Mixing
Finding the number of solutions to cubic polynomial equations
0
$begingroup$
I am looking to find the number of solutions to the following equation:
$$x^3 +0.1=10x$$
Looking at the graph of the expressions on each side of the equation I understand that there are 3 solutions.
How would I go about illustrating this point algebraically? My 1st instinct is to equate everything to 0
$$x^3 –10x + 0.1=0$$
but I would have no idea how to factorize this equation. Any tips?
factoring
asked Jan 7 at 13:49
esc1234esc1234
82
$endgroup$
|
show 3 more comments
0
$begingroup$
I am looking to find the number of solutions to the following equation:
$$x^3 +0.1=10x$$
Looking at the graph of the expressions on each side of the equation I understand that there are 3 solutions.
How would I go about illustrating this point algebraically? My 1st instinct is to equate everything to 0
$$x^3 –10x + 0.1=0$$
but I would have no idea how to factorize this equation. Any tips?
factoring
asked Jan 7 at 13:49
esc1234esc1234
82
$endgroup$
2
$begingroup$
The three roots aren't especially pleasant. If you need values for them, you'll probably want to use numerical methods. If all you want is to see that there are three roots, note that $p(-10)<0, p(0)>0,p(1)<0, p(10)>0$.
$endgroup$
– lulu
Jan 7 at 13:53
$begingroup$
so without the use of a graphing calculator or attempting to graph the equation, is the only way to find the number of solutions is to test input values from various intervals that look pertinent?
$endgroup$
– esc1234
Jan 7 at 13:58
$begingroup$
I didn't say it was the only way. But it's a good way.
$endgroup$
– lulu
Jan 7 at 13:59
$begingroup$
yeah that's okay. I'm just looking for viable methods of tackling the problem algebraically without having to depend on visual graphs as some equations may be cumbersome to graph.
$endgroup$
– esc1234
Jan 7 at 14:00
$begingroup$
Well, evaluating the function at a few points is generally a good place to start. Here you could also use calculus to find the local max and min, but I don't see that as any easier. Descartes' Rule of Signs used to be a popular method but it has largely fallen out of favor as computational devices have become easy to use.
$endgroup$
– lulu
Jan 7 at 14:03
|
show 3 more comments
0
0
0
$begingroup$
I am looking to find the number of solutions to the following equation:
$$x^3 +0.1=10x$$
Looking at the graph of the expressions on each side of the equation I understand that there are 3 solutions.
How would I go about illustrating this point algebraically? My 1st instinct is to equate everything to 0
$$x^3 –10x + 0.1=0$$
but I would have no idea how to factorize this equation. Any tips?
factoring
asked Jan 7 at 13:49
esc1234esc1234
82
$endgroup$
I am looking to find the number of solutions to the following equation:
$$x^3 +0.1=10x$$
Looking at the graph of the expressions on each side of the equation I understand that there are 3 solutions.
How would I go about illustrating this point algebraically? My 1st instinct is to equate everything to 0
$$x^3 –10x + 0.1=0$$
but I would have no idea how to factorize this equation. Any tips?
factoring
factoring
asked Jan 7 at 13:49
esc1234esc1234
82
asked Jan 7 at 13:49
esc1234esc1234
82
asked Jan 7 at 13:49
esc1234esc1234
82
asked Jan 7 at 13:49
esc1234esc1234
82
asked Jan 7 at 13:49
esc1234esc1234
82
82
2
$begingroup$
The three roots aren't especially pleasant. If you need values for them, you'll probably want to use numerical methods. If all you want is to see that there are three roots, note that $p(-10)<0, p(0)>0,p(1)<0, p(10)>0$.
$endgroup$
– lulu
Jan 7 at 13:53
$begingroup$
so without the use of a graphing calculator or attempting to graph the equation, is the only way to find the number of solutions is to test input values from various intervals that look pertinent?
$endgroup$
– esc1234
Jan 7 at 13:58
$begingroup$
I didn't say it was the only way. But it's a good way.
$endgroup$
– lulu
Jan 7 at 13:59
$begingroup$
yeah that's okay. I'm just looking for viable methods of tackling the problem algebraically without having to depend on visual graphs as some equations may be cumbersome to graph.
$endgroup$
– esc1234
Jan 7 at 14:00
$begingroup$
Well, evaluating the function at a few points is generally a good place to start. Here you could also use calculus to find the local max and min, but I don't see that as any easier. Descartes' Rule of Signs used to be a popular method but it has largely fallen out of favor as computational devices have become easy to use.
$endgroup$
– lulu
Jan 7 at 14:03
|
show 3 more comments
2
$begingroup$
The three roots aren't especially pleasant. If you need values for them, you'll probably want to use numerical methods. If all you want is to see that there are three roots, note that $p(-10)<0, p(0)>0,p(1)<0, p(10)>0$.
$endgroup$
– lulu
Jan 7 at 13:53
$begingroup$
so without the use of a graphing calculator or attempting to graph the equation, is the only way to find the number of solutions is to test input values from various intervals that look pertinent?
$endgroup$
– esc1234
Jan 7 at 13:58
$begingroup$
I didn't say it was the only way. But it's a good way.
$endgroup$
– lulu
Jan 7 at 13:59
$begingroup$
yeah that's okay. I'm just looking for viable methods of tackling the problem algebraically without having to depend on visual graphs as some equations may be cumbersome to graph.
$endgroup$
– esc1234
Jan 7 at 14:00
$begingroup$
Well, evaluating the function at a few points is generally a good place to start. Here you could also use calculus to find the local max and min, but I don't see that as any easier. Descartes' Rule of Signs used to be a popular method but it has largely fallen out of favor as computational devices have become easy to use.
$endgroup$
– lulu
Jan 7 at 14:03
2
2
$begingroup$
The three roots aren't especially pleasant. If you need values for them, you'll probably want to use numerical methods. If all you want is to see that there are three roots, note that $p(-10)<0, p(0)>0,p(1)<0, p(10)>0$.
$endgroup$
– lulu
Jan 7 at 13:53
$begingroup$
The three roots aren't especially pleasant. If you need values for them, you'll probably want to use numerical methods. If all you want is to see that there are three roots, note that $p(-10)<0, p(0)>0,p(1)<0, p(10)>0$.
$endgroup$
– lulu
Jan 7 at 13:53
$begingroup$
so without the use of a graphing calculator or attempting to graph the equation, is the only way to find the number of solutions is to test input values from various intervals that look pertinent?
$endgroup$
– esc1234
Jan 7 at 13:58
$begingroup$
so without the use of a graphing calculator or attempting to graph the equation, is the only way to find the number of solutions is to test input values from various intervals that look pertinent?
$endgroup$
– esc1234
Jan 7 at 13:58
$begingroup$
I didn't say it was the only way. But it's a good way.
$endgroup$
– lulu
Jan 7 at 13:59
$begingroup$
I didn't say it was the only way. But it's a good way.
$endgroup$
– lulu
Jan 7 at 13:59
$begingroup$
yeah that's okay. I'm just looking for viable methods of tackling the problem algebraically without having to depend on visual graphs as some equations may be cumbersome to graph.
$endgroup$
– esc1234
Jan 7 at 14:00
$begingroup$
yeah that's okay. I'm just looking for viable methods of tackling the problem algebraically without having to depend on visual graphs as some equations may be cumbersome to graph.
$endgroup$
– esc1234
Jan 7 at 14:00
$begingroup$
Well, evaluating the function at a few points is generally a good place to start. Here you could also use calculus to find the local max and min, but I don't see that as any easier. Descartes' Rule of Signs used to be a popular method but it has largely fallen out of favor as computational devices have become easy to use.
$endgroup$
– lulu
Jan 7 at 14:03
$begingroup$
Well, evaluating the function at a few points is generally a good place to start. Here you could also use calculus to find the local max and min, but I don't see that as any easier. Descartes' Rule of Signs used to be a popular method but it has largely fallen out of favor as computational devices have become easy to use.
$endgroup$
– lulu
Jan 7 at 14:03
|
show 3 more comments
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2
$begingroup$
The three roots aren't especially pleasant. If you need values for them, you'll probably want to use numerical methods. If all you want is to see that there are three roots, note that $p(-10)<0, p(0)>0,p(1)<0, p(10)>0$.
$endgroup$
– lulu
Jan 7 at 13:53
$begingroup$
so without the use of a graphing calculator or attempting to graph the equation, is the only way to find the number of solutions is to test input values from various intervals that look pertinent?
$endgroup$
– esc1234
Jan 7 at 13:58
$begingroup$
I didn't say it was the only way. But it's a good way.
$endgroup$
– lulu
Jan 7 at 13:59
$begingroup$
yeah that's okay. I'm just looking for viable methods of tackling the problem algebraically without having to depend on visual graphs as some equations may be cumbersome to graph.
$endgroup$
– esc1234
Jan 7 at 14:00
$begingroup$
Well, evaluating the function at a few points is generally a good place to start. Here you could also use calculus to find the local max and min, but I don't see that as any easier. Descartes' Rule of Signs used to be a popular method but it has largely fallen out of favor as computational devices have become easy to use.
$endgroup$
– lulu
Jan 7 at 14:03