Mixing
Solve the magnitudes of the solutions
$begingroup$
I was wondering how I could do these two problems
x^2 + 1=0
x^3 + 1 = 0
I m trying to find the roots and magnitudes of these questions. Thank you
polynomials
asked Feb 3 at 2:58
Dylan OngDylan Ong
32
$endgroup$
add a comment |
$begingroup$
I was wondering how I could do these two problems
x^2 + 1=0
x^3 + 1 = 0
I m trying to find the roots and magnitudes of these questions. Thank you
polynomials
asked Feb 3 at 2:58
Dylan OngDylan Ong
32
$endgroup$
add a comment |
$begingroup$
I was wondering how I could do these two problems
x^2 + 1=0
x^3 + 1 = 0
I m trying to find the roots and magnitudes of these questions. Thank you
polynomials
asked Feb 3 at 2:58
Dylan OngDylan Ong
32
$endgroup$
I was wondering how I could do these two problems
x^2 + 1=0
x^3 + 1 = 0
I m trying to find the roots and magnitudes of these questions. Thank you
polynomials
polynomials
asked Feb 3 at 2:58
Dylan OngDylan Ong
32
asked Feb 3 at 2:58
Dylan OngDylan Ong
32
asked Feb 3 at 2:58
Dylan OngDylan Ong
32
asked Feb 3 at 2:58
Dylan OngDylan Ong
32
asked Feb 3 at 2:58
Dylan OngDylan Ong
32
32
add a comment |
add a comment |
1 Answer
1
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$begingroup$
$$mbox{x^2=-1}$$
$$mbox{x = i and -i (These are complex roots)}$$
The Magnitude is $sqrt{(-1)^2} = 1$
$$mbox{x^3 = 1}$$
$$mbox{x = 1}$$
The Magnitude is simply $1$
The general formula for finding the magnitude of a complex number:
$|a + bi| = sqrt{a^2 + b^2}$
answered Feb 3 at 3:12
KevinKevin
497
$endgroup$
$begingroup$
I got that. How do I find the magnitude
$endgroup$
– Dylan Ong
Feb 3 at 3:13
$begingroup$
This might help quora.com/…
$endgroup$
– Kevin
Feb 3 at 3:17
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$mbox{x^2=-1}$$
$$mbox{x = i and -i (These are complex roots)}$$
The Magnitude is $sqrt{(-1)^2} = 1$
$$mbox{x^3 = 1}$$
$$mbox{x = 1}$$
The Magnitude is simply $1$
The general formula for finding the magnitude of a complex number:
$|a + bi| = sqrt{a^2 + b^2}$
answered Feb 3 at 3:12
KevinKevin
497
$endgroup$
$begingroup$
I got that. How do I find the magnitude
$endgroup$
– Dylan Ong
Feb 3 at 3:13
$begingroup$
This might help quora.com/…
$endgroup$
– Kevin
Feb 3 at 3:17
add a comment |
$begingroup$
$$mbox{x^2=-1}$$
$$mbox{x = i and -i (These are complex roots)}$$
The Magnitude is $sqrt{(-1)^2} = 1$
$$mbox{x^3 = 1}$$
$$mbox{x = 1}$$
The Magnitude is simply $1$
The general formula for finding the magnitude of a complex number:
$|a + bi| = sqrt{a^2 + b^2}$
answered Feb 3 at 3:12
KevinKevin
497
$endgroup$
$begingroup$
I got that. How do I find the magnitude
$endgroup$
– Dylan Ong
Feb 3 at 3:13
$begingroup$
This might help quora.com/…
$endgroup$
– Kevin
Feb 3 at 3:17
add a comment |
$begingroup$
$$mbox{x^2=-1}$$
$$mbox{x = i and -i (These are complex roots)}$$
The Magnitude is $sqrt{(-1)^2} = 1$
$$mbox{x^3 = 1}$$
$$mbox{x = 1}$$
The Magnitude is simply $1$
The general formula for finding the magnitude of a complex number:
$|a + bi| = sqrt{a^2 + b^2}$
answered Feb 3 at 3:12
KevinKevin
497
$endgroup$
$$mbox{x^2=-1}$$
$$mbox{x = i and -i (These are complex roots)}$$
The Magnitude is $sqrt{(-1)^2} = 1$
$$mbox{x^3 = 1}$$
$$mbox{x = 1}$$
The Magnitude is simply $1$
The general formula for finding the magnitude of a complex number:
$|a + bi| = sqrt{a^2 + b^2}$
answered Feb 3 at 3:12
KevinKevin
497
edited Feb 3 at 3:21
answered Feb 3 at 3:12
KevinKevin
497
answered Feb 3 at 3:12
KevinKevin
497
answered Feb 3 at 3:12
KevinKevin
497
497
$begingroup$
I got that. How do I find the magnitude
$endgroup$
– Dylan Ong
Feb 3 at 3:13
$begingroup$
This might help quora.com/…
$endgroup$
– Kevin
Feb 3 at 3:17
add a comment |
$begingroup$
I got that. How do I find the magnitude
$endgroup$
– Dylan Ong
Feb 3 at 3:13
$begingroup$
This might help quora.com/…
$endgroup$
– Kevin
Feb 3 at 3:17
$begingroup$
I got that. How do I find the magnitude
$endgroup$
– Dylan Ong
Feb 3 at 3:13
$begingroup$
I got that. How do I find the magnitude
$endgroup$
– Dylan Ong
Feb 3 at 3:13
$begingroup$
This might help quora.com/…
$endgroup$
– Kevin
Feb 3 at 3:17
$begingroup$
This might help quora.com/…
$endgroup$
– Kevin
Feb 3 at 3:17
add a comment |
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