Silly Question on Definition of real solutions and real roots
$begingroup$
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
asked Dec 19 '18 at 16:45
jayant98jayant98
552116
$endgroup$
add a comment |
$begingroup$
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
asked Dec 19 '18 at 16:45
jayant98jayant98
552116
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 19 '18 at 16:45
jayant98jayant98
552116
$endgroup$
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
asked Dec 19 '18 at 16:45
jayant98jayant98
552116
asked Dec 19 '18 at 16:45
jayant98jayant98
552116
edited Dec 19 '18 at 17:09
jayant98
asked Dec 19 '18 at 16:45
jayant98jayant98
552116
asked Dec 19 '18 at 16:45
jayant98jayant98
552116
asked Dec 19 '18 at 16:45
jayant98jayant98
552116
552116
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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".
answered Dec 19 '18 at 17:28
Patrick StevensPatrick Stevens
28.6k52874
$endgroup$
add a comment |
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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".
answered Dec 19 '18 at 17:28
Patrick StevensPatrick Stevens
28.6k52874
$endgroup$
add a comment |
$begingroup$
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".
answered Dec 19 '18 at 17:28
Patrick StevensPatrick Stevens
28.6k52874
$endgroup$
add a comment |
$begingroup$
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".
answered Dec 19 '18 at 17:28
Patrick StevensPatrick Stevens
28.6k52874
$endgroup$
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".
answered Dec 19 '18 at 17:28
Patrick StevensPatrick Stevens
28.6k52874
answered Dec 19 '18 at 17:28
Patrick StevensPatrick Stevens
28.6k52874
answered Dec 19 '18 at 17:28
Patrick StevensPatrick Stevens
28.6k52874
answered Dec 19 '18 at 17:28
Patrick StevensPatrick Stevens
28.6k52874
28.6k52874
add a comment |
add a comment |
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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