Polynomial roots?












1












$begingroup$



For which real values of parameter $a$ both roots of polynom $f(x)=(a+1)x^2 + 2ax + a +3$ are positive numbers.




In solution they give 3 conditions that have to be satisfied.



1) $(a+1)f(0)>0$



2) $D>0$



3) $x_{0}>0$



First 2 i understand but not last one.



I dont know what $x_{0}>0$ is . When i calculate first 2 conditions i get



$ain left langle -infty,-3 right rangle cup left langle -1,-3/4 right rangle$



But books final solution is only $ left langle -1,-3/4 right rangle$






polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Neither do I, as $x_0$ is not defined.
    $endgroup$
    – Yuriy S
    Dec 29 '18 at 15:41










  • $begingroup$
    What is $;x_0;$ ...?
    $endgroup$
    – DonAntonio
    Dec 29 '18 at 15:42










  • $begingroup$
    The basic condition is $a ne -1$, otherwise it does not make sense to speak about both roots. Now you can solve the quadratic equation.
    $endgroup$
    – Paul Frost
    Dec 29 '18 at 16:49










  • $begingroup$
    If you don't know what $;x_0;$ is, how can you expect we'll know? Guessing?
    $endgroup$
    – DonAntonio
    Dec 29 '18 at 17:55










  • $begingroup$
    @Ivan be careful, the brackets have to be open (-1,-3/4).
    $endgroup$
    – user376343
    Dec 29 '18 at 20:09
















1












$begingroup$



For which real values of parameter $a$ both roots of polynom $f(x)=(a+1)x^2 + 2ax + a +3$ are positive numbers.




In solution they give 3 conditions that have to be satisfied.



1) $(a+1)f(0)>0$



2) $D>0$



3) $x_{0}>0$



First 2 i understand but not last one.



I dont know what $x_{0}>0$ is . When i calculate first 2 conditions i get



$ain left langle -infty,-3 right rangle cup left langle -1,-3/4 right rangle$



But books final solution is only $ left langle -1,-3/4 right rangle$






polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Neither do I, as $x_0$ is not defined.
    $endgroup$
    – Yuriy S
    Dec 29 '18 at 15:41










  • $begingroup$
    What is $;x_0;$ ...?
    $endgroup$
    – DonAntonio
    Dec 29 '18 at 15:42










  • $begingroup$
    The basic condition is $a ne -1$, otherwise it does not make sense to speak about both roots. Now you can solve the quadratic equation.
    $endgroup$
    – Paul Frost
    Dec 29 '18 at 16:49










  • $begingroup$
    If you don't know what $;x_0;$ is, how can you expect we'll know? Guessing?
    $endgroup$
    – DonAntonio
    Dec 29 '18 at 17:55










  • $begingroup$
    @Ivan be careful, the brackets have to be open (-1,-3/4).
    $endgroup$
    – user376343
    Dec 29 '18 at 20:09














1












1








1


0



$begingroup$



For which real values of parameter $a$ both roots of polynom $f(x)=(a+1)x^2 + 2ax + a +3$ are positive numbers.




In solution they give 3 conditions that have to be satisfied.



1) $(a+1)f(0)>0$



2) $D>0$



3) $x_{0}>0$



First 2 i understand but not last one.



I dont know what $x_{0}>0$ is . When i calculate first 2 conditions i get



$ain left langle -infty,-3 right rangle cup left langle -1,-3/4 right rangle$



But books final solution is only $ left langle -1,-3/4 right rangle$






polynomials







share|cite|improve this question











$endgroup$





For which real values of parameter $a$ both roots of polynom $f(x)=(a+1)x^2 + 2ax + a +3$ are positive numbers.




In solution they give 3 conditions that have to be satisfied.



1) $(a+1)f(0)>0$



2) $D>0$



3) $x_{0}>0$



First 2 i understand but not last one.



I dont know what $x_{0}>0$ is . When i calculate first 2 conditions i get



$ain left langle -infty,-3 right rangle cup left langle -1,-3/4 right rangle$



But books final solution is only $ left langle -1,-3/4 right rangle$






polynomials




polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 16:12







Ivan Ljilja

















asked Dec 29 '18 at 15:39









Ivan LjiljaIvan Ljilja

205




205








  • 1




    $begingroup$
    Neither do I, as $x_0$ is not defined.
    $endgroup$
    – Yuriy S
    Dec 29 '18 at 15:41










  • $begingroup$
    What is $;x_0;$ ...?
    $endgroup$
    – DonAntonio
    Dec 29 '18 at 15:42










  • $begingroup$
    The basic condition is $a ne -1$, otherwise it does not make sense to speak about both roots. Now you can solve the quadratic equation.
    $endgroup$
    – Paul Frost
    Dec 29 '18 at 16:49










  • $begingroup$
    If you don't know what $;x_0;$ is, how can you expect we'll know? Guessing?
    $endgroup$
    – DonAntonio
    Dec 29 '18 at 17:55










  • $begingroup$
    @Ivan be careful, the brackets have to be open (-1,-3/4).
    $endgroup$
    – user376343
    Dec 29 '18 at 20:09














  • 1




    $begingroup$
    Neither do I, as $x_0$ is not defined.
    $endgroup$
    – Yuriy S
    Dec 29 '18 at 15:41










  • $begingroup$
    What is $;x_0;$ ...?
    $endgroup$
    – DonAntonio
    Dec 29 '18 at 15:42










  • $begingroup$
    The basic condition is $a ne -1$, otherwise it does not make sense to speak about both roots. Now you can solve the quadratic equation.
    $endgroup$
    – Paul Frost
    Dec 29 '18 at 16:49










  • $begingroup$
    If you don't know what $;x_0;$ is, how can you expect we'll know? Guessing?
    $endgroup$
    – DonAntonio
    Dec 29 '18 at 17:55










  • $begingroup$
    @Ivan be careful, the brackets have to be open (-1,-3/4).
    $endgroup$
    – user376343
    Dec 29 '18 at 20:09








1




1




$begingroup$
Neither do I, as $x_0$ is not defined.
$endgroup$
– Yuriy S
Dec 29 '18 at 15:41




$begingroup$
Neither do I, as $x_0$ is not defined.
$endgroup$
– Yuriy S
Dec 29 '18 at 15:41












$begingroup$
What is $;x_0;$ ...?
$endgroup$
– DonAntonio
Dec 29 '18 at 15:42




$begingroup$
What is $;x_0;$ ...?
$endgroup$
– DonAntonio
Dec 29 '18 at 15:42












$begingroup$
The basic condition is $a ne -1$, otherwise it does not make sense to speak about both roots. Now you can solve the quadratic equation.
$endgroup$
– Paul Frost
Dec 29 '18 at 16:49




$begingroup$
The basic condition is $a ne -1$, otherwise it does not make sense to speak about both roots. Now you can solve the quadratic equation.
$endgroup$
– Paul Frost
Dec 29 '18 at 16:49












$begingroup$
If you don't know what $;x_0;$ is, how can you expect we'll know? Guessing?
$endgroup$
– DonAntonio
Dec 29 '18 at 17:55




$begingroup$
If you don't know what $;x_0;$ is, how can you expect we'll know? Guessing?
$endgroup$
– DonAntonio
Dec 29 '18 at 17:55












$begingroup$
@Ivan be careful, the brackets have to be open (-1,-3/4).
$endgroup$
– user376343
Dec 29 '18 at 20:09




$begingroup$
@Ivan be careful, the brackets have to be open (-1,-3/4).
$endgroup$
– user376343
Dec 29 '18 at 20:09










2 Answers
2






active

oldest

votes


















2












$begingroup$

The discriminant must be positive:



$$Delta=4a^2-4(a+1)(a+3)>0implies-16a-12>0implies a<-frac{12}{16}=-frac34$$



Both roots, say $;x_1,x_2;$ , positive:



$$begin{cases}0<x_1x_2=cfrac{a+3}{a+1}iff a<-3;;text{or};;a>-1\{}\text{And}\{}{}\
0<x_1+x_2=-frac{2a}{a+1}impliesfrac a{a+1}<0iff -1<a<0end{cases}$$



Now put things together and do a little mathematics here.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Discriminant...
    $endgroup$
    – user376343
    Dec 29 '18 at 20:07



















0












$begingroup$

It must be $$-frac{a}{a+1}+frac{sqrt{-4a-3}}{a+1}>0$$ and $$-frac{a}{a+1}-frac{sqrt{-4a-3}}{a+1}>0$$ and $$-4a-3>0$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The discriminant must be positive:



    $$Delta=4a^2-4(a+1)(a+3)>0implies-16a-12>0implies a<-frac{12}{16}=-frac34$$



    Both roots, say $;x_1,x_2;$ , positive:



    $$begin{cases}0<x_1x_2=cfrac{a+3}{a+1}iff a<-3;;text{or};;a>-1\{}\text{And}\{}{}\
    0<x_1+x_2=-frac{2a}{a+1}impliesfrac a{a+1}<0iff -1<a<0end{cases}$$



    Now put things together and do a little mathematics here.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Discriminant...
      $endgroup$
      – user376343
      Dec 29 '18 at 20:07
















    2












    $begingroup$

    The discriminant must be positive:



    $$Delta=4a^2-4(a+1)(a+3)>0implies-16a-12>0implies a<-frac{12}{16}=-frac34$$



    Both roots, say $;x_1,x_2;$ , positive:



    $$begin{cases}0<x_1x_2=cfrac{a+3}{a+1}iff a<-3;;text{or};;a>-1\{}\text{And}\{}{}\
    0<x_1+x_2=-frac{2a}{a+1}impliesfrac a{a+1}<0iff -1<a<0end{cases}$$



    Now put things together and do a little mathematics here.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Discriminant...
      $endgroup$
      – user376343
      Dec 29 '18 at 20:07














    2












    2








    2





    $begingroup$

    The discriminant must be positive:



    $$Delta=4a^2-4(a+1)(a+3)>0implies-16a-12>0implies a<-frac{12}{16}=-frac34$$



    Both roots, say $;x_1,x_2;$ , positive:



    $$begin{cases}0<x_1x_2=cfrac{a+3}{a+1}iff a<-3;;text{or};;a>-1\{}\text{And}\{}{}\
    0<x_1+x_2=-frac{2a}{a+1}impliesfrac a{a+1}<0iff -1<a<0end{cases}$$



    Now put things together and do a little mathematics here.






    share|cite|improve this answer











    $endgroup$



    The discriminant must be positive:



    $$Delta=4a^2-4(a+1)(a+3)>0implies-16a-12>0implies a<-frac{12}{16}=-frac34$$



    Both roots, say $;x_1,x_2;$ , positive:



    $$begin{cases}0<x_1x_2=cfrac{a+3}{a+1}iff a<-3;;text{or};;a>-1\{}\text{And}\{}{}\
    0<x_1+x_2=-frac{2a}{a+1}impliesfrac a{a+1}<0iff -1<a<0end{cases}$$



    Now put things together and do a little mathematics here.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 30 '18 at 15:33

























    answered Dec 29 '18 at 17:57









    DonAntonioDonAntonio

    180k1494233




    180k1494233












    • $begingroup$
      Discriminant...
      $endgroup$
      – user376343
      Dec 29 '18 at 20:07


















    • $begingroup$
      Discriminant...
      $endgroup$
      – user376343
      Dec 29 '18 at 20:07
















    $begingroup$
    Discriminant...
    $endgroup$
    – user376343
    Dec 29 '18 at 20:07




    $begingroup$
    Discriminant...
    $endgroup$
    – user376343
    Dec 29 '18 at 20:07











    0












    $begingroup$

    It must be $$-frac{a}{a+1}+frac{sqrt{-4a-3}}{a+1}>0$$ and $$-frac{a}{a+1}-frac{sqrt{-4a-3}}{a+1}>0$$ and $$-4a-3>0$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      It must be $$-frac{a}{a+1}+frac{sqrt{-4a-3}}{a+1}>0$$ and $$-frac{a}{a+1}-frac{sqrt{-4a-3}}{a+1}>0$$ and $$-4a-3>0$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        It must be $$-frac{a}{a+1}+frac{sqrt{-4a-3}}{a+1}>0$$ and $$-frac{a}{a+1}-frac{sqrt{-4a-3}}{a+1}>0$$ and $$-4a-3>0$$






        share|cite|improve this answer









        $endgroup$



        It must be $$-frac{a}{a+1}+frac{sqrt{-4a-3}}{a+1}>0$$ and $$-frac{a}{a+1}-frac{sqrt{-4a-3}}{a+1}>0$$ and $$-4a-3>0$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 15:53









        Dr. Sonnhard GraubnerDr. Sonnhard Graubner

        78k42866




        78k42866






























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