Solve the inequality $|x^2-2|<1$












1












$begingroup$


I'm trying to solve inequalities that include absolute values. The way we solved these questions in class is different than what I am used to. The teacher would take the positive and negative possibilities of the inequality and treat them separately. I am trying to solve this as a union of intervals.



$$|x^2-2|<1$$



Normally I would just write this as:
$$-sqrt{3} < x < -1 text{ and } 1 < x < sqrt{3}$$



However this is not correct I don't believe.






inequality







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    That looks correct to me. when you break it out, you will have 1<x^2<3 , then -3<x<3 and x<-1 and x>1. When you combine, you will get your answer.
    $endgroup$
    – Peter
    Oct 3 '13 at 21:20
















1












$begingroup$


I'm trying to solve inequalities that include absolute values. The way we solved these questions in class is different than what I am used to. The teacher would take the positive and negative possibilities of the inequality and treat them separately. I am trying to solve this as a union of intervals.



$$|x^2-2|<1$$



Normally I would just write this as:
$$-sqrt{3} < x < -1 text{ and } 1 < x < sqrt{3}$$



However this is not correct I don't believe.






inequality







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    That looks correct to me. when you break it out, you will have 1<x^2<3 , then -3<x<3 and x<-1 and x>1. When you combine, you will get your answer.
    $endgroup$
    – Peter
    Oct 3 '13 at 21:20














1












1








1





$begingroup$


I'm trying to solve inequalities that include absolute values. The way we solved these questions in class is different than what I am used to. The teacher would take the positive and negative possibilities of the inequality and treat them separately. I am trying to solve this as a union of intervals.



$$|x^2-2|<1$$



Normally I would just write this as:
$$-sqrt{3} < x < -1 text{ and } 1 < x < sqrt{3}$$



However this is not correct I don't believe.






inequality







share|cite|improve this question











$endgroup$




I'm trying to solve inequalities that include absolute values. The way we solved these questions in class is different than what I am used to. The teacher would take the positive and negative possibilities of the inequality and treat them separately. I am trying to solve this as a union of intervals.



$$|x^2-2|<1$$



Normally I would just write this as:
$$-sqrt{3} < x < -1 text{ and } 1 < x < sqrt{3}$$



However this is not correct I don't believe.






inequality




inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 17:35









Larry

2,53031131




2,53031131










asked Oct 3 '13 at 21:12









Ryan JamesRyan James

91




91








  • 3




    $begingroup$
    That looks correct to me. when you break it out, you will have 1<x^2<3 , then -3<x<3 and x<-1 and x>1. When you combine, you will get your answer.
    $endgroup$
    – Peter
    Oct 3 '13 at 21:20














  • 3




    $begingroup$
    That looks correct to me. when you break it out, you will have 1<x^2<3 , then -3<x<3 and x<-1 and x>1. When you combine, you will get your answer.
    $endgroup$
    – Peter
    Oct 3 '13 at 21:20








3




3




$begingroup$
That looks correct to me. when you break it out, you will have 1<x^2<3 , then -3<x<3 and x<-1 and x>1. When you combine, you will get your answer.
$endgroup$
– Peter
Oct 3 '13 at 21:20




$begingroup$
That looks correct to me. when you break it out, you will have 1<x^2<3 , then -3<x<3 and x<-1 and x>1. When you combine, you will get your answer.
$endgroup$
– Peter
Oct 3 '13 at 21:20










3 Answers
3






active

oldest

votes


















6












$begingroup$

Hints:



$$|x^2-2|<1iff -1<x^2-2<1iff 1<x^2<3ldots$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    $$|x^2-2|<1iff-1< x^2-2 <1$$
    $$-1< x^2-2 <1iff1<x^2<3$$
    $$1<x^2<3iff x^2-1>0space &space x^2-3<0$$
    $$x^2-1>0space &space x^2-3<0 iff xin((-infty,-1)cup(1,infty))cap(-sqrt 3,sqrt 3)$$
    $$ xin((-infty,-1)cup(1,infty))cap(-sqrt 3,sqrt 3)iff xin(-sqrt3,-1)cup(1,sqrt3)$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Change the intersection symbol in your last line to a union one, or else that denotes the empty set...
      $endgroup$
      – DonAntonio
      Oct 3 '13 at 21:28












    • $begingroup$
      Thanks for this. Can you explain the step when you get the inequality of: x^2-1>0
      $endgroup$
      – Ryan James
      Oct 3 '13 at 21:29












    • $begingroup$
      $ddotsmile$ +1
      $endgroup$
      – mrs
      Oct 6 '13 at 16:06



















    1












    $begingroup$

    $|x^2-2|<1$



    $-1<x^2-2<1$



    $1<x^2$ or $x^2<3$



    $1<x$, $-1>x$, or $-sqrt{3}<x<sqrt{3}$



    Therefore,
    $-sqrt{3}<x<-1$ and $1<x<sqrt{3}$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Your last line's missing a minus sign to the left of the leftmost $;sqrt3;$
      $endgroup$
      – DonAntonio
      Oct 3 '13 at 21:31












    • $begingroup$
      Instead of an 'or' in the third line, you should have an 'and'.
      $endgroup$
      – robjohn♦
      Oct 3 '13 at 21:33












    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    Hints:



    $$|x^2-2|<1iff -1<x^2-2<1iff 1<x^2<3ldots$$






    share|cite|improve this answer









    $endgroup$


















      6












      $begingroup$

      Hints:



      $$|x^2-2|<1iff -1<x^2-2<1iff 1<x^2<3ldots$$






      share|cite|improve this answer









      $endgroup$
















        6












        6








        6





        $begingroup$

        Hints:



        $$|x^2-2|<1iff -1<x^2-2<1iff 1<x^2<3ldots$$






        share|cite|improve this answer









        $endgroup$



        Hints:



        $$|x^2-2|<1iff -1<x^2-2<1iff 1<x^2<3ldots$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 3 '13 at 21:21









        DonAntonioDonAntonio

        180k1494233




        180k1494233























            2












            $begingroup$

            $$|x^2-2|<1iff-1< x^2-2 <1$$
            $$-1< x^2-2 <1iff1<x^2<3$$
            $$1<x^2<3iff x^2-1>0space &space x^2-3<0$$
            $$x^2-1>0space &space x^2-3<0 iff xin((-infty,-1)cup(1,infty))cap(-sqrt 3,sqrt 3)$$
            $$ xin((-infty,-1)cup(1,infty))cap(-sqrt 3,sqrt 3)iff xin(-sqrt3,-1)cup(1,sqrt3)$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Change the intersection symbol in your last line to a union one, or else that denotes the empty set...
              $endgroup$
              – DonAntonio
              Oct 3 '13 at 21:28












            • $begingroup$
              Thanks for this. Can you explain the step when you get the inequality of: x^2-1>0
              $endgroup$
              – Ryan James
              Oct 3 '13 at 21:29












            • $begingroup$
              $ddotsmile$ +1
              $endgroup$
              – mrs
              Oct 6 '13 at 16:06
















            2












            $begingroup$

            $$|x^2-2|<1iff-1< x^2-2 <1$$
            $$-1< x^2-2 <1iff1<x^2<3$$
            $$1<x^2<3iff x^2-1>0space &space x^2-3<0$$
            $$x^2-1>0space &space x^2-3<0 iff xin((-infty,-1)cup(1,infty))cap(-sqrt 3,sqrt 3)$$
            $$ xin((-infty,-1)cup(1,infty))cap(-sqrt 3,sqrt 3)iff xin(-sqrt3,-1)cup(1,sqrt3)$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Change the intersection symbol in your last line to a union one, or else that denotes the empty set...
              $endgroup$
              – DonAntonio
              Oct 3 '13 at 21:28












            • $begingroup$
              Thanks for this. Can you explain the step when you get the inequality of: x^2-1>0
              $endgroup$
              – Ryan James
              Oct 3 '13 at 21:29












            • $begingroup$
              $ddotsmile$ +1
              $endgroup$
              – mrs
              Oct 6 '13 at 16:06














            2












            2








            2





            $begingroup$

            $$|x^2-2|<1iff-1< x^2-2 <1$$
            $$-1< x^2-2 <1iff1<x^2<3$$
            $$1<x^2<3iff x^2-1>0space &space x^2-3<0$$
            $$x^2-1>0space &space x^2-3<0 iff xin((-infty,-1)cup(1,infty))cap(-sqrt 3,sqrt 3)$$
            $$ xin((-infty,-1)cup(1,infty))cap(-sqrt 3,sqrt 3)iff xin(-sqrt3,-1)cup(1,sqrt3)$$






            share|cite|improve this answer











            $endgroup$



            $$|x^2-2|<1iff-1< x^2-2 <1$$
            $$-1< x^2-2 <1iff1<x^2<3$$
            $$1<x^2<3iff x^2-1>0space &space x^2-3<0$$
            $$x^2-1>0space &space x^2-3<0 iff xin((-infty,-1)cup(1,infty))cap(-sqrt 3,sqrt 3)$$
            $$ xin((-infty,-1)cup(1,infty))cap(-sqrt 3,sqrt 3)iff xin(-sqrt3,-1)cup(1,sqrt3)$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Oct 3 '13 at 21:56

























            answered Oct 3 '13 at 21:25









            Adi DaniAdi Dani

            15.3k32246




            15.3k32246












            • $begingroup$
              Change the intersection symbol in your last line to a union one, or else that denotes the empty set...
              $endgroup$
              – DonAntonio
              Oct 3 '13 at 21:28












            • $begingroup$
              Thanks for this. Can you explain the step when you get the inequality of: x^2-1>0
              $endgroup$
              – Ryan James
              Oct 3 '13 at 21:29












            • $begingroup$
              $ddotsmile$ +1
              $endgroup$
              – mrs
              Oct 6 '13 at 16:06


















            • $begingroup$
              Change the intersection symbol in your last line to a union one, or else that denotes the empty set...
              $endgroup$
              – DonAntonio
              Oct 3 '13 at 21:28












            • $begingroup$
              Thanks for this. Can you explain the step when you get the inequality of: x^2-1>0
              $endgroup$
              – Ryan James
              Oct 3 '13 at 21:29












            • $begingroup$
              $ddotsmile$ +1
              $endgroup$
              – mrs
              Oct 6 '13 at 16:06
















            $begingroup$
            Change the intersection symbol in your last line to a union one, or else that denotes the empty set...
            $endgroup$
            – DonAntonio
            Oct 3 '13 at 21:28






            $begingroup$
            Change the intersection symbol in your last line to a union one, or else that denotes the empty set...
            $endgroup$
            – DonAntonio
            Oct 3 '13 at 21:28














            $begingroup$
            Thanks for this. Can you explain the step when you get the inequality of: x^2-1>0
            $endgroup$
            – Ryan James
            Oct 3 '13 at 21:29






            $begingroup$
            Thanks for this. Can you explain the step when you get the inequality of: x^2-1>0
            $endgroup$
            – Ryan James
            Oct 3 '13 at 21:29














            $begingroup$
            $ddotsmile$ +1
            $endgroup$
            – mrs
            Oct 6 '13 at 16:06




            $begingroup$
            $ddotsmile$ +1
            $endgroup$
            – mrs
            Oct 6 '13 at 16:06











            1












            $begingroup$

            $|x^2-2|<1$



            $-1<x^2-2<1$



            $1<x^2$ or $x^2<3$



            $1<x$, $-1>x$, or $-sqrt{3}<x<sqrt{3}$



            Therefore,
            $-sqrt{3}<x<-1$ and $1<x<sqrt{3}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Your last line's missing a minus sign to the left of the leftmost $;sqrt3;$
              $endgroup$
              – DonAntonio
              Oct 3 '13 at 21:31












            • $begingroup$
              Instead of an 'or' in the third line, you should have an 'and'.
              $endgroup$
              – robjohn♦
              Oct 3 '13 at 21:33
















            1












            $begingroup$

            $|x^2-2|<1$



            $-1<x^2-2<1$



            $1<x^2$ or $x^2<3$



            $1<x$, $-1>x$, or $-sqrt{3}<x<sqrt{3}$



            Therefore,
            $-sqrt{3}<x<-1$ and $1<x<sqrt{3}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Your last line's missing a minus sign to the left of the leftmost $;sqrt3;$
              $endgroup$
              – DonAntonio
              Oct 3 '13 at 21:31












            • $begingroup$
              Instead of an 'or' in the third line, you should have an 'and'.
              $endgroup$
              – robjohn♦
              Oct 3 '13 at 21:33














            1












            1








            1





            $begingroup$

            $|x^2-2|<1$



            $-1<x^2-2<1$



            $1<x^2$ or $x^2<3$



            $1<x$, $-1>x$, or $-sqrt{3}<x<sqrt{3}$



            Therefore,
            $-sqrt{3}<x<-1$ and $1<x<sqrt{3}$






            share|cite|improve this answer











            $endgroup$



            $|x^2-2|<1$



            $-1<x^2-2<1$



            $1<x^2$ or $x^2<3$



            $1<x$, $-1>x$, or $-sqrt{3}<x<sqrt{3}$



            Therefore,
            $-sqrt{3}<x<-1$ and $1<x<sqrt{3}$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Oct 3 '13 at 21:40

























            answered Oct 3 '13 at 21:22









            Math.StackExchangeMath.StackExchange

            2,402921




            2,402921












            • $begingroup$
              Your last line's missing a minus sign to the left of the leftmost $;sqrt3;$
              $endgroup$
              – DonAntonio
              Oct 3 '13 at 21:31












            • $begingroup$
              Instead of an 'or' in the third line, you should have an 'and'.
              $endgroup$
              – robjohn♦
              Oct 3 '13 at 21:33


















            • $begingroup$
              Your last line's missing a minus sign to the left of the leftmost $;sqrt3;$
              $endgroup$
              – DonAntonio
              Oct 3 '13 at 21:31












            • $begingroup$
              Instead of an 'or' in the third line, you should have an 'and'.
              $endgroup$
              – robjohn♦
              Oct 3 '13 at 21:33
















            $begingroup$
            Your last line's missing a minus sign to the left of the leftmost $;sqrt3;$
            $endgroup$
            – DonAntonio
            Oct 3 '13 at 21:31






            $begingroup$
            Your last line's missing a minus sign to the left of the leftmost $;sqrt3;$
            $endgroup$
            – DonAntonio
            Oct 3 '13 at 21:31














            $begingroup$
            Instead of an 'or' in the third line, you should have an 'and'.
            $endgroup$
            – robjohn♦
            Oct 3 '13 at 21:33




            $begingroup$
            Instead of an 'or' in the third line, you should have an 'and'.
            $endgroup$
            – robjohn♦
            Oct 3 '13 at 21:33


















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