solve-in-R-x-3-5-x-

Question Number 86426 by M±th+et£s last updated on 28/Mar/20

s o l v e i n R

x^{3} - 5 = [x]

Answered by mr W last updated on 28/Mar/20

l e t x = n + δ

{(n + δ)}^{3} - 5 = [x] = n

{(n + δ)}^{3} - n = 5

w i t h n = 0, L H S < 1

w i t h n = 2, L H S ⩾ 6

\Rightarrow n = 1

{(1 + δ)}^{3} - 1 = 5

{(1 + δ)}^{3} = 6

δ = \sqrt[3]{6} - 1

\Rightarrow x = n + δ = \sqrt[3]{6} \approx 1.81712

Commented by M±th+et£s last updated on 28/Mar/20

g o d b l e s s y o u s i r

Answered by MJS last updated on 28/Mar/20

x^{3} = 5 + [x]

- 8 ⩽ x^{3} < - 1 \Rightarrow 5 + [x] = 3

- 1 ⩽ x^{3} < 0 \Rightarrow 5 + [x] = 4

0 ⩽ x^{3} < 1 \Rightarrow 5 + [x] = 5

1 ⩽ x^{3} < 8 \Rightarrow 5 + [x] = 6 \Rightarrow x^{3} = 6 \Leftrightarrow x = \sqrt[3]{6}

8 ⩽ x^{3} < 27 \Rightarrow 5 + [x] = 7

Commented by M±th+et£s last updated on 28/Mar/20

g o d b l e s s y o u s i r

Commented by M±th+et£s last updated on 28/Mar/20

s i r w h y - 8 ⩽ x < - 1

Commented by M±th+et£s last updated on 28/Mar/20

s o r r y x^{3}

Commented by MJS last updated on 28/Mar/20

- 8 ⩽ x^{3} < - 1 \Leftrightarrow - 2 ⩽ x < - 1 \Rightarrow [x] = - 2

generally

\forall a \in Z : a ⩽ x < a + 1 \Rightarrow [x] = a

and a ⩽ x < a + 1 \Leftrightarrow a^{3} ⩽ x^{3} < {(a + 1)}^{3}

Commented by M±th+et£s last updated on 28/Mar/20

t h a n k y o u s i r