x + 6 = 15 n \land x < 0

x = \frac{7}{4} - \frac{\sqrt{120 n + 1}}{4}

- 1 < x < 0 \Rightarrow \frac{2}{5} < n < 1 \Rightarrow no n exists

x ⩽ - 1

n \in N^{★} : f (- n ⩽ x < - n + 1) = n

2 x^{2} - 7 x + 6 = 15 n \land x < 0

x = \frac{7}{4} - \frac{\sqrt{120 n + 1}}{4}

x ⩽ - 1 \Rightarrow n ⩾ 1

- n ⩽ \frac{7}{4} - \frac{\sqrt{120 n + 1}}{4} < - n + 1

4 n + 3 < \sqrt{120 n + 1} ⩽ 4 n + 7

{(4 n + 3)}^{2} < 120 n + 1 ⩽ {(4 n + 7)}^{2}

n^{2} - 6 n + \frac{1}{2} < 0 ⩽ n^{2} - 4 n + 3

n^{2} - 6 n + \frac{1}{2} < 0 \Rightarrow 1 ⩽ n ⩽ 5

0 ⩽ n^{2} - 4 n + 3 \Rightarrow n = 1 \lor n ⩾ 3

\Rightarrow n = 1 \lor 3 ⩽ n ⩽ 5

x = \frac{7}{4} - \frac{\sqrt{120 n + 1}}{4}

n = 1 \Rightarrow - 1 ⩽ x < 0 \Rightarrow x = - 1

n = 3 \Rightarrow - 3 ⩽ x < - 2 \Rightarrow x = - 3

n = 4 \Rightarrow - 4 ⩽ x < - 3 \Rightarrow x = \frac{7}{4} - \frac{\sqrt{481}}{4}

n = 5 \Rightarrow - 5 ⩽ x < - 4 \Rightarrow x = \frac{7}{4} - \frac{\sqrt{601}}{4}

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