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Find the number of possible solution of |[x]-2x|=4
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Find the number of possible values of x of the equation |[x]-2x|=4.
|x| represent absolute value of x.
[x] represent greatest integer lesser than x.
I plotted the curve in desmos.com and got the following values x=3.5,4 & x=-4,-4.5 but got the idea of the values after having a look at the curve. Can someone suggest ways of finding value directly.
absolute-value floor-function
edited Jul 23 at 11:48
Harry Peter
5,45311438
asked Jul 21 at 2:06
Samar Imam Zaidi
1,070316
add a comment |Â
up vote
0
down vote
favorite
Find the number of possible values of x of the equation |[x]-2x|=4.
|x| represent absolute value of x.
[x] represent greatest integer lesser than x.
I plotted the curve in desmos.com and got the following values x=3.5,4 & x=-4,-4.5 but got the idea of the values after having a look at the curve. Can someone suggest ways of finding value directly.
absolute-value floor-function
edited Jul 23 at 11:48
Harry Peter
5,45311438
asked Jul 21 at 2:06
Samar Imam Zaidi
1,070316
Since $|[x]-2x|$ is equal to either $[x]-2x$ or $2x-[x]$ and the result must be an integer, then $2x$ is an integer. This means that either $x$ is an integer, or an integer $+1/2$. If $x$ is an integer then $[x]=x$ and the equation becomes $|x|=4$. So, $x=pm4$. If $x+1/2$ is an integer, then $[x]=x-1/2$ and the equation becomes $|x+1/2|=4$. So, $x=pm4-1/2$.
– user577471
Jul 21 at 2:14
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find the number of possible values of x of the equation |[x]-2x|=4.
|x| represent absolute value of x.
[x] represent greatest integer lesser than x.
I plotted the curve in desmos.com and got the following values x=3.5,4 & x=-4,-4.5 but got the idea of the values after having a look at the curve. Can someone suggest ways of finding value directly.
absolute-value floor-function
edited Jul 23 at 11:48
Harry Peter
5,45311438
asked Jul 21 at 2:06
Samar Imam Zaidi
1,070316
Find the number of possible values of x of the equation |[x]-2x|=4.
|x| represent absolute value of x.
[x] represent greatest integer lesser than x.
I plotted the curve in desmos.com and got the following values x=3.5,4 & x=-4,-4.5 but got the idea of the values after having a look at the curve. Can someone suggest ways of finding value directly.
absolute-value floor-function
edited Jul 23 at 11:48
Harry Peter
5,45311438
asked Jul 21 at 2:06
Samar Imam Zaidi
1,070316
edited Jul 23 at 11:48
Harry Peter
5,45311438
edited Jul 23 at 11:48
Harry Peter
5,45311438
edited Jul 23 at 11:48
Harry Peter
5,45311438
5,45311438
asked Jul 21 at 2:06
Samar Imam Zaidi
1,070316
asked Jul 21 at 2:06
Samar Imam Zaidi
1,070316
asked Jul 21 at 2:06
Samar Imam Zaidi
1,070316
1,070316
Since $|[x]-2x|$ is equal to either $[x]-2x$ or $2x-[x]$ and the result must be an integer, then $2x$ is an integer. This means that either $x$ is an integer, or an integer $+1/2$. If $x$ is an integer then $[x]=x$ and the equation becomes $|x|=4$. So, $x=pm4$. If $x+1/2$ is an integer, then $[x]=x-1/2$ and the equation becomes $|x+1/2|=4$. So, $x=pm4-1/2$.
– user577471
Jul 21 at 2:14
add a comment |Â
Since $|[x]-2x|$ is equal to either $[x]-2x$ or $2x-[x]$ and the result must be an integer, then $2x$ is an integer. This means that either $x$ is an integer, or an integer $+1/2$. If $x$ is an integer then $[x]=x$ and the equation becomes $|x|=4$. So, $x=pm4$. If $x+1/2$ is an integer, then $[x]=x-1/2$ and the equation becomes $|x+1/2|=4$. So, $x=pm4-1/2$.
– user577471
Jul 21 at 2:14
Since $|[x]-2x|$ is equal to either $[x]-2x$ or $2x-[x]$ and the result must be an integer, then $2x$ is an integer. This means that either $x$ is an integer, or an integer $+1/2$. If $x$ is an integer then $[x]=x$ and the equation becomes $|x|=4$. So, $x=pm4$. If $x+1/2$ is an integer, then $[x]=x-1/2$ and the equation becomes $|x+1/2|=4$. So, $x=pm4-1/2$.
– user577471
Jul 21 at 2:14
Since $|[x]-2x|$ is equal to either $[x]-2x$ or $2x-[x]$ and the result must be an integer, then $2x$ is an integer. This means that either $x$ is an integer, or an integer $+1/2$. If $x$ is an integer then $[x]=x$ and the equation becomes $|x|=4$. So, $x=pm4$. If $x+1/2$ is an integer, then $[x]=x-1/2$ and the equation becomes $|x+1/2|=4$. So, $x=pm4-1/2$.
– user577471
Jul 21 at 2:14
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Take a deep breath and do it.
Note $[x]$ is an integer and $4$ is an integer so $2x$ is an integer so either $x$ is an integer or $x = n + frac 12$ for some integer $n$.
If $x$ is an integer then $|[x] - 2x| = |x-2x| = |-x| = 4$ and $x = pm 4$.
If $x = n+frac 12$ then $|[x] + 2(x)| = |n - 2n - 1| = |-n-1| = 4$ so $n+1 =pm4$ and $n = 3$ or $n = -5$ and so $x = 3frac 12$ or $-4frac 12$.
So solutions are $x = -4frac 12, -4, 3, 3frac 12$.
answered Jul 21 at 2:44
fleablood
60.4k22575
add a comment |Â
up vote
2
down vote
We have $x=[x]+x$ where $x$ is the fractional part and $[x]=ninmathbb Z$
So $[x]-2x=n-underbrace[2n]_= 2[n]-2x=-n-underbrace2x_= 0,1=pm 4iff begincasesx=0 & n=-4text or 4\x=frac 12 & n=-5text or 3endcases$
Giving $xin-4,4,-5+frac 12,3+frac 12$
answered Jul 21 at 2:25
zwim
11k627
Thanking you for your prompt reply.
– Samar Imam Zaidi
Jul 21 at 2:37
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Take a deep breath and do it.
Note $[x]$ is an integer and $4$ is an integer so $2x$ is an integer so either $x$ is an integer or $x = n + frac 12$ for some integer $n$.
If $x$ is an integer then $|[x] - 2x| = |x-2x| = |-x| = 4$ and $x = pm 4$.
If $x = n+frac 12$ then $|[x] + 2(x)| = |n - 2n - 1| = |-n-1| = 4$ so $n+1 =pm4$ and $n = 3$ or $n = -5$ and so $x = 3frac 12$ or $-4frac 12$.
So solutions are $x = -4frac 12, -4, 3, 3frac 12$.
answered Jul 21 at 2:44
fleablood
60.4k22575
add a comment |Â
up vote
1
down vote
accepted
Take a deep breath and do it.
Note $[x]$ is an integer and $4$ is an integer so $2x$ is an integer so either $x$ is an integer or $x = n + frac 12$ for some integer $n$.
If $x$ is an integer then $|[x] - 2x| = |x-2x| = |-x| = 4$ and $x = pm 4$.
If $x = n+frac 12$ then $|[x] + 2(x)| = |n - 2n - 1| = |-n-1| = 4$ so $n+1 =pm4$ and $n = 3$ or $n = -5$ and so $x = 3frac 12$ or $-4frac 12$.
So solutions are $x = -4frac 12, -4, 3, 3frac 12$.
answered Jul 21 at 2:44
fleablood
60.4k22575
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Take a deep breath and do it.
Note $[x]$ is an integer and $4$ is an integer so $2x$ is an integer so either $x$ is an integer or $x = n + frac 12$ for some integer $n$.
If $x$ is an integer then $|[x] - 2x| = |x-2x| = |-x| = 4$ and $x = pm 4$.
If $x = n+frac 12$ then $|[x] + 2(x)| = |n - 2n - 1| = |-n-1| = 4$ so $n+1 =pm4$ and $n = 3$ or $n = -5$ and so $x = 3frac 12$ or $-4frac 12$.
So solutions are $x = -4frac 12, -4, 3, 3frac 12$.
answered Jul 21 at 2:44
fleablood
60.4k22575
Take a deep breath and do it.
Note $[x]$ is an integer and $4$ is an integer so $2x$ is an integer so either $x$ is an integer or $x = n + frac 12$ for some integer $n$.
If $x$ is an integer then $|[x] - 2x| = |x-2x| = |-x| = 4$ and $x = pm 4$.
If $x = n+frac 12$ then $|[x] + 2(x)| = |n - 2n - 1| = |-n-1| = 4$ so $n+1 =pm4$ and $n = 3$ or $n = -5$ and so $x = 3frac 12$ or $-4frac 12$.
So solutions are $x = -4frac 12, -4, 3, 3frac 12$.
answered Jul 21 at 2:44
fleablood
60.4k22575
answered Jul 21 at 2:44
fleablood
60.4k22575
answered Jul 21 at 2:44
fleablood
60.4k22575
answered Jul 21 at 2:44
fleablood
60.4k22575
60.4k22575
add a comment |Â
add a comment |Â
up vote
2
down vote
We have $x=[x]+x$ where $x$ is the fractional part and $[x]=ninmathbb Z$
So $[x]-2x=n-underbrace[2n]_= 2[n]-2x=-n-underbrace2x_= 0,1=pm 4iff begincasesx=0 & n=-4text or 4\x=frac 12 & n=-5text or 3endcases$
Giving $xin-4,4,-5+frac 12,3+frac 12$
answered Jul 21 at 2:25
zwim
11k627
Thanking you for your prompt reply.
– Samar Imam Zaidi
Jul 21 at 2:37
add a comment |Â
up vote
2
down vote
We have $x=[x]+x$ where $x$ is the fractional part and $[x]=ninmathbb Z$
So $[x]-2x=n-underbrace[2n]_= 2[n]-2x=-n-underbrace2x_= 0,1=pm 4iff begincasesx=0 & n=-4text or 4\x=frac 12 & n=-5text or 3endcases$
Giving $xin-4,4,-5+frac 12,3+frac 12$
answered Jul 21 at 2:25
zwim
11k627
Thanking you for your prompt reply.
– Samar Imam Zaidi
Jul 21 at 2:37
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We have $x=[x]+x$ where $x$ is the fractional part and $[x]=ninmathbb Z$
So $[x]-2x=n-underbrace[2n]_= 2[n]-2x=-n-underbrace2x_= 0,1=pm 4iff begincasesx=0 & n=-4text or 4\x=frac 12 & n=-5text or 3endcases$
Giving $xin-4,4,-5+frac 12,3+frac 12$
answered Jul 21 at 2:25
zwim
11k627
We have $x=[x]+x$ where $x$ is the fractional part and $[x]=ninmathbb Z$
So $[x]-2x=n-underbrace[2n]_= 2[n]-2x=-n-underbrace2x_= 0,1=pm 4iff begincasesx=0 & n=-4text or 4\x=frac 12 & n=-5text or 3endcases$
Giving $xin-4,4,-5+frac 12,3+frac 12$
answered Jul 21 at 2:25
zwim
11k627
answered Jul 21 at 2:25
zwim
11k627
answered Jul 21 at 2:25
zwim
11k627
answered Jul 21 at 2:25
zwim
11k627
11k627
Thanking you for your prompt reply.
– Samar Imam Zaidi
Jul 21 at 2:37
add a comment |Â
Thanking you for your prompt reply.
– Samar Imam Zaidi
Jul 21 at 2:37
Thanking you for your prompt reply.
– Samar Imam Zaidi
Jul 21 at 2:37
Thanking you for your prompt reply.
– Samar Imam Zaidi
Jul 21 at 2:37
add a comment |Â
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Since $|[x]-2x|$ is equal to either $[x]-2x$ or $2x-[x]$ and the result must be an integer, then $2x$ is an integer. This means that either $x$ is an integer, or an integer $+1/2$. If $x$ is an integer then $[x]=x$ and the equation becomes $|x|=4$. So, $x=pm4$. If $x+1/2$ is an integer, then $[x]=x-1/2$ and the equation becomes $|x+1/2|=4$. So, $x=pm4-1/2$.
– user577471
Jul 21 at 2:14