Symbol for “take the positive solution of this formula”
$begingroup$
I got a formula:
x=√(((−2*aks)/(r*(s-k)*v)))
Notice it is all wrapped inside a square root.
When taking the square root, there are to possible outcomes as we all know. How do I denote that one should use the positive number?
education
asked Jan 6 at 12:20
Ryan CameronRyan Cameron
697
$endgroup$
add a comment |
$begingroup$
I got a formula:
x=√(((−2*aks)/(r*(s-k)*v)))
Notice it is all wrapped inside a square root.
When taking the square root, there are to possible outcomes as we all know. How do I denote that one should use the positive number?
education
asked Jan 6 at 12:20
Ryan CameronRyan Cameron
697
$endgroup$
$begingroup$
What are the Solutions of this Problem?
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 12:21
$begingroup$
Depends on what you set the variables to be. Eg. "x_1=2, x_2=-4".
$endgroup$
– Ryan Cameron
Jan 6 at 12:21
$begingroup$
$sqrt{a}$ means the positive square root of $a$. It's a common error to somehow push the $pm$ in $pm sqrt{a}$ into the $sqrt{a}$. Your "two possible outcomes" requires you to write $pmsqrt{a}.$
$endgroup$
– B. Goddard
Jan 6 at 12:28
add a comment |
$begingroup$
I got a formula:
x=√(((−2*aks)/(r*(s-k)*v)))
Notice it is all wrapped inside a square root.
When taking the square root, there are to possible outcomes as we all know. How do I denote that one should use the positive number?
education
asked Jan 6 at 12:20
Ryan CameronRyan Cameron
697
$endgroup$
I got a formula:
x=√(((−2*aks)/(r*(s-k)*v)))
Notice it is all wrapped inside a square root.
When taking the square root, there are to possible outcomes as we all know. How do I denote that one should use the positive number?
education
education
asked Jan 6 at 12:20
Ryan CameronRyan Cameron
697
asked Jan 6 at 12:20
Ryan CameronRyan Cameron
697
edited Jan 6 at 12:27
Ryan Cameron
asked Jan 6 at 12:20
Ryan CameronRyan Cameron
697
asked Jan 6 at 12:20
Ryan CameronRyan Cameron
697
asked Jan 6 at 12:20
Ryan CameronRyan Cameron
697
697
$begingroup$
What are the Solutions of this Problem?
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 12:21
$begingroup$
Depends on what you set the variables to be. Eg. "x_1=2, x_2=-4".
$endgroup$
– Ryan Cameron
Jan 6 at 12:21
$begingroup$
$sqrt{a}$ means the positive square root of $a$. It's a common error to somehow push the $pm$ in $pm sqrt{a}$ into the $sqrt{a}$. Your "two possible outcomes" requires you to write $pmsqrt{a}.$
$endgroup$
– B. Goddard
Jan 6 at 12:28
add a comment |
$begingroup$
What are the Solutions of this Problem?
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 12:21
$begingroup$
Depends on what you set the variables to be. Eg. "x_1=2, x_2=-4".
$endgroup$
– Ryan Cameron
Jan 6 at 12:21
$begingroup$
$sqrt{a}$ means the positive square root of $a$. It's a common error to somehow push the $pm$ in $pm sqrt{a}$ into the $sqrt{a}$. Your "two possible outcomes" requires you to write $pmsqrt{a}.$
$endgroup$
– B. Goddard
Jan 6 at 12:28
$begingroup$
What are the Solutions of this Problem?
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 12:21
$begingroup$
What are the Solutions of this Problem?
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 12:21
$begingroup$
Depends on what you set the variables to be. Eg. "x_1=2, x_2=-4".
$endgroup$
– Ryan Cameron
Jan 6 at 12:21
$begingroup$
Depends on what you set the variables to be. Eg. "x_1=2, x_2=-4".
$endgroup$
– Ryan Cameron
Jan 6 at 12:21
$begingroup$
$sqrt{a}$ means the positive square root of $a$. It's a common error to somehow push the $pm$ in $pm sqrt{a}$ into the $sqrt{a}$. Your "two possible outcomes" requires you to write $pmsqrt{a}.$
$endgroup$
– B. Goddard
Jan 6 at 12:28
$begingroup$
$sqrt{a}$ means the positive square root of $a$. It's a common error to somehow push the $pm$ in $pm sqrt{a}$ into the $sqrt{a}$. Your "two possible outcomes" requires you to write $pmsqrt{a}.$
$endgroup$
– B. Goddard
Jan 6 at 12:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
That one. $sqrt{a}$ denotes the positive solution of the equation $x^2 = a$.
answered Jan 6 at 12:29
user3482749user3482749
4,3111019
$endgroup$
add a comment |
$begingroup$
Your formula does not require any change. By definition, $x = sqrt{a}$ will return only the non-negative square root, or the principal square root. You’re confusing this with $x^2 = a iff vert xvert = sqrt{a}$, which has two solutions: $x = pm sqrt{a}$.
Therefore, $x = sqrt{a}$ already implies that the returned value isn’t negative. (Note that there is no $pm$ sign in front of the square root.)
answered Jan 6 at 12:34
KM101KM101
6,0901525
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That one. $sqrt{a}$ denotes the positive solution of the equation $x^2 = a$.
answered Jan 6 at 12:29
user3482749user3482749
4,3111019
$endgroup$
add a comment |
$begingroup$
That one. $sqrt{a}$ denotes the positive solution of the equation $x^2 = a$.
answered Jan 6 at 12:29
user3482749user3482749
4,3111019
$endgroup$
add a comment |
$begingroup$
That one. $sqrt{a}$ denotes the positive solution of the equation $x^2 = a$.
answered Jan 6 at 12:29
user3482749user3482749
4,3111019
$endgroup$
That one. $sqrt{a}$ denotes the positive solution of the equation $x^2 = a$.
answered Jan 6 at 12:29
user3482749user3482749
4,3111019
answered Jan 6 at 12:29
user3482749user3482749
4,3111019
answered Jan 6 at 12:29
user3482749user3482749
4,3111019
answered Jan 6 at 12:29
user3482749user3482749
4,3111019
4,3111019
add a comment |
add a comment |
$begingroup$
Your formula does not require any change. By definition, $x = sqrt{a}$ will return only the non-negative square root, or the principal square root. You’re confusing this with $x^2 = a iff vert xvert = sqrt{a}$, which has two solutions: $x = pm sqrt{a}$.
Therefore, $x = sqrt{a}$ already implies that the returned value isn’t negative. (Note that there is no $pm$ sign in front of the square root.)
answered Jan 6 at 12:34
KM101KM101
6,0901525
$endgroup$
add a comment |
$begingroup$
Your formula does not require any change. By definition, $x = sqrt{a}$ will return only the non-negative square root, or the principal square root. You’re confusing this with $x^2 = a iff vert xvert = sqrt{a}$, which has two solutions: $x = pm sqrt{a}$.
Therefore, $x = sqrt{a}$ already implies that the returned value isn’t negative. (Note that there is no $pm$ sign in front of the square root.)
answered Jan 6 at 12:34
KM101KM101
6,0901525
$endgroup$
add a comment |
$begingroup$
Your formula does not require any change. By definition, $x = sqrt{a}$ will return only the non-negative square root, or the principal square root. You’re confusing this with $x^2 = a iff vert xvert = sqrt{a}$, which has two solutions: $x = pm sqrt{a}$.
Therefore, $x = sqrt{a}$ already implies that the returned value isn’t negative. (Note that there is no $pm$ sign in front of the square root.)
answered Jan 6 at 12:34
KM101KM101
6,0901525
$endgroup$
Your formula does not require any change. By definition, $x = sqrt{a}$ will return only the non-negative square root, or the principal square root. You’re confusing this with $x^2 = a iff vert xvert = sqrt{a}$, which has two solutions: $x = pm sqrt{a}$.
Therefore, $x = sqrt{a}$ already implies that the returned value isn’t negative. (Note that there is no $pm$ sign in front of the square root.)
answered Jan 6 at 12:34
KM101KM101
6,0901525
answered Jan 6 at 12:34
KM101KM101
6,0901525
answered Jan 6 at 12:34
KM101KM101
6,0901525
answered Jan 6 at 12:34
KM101KM101
6,0901525
6,0901525
add a comment |
add a comment |
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$begingroup$
What are the Solutions of this Problem?
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 12:21
$begingroup$
Depends on what you set the variables to be. Eg. "x_1=2, x_2=-4".
$endgroup$
– Ryan Cameron
Jan 6 at 12:21
$begingroup$
$sqrt{a}$ means the positive square root of $a$. It's a common error to somehow push the $pm$ in $pm sqrt{a}$ into the $sqrt{a}$. Your "two possible outcomes" requires you to write $pmsqrt{a}.$
$endgroup$
– B. Goddard
Jan 6 at 12:28