Solving for $g$ in a quadratic equation [on hold]
is there a way to get that g = ?
I want to get the g out . Getting nasty results like
2h = gg - g . I'm stuck . Is there a way to find square root of gg -g or is there any other solution . Thank you
algebra-precalculus
edited Jan 4 at 1:07
David G. Stork
9,96021232
asked Jan 4 at 0:35
MAtt
61
New contributor
MAtt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Eevee Trainer, Shailesh, Adrian Keister, metamorphy, stressed out Jan 4 at 2:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Adrian Keister, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
is there a way to get that g = ?
I want to get the g out . Getting nasty results like
2h = gg - g . I'm stuck . Is there a way to find square root of gg -g or is there any other solution . Thank you
algebra-precalculus
edited Jan 4 at 1:07
David G. Stork
9,96021232
asked Jan 4 at 0:35
MAtt
61
New contributor
MAtt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Eevee Trainer, Shailesh, Adrian Keister, metamorphy, stressed out Jan 4 at 2:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Adrian Keister, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
You're going to have to elaborate. Do you mean solve for $g$ in terms of $h$?
– Eevee Trainer
Jan 4 at 0:36
What's the difference between g and $g$?
– clathratus
Jan 4 at 0:41
sorry it means g * g
– MAtt
Jan 4 at 0:43
add a comment |
is there a way to get that g = ?
I want to get the g out . Getting nasty results like
2h = gg - g . I'm stuck . Is there a way to find square root of gg -g or is there any other solution . Thank you
algebra-precalculus
edited Jan 4 at 1:07
David G. Stork
9,96021232
asked Jan 4 at 0:35
MAtt
61
New contributor
MAtt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
is there a way to get that g = ?
I want to get the g out . Getting nasty results like
2h = gg - g . I'm stuck . Is there a way to find square root of gg -g or is there any other solution . Thank you
algebra-precalculus
algebra-precalculus
edited Jan 4 at 1:07
David G. Stork
9,96021232
asked Jan 4 at 0:35
MAtt
61
New contributor
MAtt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 4 at 1:07
David G. Stork
9,96021232
asked Jan 4 at 0:35
MAtt
61
New contributor
MAtt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 4 at 1:07
David G. Stork
9,96021232
edited Jan 4 at 1:07
David G. Stork
9,96021232
edited Jan 4 at 1:07
David G. Stork
9,96021232
9,96021232
asked Jan 4 at 0:35
MAtt
61
New contributor
MAtt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 0:35
MAtt
61
asked Jan 4 at 0:35
MAtt
61
61
New contributor
MAtt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
MAtt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
MAtt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Eevee Trainer, Shailesh, Adrian Keister, metamorphy, stressed out Jan 4 at 2:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Adrian Keister, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Eevee Trainer, Shailesh, Adrian Keister, metamorphy, stressed out Jan 4 at 2:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Adrian Keister, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
You're going to have to elaborate. Do you mean solve for $g$ in terms of $h$?
– Eevee Trainer
Jan 4 at 0:36
What's the difference between g and $g$?
– clathratus
Jan 4 at 0:41
sorry it means g * g
– MAtt
Jan 4 at 0:43
add a comment |
You're going to have to elaborate. Do you mean solve for $g$ in terms of $h$?
– Eevee Trainer
Jan 4 at 0:36
What's the difference between g and $g$?
– clathratus
Jan 4 at 0:41
sorry it means g * g
– MAtt
Jan 4 at 0:43
You're going to have to elaborate. Do you mean solve for $g$ in terms of $h$?
– Eevee Trainer
Jan 4 at 0:36
You're going to have to elaborate. Do you mean solve for $g$ in terms of $h$?
– Eevee Trainer
Jan 4 at 0:36
What's the difference between g and $g$?
– clathratus
Jan 4 at 0:41
What's the difference between g and $g$?
– clathratus
Jan 4 at 0:41
sorry it means g * g
– MAtt
Jan 4 at 0:43
sorry it means g * g
– MAtt
Jan 4 at 0:43
add a comment |
2 Answers
2
active
oldest
votes
$g^2-g-2h=0$. This is called a quadratic equation and there is a standard formula for finding a solution to this type of equation.
If the equation is $ax^2+bx+c=0$, the solutions are given by:
$x = frac{-b + sqrt{b^2-4ac}}{2a}$ and $x = frac{-b - sqrt{b^2-4ac}}{2a}$.
In your case, $a=1, b=-1, c=-2h$, so the solutions are:
$g = frac{1 + sqrt{1+8h}}{2}$ and $g = frac{1 - sqrt{1+8h}}{2}$.
You will have to constrain the problem further to determine if any or both of the above solutions are valid.
answered Jan 4 at 0:41
Aditya Dua
86418
Now i'm in home . Thanks a lot !
– MAtt
Jan 4 at 0:47
add a comment |
Multiply out to obtain the quadratic:
$$h = {g^2 - g over 2}$$
so $$g^2 - 2 h - g = 0$$.
Use the quadratic formula:
$$g = frac{1}{2} left( 1pm sqrt{8 h+1} right)$$
answered Jan 4 at 0:40
David G. Stork
9,96021232
that is what i was looking for , could you please explain this shortly
– MAtt
Jan 4 at 0:41
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$g^2-g-2h=0$. This is called a quadratic equation and there is a standard formula for finding a solution to this type of equation.
If the equation is $ax^2+bx+c=0$, the solutions are given by:
$x = frac{-b + sqrt{b^2-4ac}}{2a}$ and $x = frac{-b - sqrt{b^2-4ac}}{2a}$.
In your case, $a=1, b=-1, c=-2h$, so the solutions are:
$g = frac{1 + sqrt{1+8h}}{2}$ and $g = frac{1 - sqrt{1+8h}}{2}$.
You will have to constrain the problem further to determine if any or both of the above solutions are valid.
answered Jan 4 at 0:41
Aditya Dua
86418
Now i'm in home . Thanks a lot !
– MAtt
Jan 4 at 0:47
add a comment |
$g^2-g-2h=0$. This is called a quadratic equation and there is a standard formula for finding a solution to this type of equation.
If the equation is $ax^2+bx+c=0$, the solutions are given by:
$x = frac{-b + sqrt{b^2-4ac}}{2a}$ and $x = frac{-b - sqrt{b^2-4ac}}{2a}$.
In your case, $a=1, b=-1, c=-2h$, so the solutions are:
$g = frac{1 + sqrt{1+8h}}{2}$ and $g = frac{1 - sqrt{1+8h}}{2}$.
You will have to constrain the problem further to determine if any or both of the above solutions are valid.
answered Jan 4 at 0:41
Aditya Dua
86418
Now i'm in home . Thanks a lot !
– MAtt
Jan 4 at 0:47
add a comment |
$g^2-g-2h=0$. This is called a quadratic equation and there is a standard formula for finding a solution to this type of equation.
If the equation is $ax^2+bx+c=0$, the solutions are given by:
$x = frac{-b + sqrt{b^2-4ac}}{2a}$ and $x = frac{-b - sqrt{b^2-4ac}}{2a}$.
In your case, $a=1, b=-1, c=-2h$, so the solutions are:
$g = frac{1 + sqrt{1+8h}}{2}$ and $g = frac{1 - sqrt{1+8h}}{2}$.
You will have to constrain the problem further to determine if any or both of the above solutions are valid.
answered Jan 4 at 0:41
Aditya Dua
86418
$g^2-g-2h=0$. This is called a quadratic equation and there is a standard formula for finding a solution to this type of equation.
If the equation is $ax^2+bx+c=0$, the solutions are given by:
$x = frac{-b + sqrt{b^2-4ac}}{2a}$ and $x = frac{-b - sqrt{b^2-4ac}}{2a}$.
In your case, $a=1, b=-1, c=-2h$, so the solutions are:
$g = frac{1 + sqrt{1+8h}}{2}$ and $g = frac{1 - sqrt{1+8h}}{2}$.
You will have to constrain the problem further to determine if any or both of the above solutions are valid.
answered Jan 4 at 0:41
Aditya Dua
86418
answered Jan 4 at 0:41
Aditya Dua
86418
answered Jan 4 at 0:41
Aditya Dua
86418
answered Jan 4 at 0:41
Aditya Dua
86418
86418
Now i'm in home . Thanks a lot !
– MAtt
Jan 4 at 0:47
add a comment |
Now i'm in home . Thanks a lot !
– MAtt
Jan 4 at 0:47
Now i'm in home . Thanks a lot !
– MAtt
Jan 4 at 0:47
Now i'm in home . Thanks a lot !
– MAtt
Jan 4 at 0:47
add a comment |
Multiply out to obtain the quadratic:
$$h = {g^2 - g over 2}$$
so $$g^2 - 2 h - g = 0$$.
Use the quadratic formula:
$$g = frac{1}{2} left( 1pm sqrt{8 h+1} right)$$
answered Jan 4 at 0:40
David G. Stork
9,96021232
that is what i was looking for , could you please explain this shortly
– MAtt
Jan 4 at 0:41
add a comment |
Multiply out to obtain the quadratic:
$$h = {g^2 - g over 2}$$
so $$g^2 - 2 h - g = 0$$.
Use the quadratic formula:
$$g = frac{1}{2} left( 1pm sqrt{8 h+1} right)$$
answered Jan 4 at 0:40
David G. Stork
9,96021232
that is what i was looking for , could you please explain this shortly
– MAtt
Jan 4 at 0:41
add a comment |
Multiply out to obtain the quadratic:
$$h = {g^2 - g over 2}$$
so $$g^2 - 2 h - g = 0$$.
Use the quadratic formula:
$$g = frac{1}{2} left( 1pm sqrt{8 h+1} right)$$
answered Jan 4 at 0:40
David G. Stork
9,96021232
Multiply out to obtain the quadratic:
$$h = {g^2 - g over 2}$$
so $$g^2 - 2 h - g = 0$$.
Use the quadratic formula:
$$g = frac{1}{2} left( 1pm sqrt{8 h+1} right)$$
answered Jan 4 at 0:40
David G. Stork
9,96021232
edited Jan 4 at 0:49
answered Jan 4 at 0:40
David G. Stork
9,96021232
answered Jan 4 at 0:40
David G. Stork
9,96021232
answered Jan 4 at 0:40
David G. Stork
9,96021232
9,96021232
that is what i was looking for , could you please explain this shortly
– MAtt
Jan 4 at 0:41
add a comment |
that is what i was looking for , could you please explain this shortly
– MAtt
Jan 4 at 0:41
that is what i was looking for , could you please explain this shortly
– MAtt
Jan 4 at 0:41
that is what i was looking for , could you please explain this shortly
– MAtt
Jan 4 at 0:41
add a comment |
You're going to have to elaborate. Do you mean solve for $g$ in terms of $h$?
– Eevee Trainer
Jan 4 at 0:36
What's the difference between g and $g$?
– clathratus
Jan 4 at 0:41
sorry it means g * g
– MAtt
Jan 4 at 0:43