conditions of an inequation












0












$begingroup$


I need to find the solutions of this inequation.



frac{1-sqrt{1-4x^{2}}}{x}<3



I put the conditions 1-4x^{2}geq 0 and xneq 0 then I found x for frac{1-sqrt{1-4x^{2}}}{x}-3<0and I obtained xin (-infty ,0)which should be intersected with x from the conditions and I obtain in the final xin [-frac{1}{2},0] but it's not the right answer.






linear-algebra functions







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I need to find the solutions of this inequation.



    frac{1-sqrt{1-4x^{2}}}{x}<3



    I put the conditions 1-4x^{2}geq 0 and xneq 0 then I found x for frac{1-sqrt{1-4x^{2}}}{x}-3<0and I obtained xin (-infty ,0)which should be intersected with x from the conditions and I obtain in the final xin [-frac{1}{2},0] but it's not the right answer.






    linear-algebra functions







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I need to find the solutions of this inequation.



      frac{1-sqrt{1-4x^{2}}}{x}<3



      I put the conditions 1-4x^{2}geq 0 and xneq 0 then I found x for frac{1-sqrt{1-4x^{2}}}{x}-3<0and I obtained xin (-infty ,0)which should be intersected with x from the conditions and I obtain in the final xin [-frac{1}{2},0] but it's not the right answer.






      linear-algebra functions







      share|cite|improve this question









      $endgroup$




      I need to find the solutions of this inequation.



      frac{1-sqrt{1-4x^{2}}}{x}<3



      I put the conditions 1-4x^{2}geq 0 and xneq 0 then I found x for frac{1-sqrt{1-4x^{2}}}{x}-3<0and I obtained xin (-infty ,0)which should be intersected with x from the conditions and I obtain in the final xin [-frac{1}{2},0] but it's not the right answer.






      linear-algebra functions




      linear-algebra functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 19 at 18:15









      Vali ROVali RO

      686




      686






















          1 Answer
          1






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          2












          $begingroup$

          For the square root you have the condition $$1geq4x^2$$ this means $$-frac{1}{2}le xle frac{1}{2}$$ and now you must consider the cases $$x<0$$ or $$x>0$$ Good luck!
          Finally you will get $$-frac{1}{2}le xle frac{1}{2}$$ and $$xne 0$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the response.I got the right answer but I don't understand why I have to consider the cases x<0 and x>0.From that square root I understood that x is in [-1/2,1/2] interval.
            $endgroup$
            – Vali RO
            Jan 19 at 18:31










          • $begingroup$
            And $$xne 0$$ since $x$ is in the denominator.
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 19 at 18:32










          • $begingroup$
            I understood, thanks a lot!
            $endgroup$
            – Vali RO
            Jan 19 at 18:34






          • 1




            $begingroup$
            Nice, wish you a nice day!
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 19 at 18:39











          Your Answer





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          1 Answer
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          active

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          2












          $begingroup$

          For the square root you have the condition $$1geq4x^2$$ this means $$-frac{1}{2}le xle frac{1}{2}$$ and now you must consider the cases $$x<0$$ or $$x>0$$ Good luck!
          Finally you will get $$-frac{1}{2}le xle frac{1}{2}$$ and $$xne 0$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the response.I got the right answer but I don't understand why I have to consider the cases x<0 and x>0.From that square root I understood that x is in [-1/2,1/2] interval.
            $endgroup$
            – Vali RO
            Jan 19 at 18:31










          • $begingroup$
            And $$xne 0$$ since $x$ is in the denominator.
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 19 at 18:32










          • $begingroup$
            I understood, thanks a lot!
            $endgroup$
            – Vali RO
            Jan 19 at 18:34






          • 1




            $begingroup$
            Nice, wish you a nice day!
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 19 at 18:39
















          2












          $begingroup$

          For the square root you have the condition $$1geq4x^2$$ this means $$-frac{1}{2}le xle frac{1}{2}$$ and now you must consider the cases $$x<0$$ or $$x>0$$ Good luck!
          Finally you will get $$-frac{1}{2}le xle frac{1}{2}$$ and $$xne 0$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the response.I got the right answer but I don't understand why I have to consider the cases x<0 and x>0.From that square root I understood that x is in [-1/2,1/2] interval.
            $endgroup$
            – Vali RO
            Jan 19 at 18:31










          • $begingroup$
            And $$xne 0$$ since $x$ is in the denominator.
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 19 at 18:32










          • $begingroup$
            I understood, thanks a lot!
            $endgroup$
            – Vali RO
            Jan 19 at 18:34






          • 1




            $begingroup$
            Nice, wish you a nice day!
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 19 at 18:39














          2












          2








          2





          $begingroup$

          For the square root you have the condition $$1geq4x^2$$ this means $$-frac{1}{2}le xle frac{1}{2}$$ and now you must consider the cases $$x<0$$ or $$x>0$$ Good luck!
          Finally you will get $$-frac{1}{2}le xle frac{1}{2}$$ and $$xne 0$$






          share|cite|improve this answer









          $endgroup$



          For the square root you have the condition $$1geq4x^2$$ this means $$-frac{1}{2}le xle frac{1}{2}$$ and now you must consider the cases $$x<0$$ or $$x>0$$ Good luck!
          Finally you will get $$-frac{1}{2}le xle frac{1}{2}$$ and $$xne 0$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 19 at 18:22









          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

          76.7k42866




          76.7k42866












          • $begingroup$
            Thank you for the response.I got the right answer but I don't understand why I have to consider the cases x<0 and x>0.From that square root I understood that x is in [-1/2,1/2] interval.
            $endgroup$
            – Vali RO
            Jan 19 at 18:31










          • $begingroup$
            And $$xne 0$$ since $x$ is in the denominator.
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 19 at 18:32










          • $begingroup$
            I understood, thanks a lot!
            $endgroup$
            – Vali RO
            Jan 19 at 18:34






          • 1




            $begingroup$
            Nice, wish you a nice day!
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 19 at 18:39


















          • $begingroup$
            Thank you for the response.I got the right answer but I don't understand why I have to consider the cases x<0 and x>0.From that square root I understood that x is in [-1/2,1/2] interval.
            $endgroup$
            – Vali RO
            Jan 19 at 18:31










          • $begingroup$
            And $$xne 0$$ since $x$ is in the denominator.
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 19 at 18:32










          • $begingroup$
            I understood, thanks a lot!
            $endgroup$
            – Vali RO
            Jan 19 at 18:34






          • 1




            $begingroup$
            Nice, wish you a nice day!
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 19 at 18:39
















          $begingroup$
          Thank you for the response.I got the right answer but I don't understand why I have to consider the cases x<0 and x>0.From that square root I understood that x is in [-1/2,1/2] interval.
          $endgroup$
          – Vali RO
          Jan 19 at 18:31




          $begingroup$
          Thank you for the response.I got the right answer but I don't understand why I have to consider the cases x<0 and x>0.From that square root I understood that x is in [-1/2,1/2] interval.
          $endgroup$
          – Vali RO
          Jan 19 at 18:31












          $begingroup$
          And $$xne 0$$ since $x$ is in the denominator.
          $endgroup$
          – Dr. Sonnhard Graubner
          Jan 19 at 18:32




          $begingroup$
          And $$xne 0$$ since $x$ is in the denominator.
          $endgroup$
          – Dr. Sonnhard Graubner
          Jan 19 at 18:32












          $begingroup$
          I understood, thanks a lot!
          $endgroup$
          – Vali RO
          Jan 19 at 18:34




          $begingroup$
          I understood, thanks a lot!
          $endgroup$
          – Vali RO
          Jan 19 at 18:34




          1




          1




          $begingroup$
          Nice, wish you a nice day!
          $endgroup$
          – Dr. Sonnhard Graubner
          Jan 19 at 18:39




          $begingroup$
          Nice, wish you a nice day!
          $endgroup$
          – Dr. Sonnhard Graubner
          Jan 19 at 18:39


















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