Solving $ 2< x^2 -[x]<5$












5












$begingroup$


How to solve inequalities in which we have quadratic terms and greatest integer function.



$$ 2< x^2 -[x]<5$$.
[.] is greatest integer function.
Do we need to break into the cases as [0,1), [1,2) and so on?






functions inequality ceiling-function







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$endgroup$












  • $begingroup$
    Yes, that is exactly what you have to do.
    $endgroup$
    – Kavi Rama Murthy
    1 hour ago
















5












$begingroup$


How to solve inequalities in which we have quadratic terms and greatest integer function.



$$ 2< x^2 -[x]<5$$.
[.] is greatest integer function.
Do we need to break into the cases as [0,1), [1,2) and so on?






functions inequality ceiling-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, that is exactly what you have to do.
    $endgroup$
    – Kavi Rama Murthy
    1 hour ago














5












5








5


1



$begingroup$


How to solve inequalities in which we have quadratic terms and greatest integer function.



$$ 2< x^2 -[x]<5$$.
[.] is greatest integer function.
Do we need to break into the cases as [0,1), [1,2) and so on?






functions inequality ceiling-function







share|cite|improve this question











$endgroup$




How to solve inequalities in which we have quadratic terms and greatest integer function.



$$ 2< x^2 -[x]<5$$.
[.] is greatest integer function.
Do we need to break into the cases as [0,1), [1,2) and so on?






functions inequality ceiling-function




functions inequality ceiling-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 mins ago









YuiTo Cheng

2,4064937




2,4064937










asked 1 hour ago









mavericmaveric

92412




92412












  • $begingroup$
    Yes, that is exactly what you have to do.
    $endgroup$
    – Kavi Rama Murthy
    1 hour ago


















  • $begingroup$
    Yes, that is exactly what you have to do.
    $endgroup$
    – Kavi Rama Murthy
    1 hour ago
















$begingroup$
Yes, that is exactly what you have to do.
$endgroup$
– Kavi Rama Murthy
1 hour ago




$begingroup$
Yes, that is exactly what you have to do.
$endgroup$
– Kavi Rama Murthy
1 hour ago










2 Answers
2






active

oldest

votes


















6












$begingroup$

Hint: use $$x-1<[x]leq x$$ and then solve some quadratic inequalites like



$$ x^2-5<[x] implies x^2-5<ximplies x^2-x-6<0$$



so $xin (-2,3)$ and so on...




  • If $xin (-2,-1)$ then $[x]=-2 $ so $0<x^2<3$ so $-sqrt{3}<x<sqrt{3}$ so $xin(-sqrt{3},-1)$

  • If $xin [-1,0)$ ...






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    While breaking up into integer intervals always works, sometimes it is too much work.



    Let's say you instead had $$
    213 < x^2 - lfloor x rfloor < 505 $$
    You really don't want to consider $24$ cases individually.



    So you get clever and change this to a pair of simultaneous inequalities: Let $x = n + y$ with $n in Bbb Z$ and $0 leq y < 1$. Then the inequalities are
    $$
    left{ begin{array}{l} 213 < (n+y)^2 - n < 505 \ 0 leq y < 1 end{array} right.
    $$
    Then $$213 < n^2 + (2y-1)n +y^2 , 0 leq y < 1 implies 213 < n^2 + (1)n + 1
    $$
    and this shows that $n geq 15$ (you can get that with one application of the quadratic formula), cutting out a lot of the work. Similarly,
    $$ n^2 + (2y-1)n +y^2 < 505 , 0 leq y < 1 implies n^2 + (-1)n + 0 < 505
    $$
    and this shows that $n < 23$.



    Finally, you can also justify only examining the allowed values of $y$ in the boundary cases ($n = 15$ and $n = 22$); in between, any value of $y$ works. (This might not be the case for cubic expressions, for example.)



    So for example, on the low end, you would need to solve for $y$ in
    $$
    213 < (15+y)^2 - 15 = 210 - 30 y + y^2 \
    y^2 -30 y -3 > 0\
    y > frac12( sqrt{912} -30 ) \
    x > (sqrt{228} - 15) + 15 implies x > sqrt{228}
    $$
    and similarly you need to consider the case of $n=22$ to get the upper limit for $x$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      Hint: use $$x-1<[x]leq x$$ and then solve some quadratic inequalites like



      $$ x^2-5<[x] implies x^2-5<ximplies x^2-x-6<0$$



      so $xin (-2,3)$ and so on...




      • If $xin (-2,-1)$ then $[x]=-2 $ so $0<x^2<3$ so $-sqrt{3}<x<sqrt{3}$ so $xin(-sqrt{3},-1)$

      • If $xin [-1,0)$ ...






      share|cite|improve this answer











      $endgroup$


















        6












        $begingroup$

        Hint: use $$x-1<[x]leq x$$ and then solve some quadratic inequalites like



        $$ x^2-5<[x] implies x^2-5<ximplies x^2-x-6<0$$



        so $xin (-2,3)$ and so on...




        • If $xin (-2,-1)$ then $[x]=-2 $ so $0<x^2<3$ so $-sqrt{3}<x<sqrt{3}$ so $xin(-sqrt{3},-1)$

        • If $xin [-1,0)$ ...






        share|cite|improve this answer











        $endgroup$
















          6












          6








          6





          $begingroup$

          Hint: use $$x-1<[x]leq x$$ and then solve some quadratic inequalites like



          $$ x^2-5<[x] implies x^2-5<ximplies x^2-x-6<0$$



          so $xin (-2,3)$ and so on...




          • If $xin (-2,-1)$ then $[x]=-2 $ so $0<x^2<3$ so $-sqrt{3}<x<sqrt{3}$ so $xin(-sqrt{3},-1)$

          • If $xin [-1,0)$ ...






          share|cite|improve this answer











          $endgroup$



          Hint: use $$x-1<[x]leq x$$ and then solve some quadratic inequalites like



          $$ x^2-5<[x] implies x^2-5<ximplies x^2-x-6<0$$



          so $xin (-2,3)$ and so on...




          • If $xin (-2,-1)$ then $[x]=-2 $ so $0<x^2<3$ so $-sqrt{3}<x<sqrt{3}$ so $xin(-sqrt{3},-1)$

          • If $xin [-1,0)$ ...







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 29 mins ago

























          answered 38 mins ago









          Maria MazurMaria Mazur

          49.9k1361125




          49.9k1361125























              4












              $begingroup$

              While breaking up into integer intervals always works, sometimes it is too much work.



              Let's say you instead had $$
              213 < x^2 - lfloor x rfloor < 505 $$
              You really don't want to consider $24$ cases individually.



              So you get clever and change this to a pair of simultaneous inequalities: Let $x = n + y$ with $n in Bbb Z$ and $0 leq y < 1$. Then the inequalities are
              $$
              left{ begin{array}{l} 213 < (n+y)^2 - n < 505 \ 0 leq y < 1 end{array} right.
              $$
              Then $$213 < n^2 + (2y-1)n +y^2 , 0 leq y < 1 implies 213 < n^2 + (1)n + 1
              $$
              and this shows that $n geq 15$ (you can get that with one application of the quadratic formula), cutting out a lot of the work. Similarly,
              $$ n^2 + (2y-1)n +y^2 < 505 , 0 leq y < 1 implies n^2 + (-1)n + 0 < 505
              $$
              and this shows that $n < 23$.



              Finally, you can also justify only examining the allowed values of $y$ in the boundary cases ($n = 15$ and $n = 22$); in between, any value of $y$ works. (This might not be the case for cubic expressions, for example.)



              So for example, on the low end, you would need to solve for $y$ in
              $$
              213 < (15+y)^2 - 15 = 210 - 30 y + y^2 \
              y^2 -30 y -3 > 0\
              y > frac12( sqrt{912} -30 ) \
              x > (sqrt{228} - 15) + 15 implies x > sqrt{228}
              $$
              and similarly you need to consider the case of $n=22$ to get the upper limit for $x$.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                While breaking up into integer intervals always works, sometimes it is too much work.



                Let's say you instead had $$
                213 < x^2 - lfloor x rfloor < 505 $$
                You really don't want to consider $24$ cases individually.



                So you get clever and change this to a pair of simultaneous inequalities: Let $x = n + y$ with $n in Bbb Z$ and $0 leq y < 1$. Then the inequalities are
                $$
                left{ begin{array}{l} 213 < (n+y)^2 - n < 505 \ 0 leq y < 1 end{array} right.
                $$
                Then $$213 < n^2 + (2y-1)n +y^2 , 0 leq y < 1 implies 213 < n^2 + (1)n + 1
                $$
                and this shows that $n geq 15$ (you can get that with one application of the quadratic formula), cutting out a lot of the work. Similarly,
                $$ n^2 + (2y-1)n +y^2 < 505 , 0 leq y < 1 implies n^2 + (-1)n + 0 < 505
                $$
                and this shows that $n < 23$.



                Finally, you can also justify only examining the allowed values of $y$ in the boundary cases ($n = 15$ and $n = 22$); in between, any value of $y$ works. (This might not be the case for cubic expressions, for example.)



                So for example, on the low end, you would need to solve for $y$ in
                $$
                213 < (15+y)^2 - 15 = 210 - 30 y + y^2 \
                y^2 -30 y -3 > 0\
                y > frac12( sqrt{912} -30 ) \
                x > (sqrt{228} - 15) + 15 implies x > sqrt{228}
                $$
                and similarly you need to consider the case of $n=22$ to get the upper limit for $x$.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  While breaking up into integer intervals always works, sometimes it is too much work.



                  Let's say you instead had $$
                  213 < x^2 - lfloor x rfloor < 505 $$
                  You really don't want to consider $24$ cases individually.



                  So you get clever and change this to a pair of simultaneous inequalities: Let $x = n + y$ with $n in Bbb Z$ and $0 leq y < 1$. Then the inequalities are
                  $$
                  left{ begin{array}{l} 213 < (n+y)^2 - n < 505 \ 0 leq y < 1 end{array} right.
                  $$
                  Then $$213 < n^2 + (2y-1)n +y^2 , 0 leq y < 1 implies 213 < n^2 + (1)n + 1
                  $$
                  and this shows that $n geq 15$ (you can get that with one application of the quadratic formula), cutting out a lot of the work. Similarly,
                  $$ n^2 + (2y-1)n +y^2 < 505 , 0 leq y < 1 implies n^2 + (-1)n + 0 < 505
                  $$
                  and this shows that $n < 23$.



                  Finally, you can also justify only examining the allowed values of $y$ in the boundary cases ($n = 15$ and $n = 22$); in between, any value of $y$ works. (This might not be the case for cubic expressions, for example.)



                  So for example, on the low end, you would need to solve for $y$ in
                  $$
                  213 < (15+y)^2 - 15 = 210 - 30 y + y^2 \
                  y^2 -30 y -3 > 0\
                  y > frac12( sqrt{912} -30 ) \
                  x > (sqrt{228} - 15) + 15 implies x > sqrt{228}
                  $$
                  and similarly you need to consider the case of $n=22$ to get the upper limit for $x$.






                  share|cite|improve this answer









                  $endgroup$



                  While breaking up into integer intervals always works, sometimes it is too much work.



                  Let's say you instead had $$
                  213 < x^2 - lfloor x rfloor < 505 $$
                  You really don't want to consider $24$ cases individually.



                  So you get clever and change this to a pair of simultaneous inequalities: Let $x = n + y$ with $n in Bbb Z$ and $0 leq y < 1$. Then the inequalities are
                  $$
                  left{ begin{array}{l} 213 < (n+y)^2 - n < 505 \ 0 leq y < 1 end{array} right.
                  $$
                  Then $$213 < n^2 + (2y-1)n +y^2 , 0 leq y < 1 implies 213 < n^2 + (1)n + 1
                  $$
                  and this shows that $n geq 15$ (you can get that with one application of the quadratic formula), cutting out a lot of the work. Similarly,
                  $$ n^2 + (2y-1)n +y^2 < 505 , 0 leq y < 1 implies n^2 + (-1)n + 0 < 505
                  $$
                  and this shows that $n < 23$.



                  Finally, you can also justify only examining the allowed values of $y$ in the boundary cases ($n = 15$ and $n = 22$); in between, any value of $y$ works. (This might not be the case for cubic expressions, for example.)



                  So for example, on the low end, you would need to solve for $y$ in
                  $$
                  213 < (15+y)^2 - 15 = 210 - 30 y + y^2 \
                  y^2 -30 y -3 > 0\
                  y > frac12( sqrt{912} -30 ) \
                  x > (sqrt{228} - 15) + 15 implies x > sqrt{228}
                  $$
                  and similarly you need to consider the case of $n=22$ to get the upper limit for $x$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 36 mins ago









                  Mark FischlerMark Fischler

                  34.1k12552




                  34.1k12552






























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