Solving for $g$ in a quadratic equation [closed]












0












$begingroup$


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







share|cite|improve this question











$endgroup$



closed 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.













  • $begingroup$
    You're going to have to elaborate. Do you mean solve for $g$ in terms of $h$?
    $endgroup$
    – Eevee Trainer
    Jan 4 at 0:36










  • $begingroup$
    What's the difference between g and $g$?
    $endgroup$
    – clathratus
    Jan 4 at 0:41










  • $begingroup$
    sorry it means g * g
    $endgroup$
    – MAtt
    Jan 4 at 0:43
















0












$begingroup$


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







share|cite|improve this question











$endgroup$



closed 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.













  • $begingroup$
    You're going to have to elaborate. Do you mean solve for $g$ in terms of $h$?
    $endgroup$
    – Eevee Trainer
    Jan 4 at 0:36










  • $begingroup$
    What's the difference between g and $g$?
    $endgroup$
    – clathratus
    Jan 4 at 0:41










  • $begingroup$
    sorry it means g * g
    $endgroup$
    – MAtt
    Jan 4 at 0:43














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 1:07









David G. Stork

10.5k31332




10.5k31332










asked Jan 4 at 0:35









MAttMAtt

61




61




closed 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.




closed 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.












  • $begingroup$
    You're going to have to elaborate. Do you mean solve for $g$ in terms of $h$?
    $endgroup$
    – Eevee Trainer
    Jan 4 at 0:36










  • $begingroup$
    What's the difference between g and $g$?
    $endgroup$
    – clathratus
    Jan 4 at 0:41










  • $begingroup$
    sorry it means g * g
    $endgroup$
    – MAtt
    Jan 4 at 0:43


















  • $begingroup$
    You're going to have to elaborate. Do you mean solve for $g$ in terms of $h$?
    $endgroup$
    – Eevee Trainer
    Jan 4 at 0:36










  • $begingroup$
    What's the difference between g and $g$?
    $endgroup$
    – clathratus
    Jan 4 at 0:41










  • $begingroup$
    sorry it means g * g
    $endgroup$
    – MAtt
    Jan 4 at 0:43
















$begingroup$
You're going to have to elaborate. Do you mean solve for $g$ in terms of $h$?
$endgroup$
– Eevee Trainer
Jan 4 at 0:36




$begingroup$
You're going to have to elaborate. Do you mean solve for $g$ in terms of $h$?
$endgroup$
– Eevee Trainer
Jan 4 at 0:36












$begingroup$
What's the difference between g and $g$?
$endgroup$
– clathratus
Jan 4 at 0:41




$begingroup$
What's the difference between g and $g$?
$endgroup$
– clathratus
Jan 4 at 0:41












$begingroup$
sorry it means g * g
$endgroup$
– MAtt
Jan 4 at 0:43




$begingroup$
sorry it means g * g
$endgroup$
– MAtt
Jan 4 at 0:43










2 Answers
2






active

oldest

votes


















2












$begingroup$

$g^2-g-2h=0$. 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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Now i'm in home . Thanks a lot !
    $endgroup$
    – MAtt
    Jan 4 at 0:47



















1












$begingroup$

Multiply out to obtain the quadratic:



$$h = {g^2 - g over 2}$$



so $$g^2 - 2 h - g = 0$$.



Use the quadratic formula:



$$g = frac{1}{2} left( 1pm sqrt{8 h+1} right)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    that is what i was looking for , could you please explain this shortly
    $endgroup$
    – MAtt
    Jan 4 at 0:41


















2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

$g^2-g-2h=0$. 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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Now i'm in home . Thanks a lot !
    $endgroup$
    – MAtt
    Jan 4 at 0:47
















2












$begingroup$

$g^2-g-2h=0$. 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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Now i'm in home . Thanks a lot !
    $endgroup$
    – MAtt
    Jan 4 at 0:47














2












2








2





$begingroup$

$g^2-g-2h=0$. 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.






share|cite|improve this answer









$endgroup$



$g^2-g-2h=0$. 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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 0:41









Aditya DuaAditya Dua

97918




97918












  • $begingroup$
    Now i'm in home . Thanks a lot !
    $endgroup$
    – MAtt
    Jan 4 at 0:47


















  • $begingroup$
    Now i'm in home . Thanks a lot !
    $endgroup$
    – MAtt
    Jan 4 at 0:47
















$begingroup$
Now i'm in home . Thanks a lot !
$endgroup$
– MAtt
Jan 4 at 0:47




$begingroup$
Now i'm in home . Thanks a lot !
$endgroup$
– MAtt
Jan 4 at 0:47











1












$begingroup$

Multiply out to obtain the quadratic:



$$h = {g^2 - g over 2}$$



so $$g^2 - 2 h - g = 0$$.



Use the quadratic formula:



$$g = frac{1}{2} left( 1pm sqrt{8 h+1} right)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    that is what i was looking for , could you please explain this shortly
    $endgroup$
    – MAtt
    Jan 4 at 0:41
















1












$begingroup$

Multiply out to obtain the quadratic:



$$h = {g^2 - g over 2}$$



so $$g^2 - 2 h - g = 0$$.



Use the quadratic formula:



$$g = frac{1}{2} left( 1pm sqrt{8 h+1} right)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    that is what i was looking for , could you please explain this shortly
    $endgroup$
    – MAtt
    Jan 4 at 0:41














1












1








1





$begingroup$

Multiply out to obtain the quadratic:



$$h = {g^2 - g over 2}$$



so $$g^2 - 2 h - g = 0$$.



Use the quadratic formula:



$$g = frac{1}{2} left( 1pm sqrt{8 h+1} right)$$






share|cite|improve this answer











$endgroup$



Multiply out to obtain the quadratic:



$$h = {g^2 - g over 2}$$



so $$g^2 - 2 h - g = 0$$.



Use the quadratic formula:



$$g = frac{1}{2} left( 1pm sqrt{8 h+1} right)$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 0:49

























answered Jan 4 at 0:40









David G. StorkDavid G. Stork

10.5k31332




10.5k31332












  • $begingroup$
    that is what i was looking for , could you please explain this shortly
    $endgroup$
    – MAtt
    Jan 4 at 0:41


















  • $begingroup$
    that is what i was looking for , could you please explain this shortly
    $endgroup$
    – MAtt
    Jan 4 at 0:41
















$begingroup$
that is what i was looking for , could you please explain this shortly
$endgroup$
– MAtt
Jan 4 at 0:41




$begingroup$
that is what i was looking for , could you please explain this shortly
$endgroup$
– MAtt
Jan 4 at 0:41



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