Silly Question on Definition of real solutions and real roots












1












$begingroup$


For $aneq0$, the equation $ax^2+b|x|+c=0$ has k real solutions and p real roots.




Here, my doubt is what does the solution means and how it is different from real roots?




Thanks for clearing my silly doubt.






roots quadratics







share|cite|improve this question











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  • 2




    $begingroup$
    In this case, they are the same. Not a silly question at all!
    $endgroup$
    – Don Thousand
    Dec 19 '18 at 16:54










  • $begingroup$
    Once you realized that $k=p$ always, were you wondering what the possible values of $k$ were?
    $endgroup$
    – Lubin
    Dec 19 '18 at 18:09










  • $begingroup$
    @Lubin No, actually I wanted to know if there is any difference between these two words.
    $endgroup$
    – jayant98
    Dec 19 '18 at 21:28
















1












$begingroup$


For $aneq0$, the equation $ax^2+b|x|+c=0$ has k real solutions and p real roots.




Here, my doubt is what does the solution means and how it is different from real roots?




Thanks for clearing my silly doubt.






roots quadratics







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    In this case, they are the same. Not a silly question at all!
    $endgroup$
    – Don Thousand
    Dec 19 '18 at 16:54










  • $begingroup$
    Once you realized that $k=p$ always, were you wondering what the possible values of $k$ were?
    $endgroup$
    – Lubin
    Dec 19 '18 at 18:09










  • $begingroup$
    @Lubin No, actually I wanted to know if there is any difference between these two words.
    $endgroup$
    – jayant98
    Dec 19 '18 at 21:28














1












1








1


0



$begingroup$


For $aneq0$, the equation $ax^2+b|x|+c=0$ has k real solutions and p real roots.




Here, my doubt is what does the solution means and how it is different from real roots?




Thanks for clearing my silly doubt.






roots quadratics







share|cite|improve this question











$endgroup$




For $aneq0$, the equation $ax^2+b|x|+c=0$ has k real solutions and p real roots.




Here, my doubt is what does the solution means and how it is different from real roots?




Thanks for clearing my silly doubt.






roots quadratics




roots quadratics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 17:09







jayant98

















asked Dec 19 '18 at 16:45









jayant98jayant98

605216




605216








  • 2




    $begingroup$
    In this case, they are the same. Not a silly question at all!
    $endgroup$
    – Don Thousand
    Dec 19 '18 at 16:54










  • $begingroup$
    Once you realized that $k=p$ always, were you wondering what the possible values of $k$ were?
    $endgroup$
    – Lubin
    Dec 19 '18 at 18:09










  • $begingroup$
    @Lubin No, actually I wanted to know if there is any difference between these two words.
    $endgroup$
    – jayant98
    Dec 19 '18 at 21:28














  • 2




    $begingroup$
    In this case, they are the same. Not a silly question at all!
    $endgroup$
    – Don Thousand
    Dec 19 '18 at 16:54










  • $begingroup$
    Once you realized that $k=p$ always, were you wondering what the possible values of $k$ were?
    $endgroup$
    – Lubin
    Dec 19 '18 at 18:09










  • $begingroup$
    @Lubin No, actually I wanted to know if there is any difference between these two words.
    $endgroup$
    – jayant98
    Dec 19 '18 at 21:28








2




2




$begingroup$
In this case, they are the same. Not a silly question at all!
$endgroup$
– Don Thousand
Dec 19 '18 at 16:54




$begingroup$
In this case, they are the same. Not a silly question at all!
$endgroup$
– Don Thousand
Dec 19 '18 at 16:54












$begingroup$
Once you realized that $k=p$ always, were you wondering what the possible values of $k$ were?
$endgroup$
– Lubin
Dec 19 '18 at 18:09




$begingroup$
Once you realized that $k=p$ always, were you wondering what the possible values of $k$ were?
$endgroup$
– Lubin
Dec 19 '18 at 18:09












$begingroup$
@Lubin No, actually I wanted to know if there is any difference between these two words.
$endgroup$
– jayant98
Dec 19 '18 at 21:28




$begingroup$
@Lubin No, actually I wanted to know if there is any difference between these two words.
$endgroup$
– jayant98
Dec 19 '18 at 21:28










1 Answer
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$begingroup$

They're the same.




If you want to be super pedantic – don't be – the equation has solutions, not roots.



For those who like getting into pointless arguments, there's a reasonable case that a function has roots and no solutions, whereas an equation has solutions but no roots. But that's fighting the same sort of losing battle as the one fights who argues that "data" is a plural and not a mass noun; everyone is sure to ignore you if you try and impose this as a rule of grammar. You'd have a better chance of convincing people that "awesome" means "terrible and mighty" and not "brilliant and wonderful".






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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    They're the same.




    If you want to be super pedantic – don't be – the equation has solutions, not roots.



    For those who like getting into pointless arguments, there's a reasonable case that a function has roots and no solutions, whereas an equation has solutions but no roots. But that's fighting the same sort of losing battle as the one fights who argues that "data" is a plural and not a mass noun; everyone is sure to ignore you if you try and impose this as a rule of grammar. You'd have a better chance of convincing people that "awesome" means "terrible and mighty" and not "brilliant and wonderful".






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      They're the same.




      If you want to be super pedantic – don't be – the equation has solutions, not roots.



      For those who like getting into pointless arguments, there's a reasonable case that a function has roots and no solutions, whereas an equation has solutions but no roots. But that's fighting the same sort of losing battle as the one fights who argues that "data" is a plural and not a mass noun; everyone is sure to ignore you if you try and impose this as a rule of grammar. You'd have a better chance of convincing people that "awesome" means "terrible and mighty" and not "brilliant and wonderful".






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        They're the same.




        If you want to be super pedantic – don't be – the equation has solutions, not roots.



        For those who like getting into pointless arguments, there's a reasonable case that a function has roots and no solutions, whereas an equation has solutions but no roots. But that's fighting the same sort of losing battle as the one fights who argues that "data" is a plural and not a mass noun; everyone is sure to ignore you if you try and impose this as a rule of grammar. You'd have a better chance of convincing people that "awesome" means "terrible and mighty" and not "brilliant and wonderful".






        share|cite|improve this answer









        $endgroup$



        They're the same.




        If you want to be super pedantic – don't be – the equation has solutions, not roots.



        For those who like getting into pointless arguments, there's a reasonable case that a function has roots and no solutions, whereas an equation has solutions but no roots. But that's fighting the same sort of losing battle as the one fights who argues that "data" is a plural and not a mass noun; everyone is sure to ignore you if you try and impose this as a rule of grammar. You'd have a better chance of convincing people that "awesome" means "terrible and mighty" and not "brilliant and wonderful".







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 17:28









        Patrick StevensPatrick Stevens

        28.7k52874




        28.7k52874






























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