Given-x-x-x-How-many-real-solutions-from-equation-x-x-2-1-with-10-x-10-

Question Number 37627 by Joel579 last updated on 16/Jun/18

Given {x} = x - ⌊ x ⌋

How many real solutions from equation

{x} + {x^{2}} = 1

with - 10 ⩽ x ⩽ 10 ?

Answered by ajfour last updated on 16/Jun/18

l e t x = m + h

w h e r e m \in Z a n d h = {x}

{x} + {x^{2}} = 1

\Rightarrow h + {(m + h)}^{2} - [{(m + h)}^{2}] = 1

h + m^{2} + 2 m h + h^{2} - m^{2} - [2 m h + h^{2}] = 1

\Rightarrow {2 m h + h^{2}} = 1 - h

\Rightarrow h \neq 0

\dots

Commented by Rasheed.Sindhi last updated on 17/Jun/18

N i c e b e g i n i n g!

Answered by ajfour last updated on 16/Jun/18

172 r e a l s o l u t i o n s, i t h i n k .

Answered by MJS last updated on 16/Jun/18

for x > 0 we must solve equations

x \in [0; 1 [

(x - 0) + (x^{2} - 0) = 1 \Rightarrow x^{2} + x - 1 = 0

x \in [1; 2 [

(x - 1) + (x^{2} - 1) = 1 \Rightarrow x^{2} + x - 3 = 0

(x - 1) + (x^{2} - 2) = 1 \Rightarrow x^{2} + x - 4 = 0

(x - 1) + (x^{2} - 3) = 1 \Rightarrow x^{2} + x - 5 = 0

x \in [2; 3 [

(x - 2) + (x^{2} - 4) = 1 \Rightarrow x^{2} + x - 7 = 0

\dots

(x - 2) + (x^{2} - 8) = 1 \Rightarrow x^{2} + x - 11 = 0

x \in [n; n + 1]

(x - n) + (x^{2} - n^{2}) = 1 \Rightarrow x^{2} + x - (n^{2} + n + 1) = 0

\dots

(x - n) + (x^{2} - ({(n + 1)}^{2} - 1)) = 1 \Rightarrow

\Rightarrow x^{2} + x - (n^{2} + 3 n + 1) = 0

we have 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 =

= 100 solutions in [0; 10]

for x < 0 the negative solutions of above

equations count

\Rightarrow we have no solution in [- 1; 0 [

1 solution in [- 2; - 1 [

3 solutions in [- 3; - 2 [

\dots

so we have 100 + 100 - 19 = 181 solutions

in [- 10; 10]

Commented by Rasheed.Sindhi last updated on 17/Jun/18

N ice idea of interval - wise analysis!