Mixing
[x-5][x-3]+2>[x-5]+2[x-3]. [.] is the greatest integer function. Find the total number of integral values...
$begingroup$
Find the number of integral values of x satisfying the inequality.
I tried and gave up
functions
asked Jan 29 at 20:31
Vibhu KapadiaVibhu Kapadia
1
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closed as off-topic by mrtaurho, verret, Gibbs, Lord Shark the Unknown, Leucippus Jan 29 at 22:16
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$begingroup$
Find the number of integral values of x satisfying the inequality.
I tried and gave up
functions
asked Jan 29 at 20:31
Vibhu KapadiaVibhu Kapadia
1
$endgroup$
closed as off-topic by mrtaurho, verret, Gibbs, Lord Shark the Unknown, Leucippus Jan 29 at 22:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, verret, Gibbs, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Find the number of integral values of x satisfying the inequality.
I tried and gave up
functions
asked Jan 29 at 20:31
Vibhu KapadiaVibhu Kapadia
1
$endgroup$
Find the number of integral values of x satisfying the inequality.
I tried and gave up
functions
functions
asked Jan 29 at 20:31
Vibhu KapadiaVibhu Kapadia
1
asked Jan 29 at 20:31
Vibhu KapadiaVibhu Kapadia
1
asked Jan 29 at 20:31
Vibhu KapadiaVibhu Kapadia
1
asked Jan 29 at 20:31
Vibhu KapadiaVibhu Kapadia
1
asked Jan 29 at 20:31
Vibhu KapadiaVibhu Kapadia
1
1
closed as off-topic by mrtaurho, verret, Gibbs, Lord Shark the Unknown, Leucippus Jan 29 at 22:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, verret, Gibbs, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by mrtaurho, verret, Gibbs, Lord Shark the Unknown, Leucippus Jan 29 at 22:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, verret, Gibbs, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
1
active
oldest
votes
$begingroup$
A start.
$x-1 < [x] le x$.
You want
$[x-5][x-3]+2>[x-5]+2[x-3]
$.
If $x = 5+n$,
$[x-5][x-3]+2
=[n][n+2]+2
gt (n-1)(n+1)+2
=n^2+1
$
and
$[x-5]+2[x-3]
=[n]+2[n+2]
le 3n+4
$.
Therefore,
if
$n^2+1
ge 3n+4
$,
the inequality is true.
This is
$n^2-3n-3 > 0$
or
$(n-frac32)^2
gt 3-frac94
=frac34
$
or,
using the positive square root,
$n
gt frac32+frac{sqrt{3}}{2}
=frac{3+sqrt{3}}{2}
$
which is true for
$n ge 3
$.
Therefore,
the inequality is true for
$x ge 8$.
I'll let you
work out the other cases.
answered Jan 29 at 21:04
marty cohenmarty cohen
74.9k549130
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A start.
$x-1 < [x] le x$.
You want
$[x-5][x-3]+2>[x-5]+2[x-3]
$.
If $x = 5+n$,
$[x-5][x-3]+2
=[n][n+2]+2
gt (n-1)(n+1)+2
=n^2+1
$
and
$[x-5]+2[x-3]
=[n]+2[n+2]
le 3n+4
$.
Therefore,
if
$n^2+1
ge 3n+4
$,
the inequality is true.
This is
$n^2-3n-3 > 0$
or
$(n-frac32)^2
gt 3-frac94
=frac34
$
or,
using the positive square root,
$n
gt frac32+frac{sqrt{3}}{2}
=frac{3+sqrt{3}}{2}
$
which is true for
$n ge 3
$.
Therefore,
the inequality is true for
$x ge 8$.
I'll let you
work out the other cases.
answered Jan 29 at 21:04
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
A start.
$x-1 < [x] le x$.
You want
$[x-5][x-3]+2>[x-5]+2[x-3]
$.
If $x = 5+n$,
$[x-5][x-3]+2
=[n][n+2]+2
gt (n-1)(n+1)+2
=n^2+1
$
and
$[x-5]+2[x-3]
=[n]+2[n+2]
le 3n+4
$.
Therefore,
if
$n^2+1
ge 3n+4
$,
the inequality is true.
This is
$n^2-3n-3 > 0$
or
$(n-frac32)^2
gt 3-frac94
=frac34
$
or,
using the positive square root,
$n
gt frac32+frac{sqrt{3}}{2}
=frac{3+sqrt{3}}{2}
$
which is true for
$n ge 3
$.
Therefore,
the inequality is true for
$x ge 8$.
I'll let you
work out the other cases.
answered Jan 29 at 21:04
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
A start.
$x-1 < [x] le x$.
You want
$[x-5][x-3]+2>[x-5]+2[x-3]
$.
If $x = 5+n$,
$[x-5][x-3]+2
=[n][n+2]+2
gt (n-1)(n+1)+2
=n^2+1
$
and
$[x-5]+2[x-3]
=[n]+2[n+2]
le 3n+4
$.
Therefore,
if
$n^2+1
ge 3n+4
$,
the inequality is true.
This is
$n^2-3n-3 > 0$
or
$(n-frac32)^2
gt 3-frac94
=frac34
$
or,
using the positive square root,
$n
gt frac32+frac{sqrt{3}}{2}
=frac{3+sqrt{3}}{2}
$
which is true for
$n ge 3
$.
Therefore,
the inequality is true for
$x ge 8$.
I'll let you
work out the other cases.
answered Jan 29 at 21:04
marty cohenmarty cohen
74.9k549130
$endgroup$
A start.
$x-1 < [x] le x$.
You want
$[x-5][x-3]+2>[x-5]+2[x-3]
$.
If $x = 5+n$,
$[x-5][x-3]+2
=[n][n+2]+2
gt (n-1)(n+1)+2
=n^2+1
$
and
$[x-5]+2[x-3]
=[n]+2[n+2]
le 3n+4
$.
Therefore,
if
$n^2+1
ge 3n+4
$,
the inequality is true.
This is
$n^2-3n-3 > 0$
or
$(n-frac32)^2
gt 3-frac94
=frac34
$
or,
using the positive square root,
$n
gt frac32+frac{sqrt{3}}{2}
=frac{3+sqrt{3}}{2}
$
which is true for
$n ge 3
$.
Therefore,
the inequality is true for
$x ge 8$.
I'll let you
work out the other cases.
answered Jan 29 at 21:04
marty cohenmarty cohen
74.9k549130
answered Jan 29 at 21:04
marty cohenmarty cohen
74.9k549130
answered Jan 29 at 21:04
marty cohenmarty cohen
74.9k549130
answered Jan 29 at 21:04
marty cohenmarty cohen
74.9k549130
74.9k549130
add a comment |
add a comment |
