Is it possible to isolate the variable X in this equation? [closed]












1












$begingroup$


$P = dfrac{1}{sqrt{x^2+xac+a^2}} + dfrac{1}{sqrt{x^2+xbc+b^2}}$






algebra-precalculus quadratics







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closed as off-topic by José Carlos Santos, Cesareo, clathratus, Eevee Trainer, user91500 Dec 30 '18 at 6:59


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." – José Carlos Santos, Cesareo, clathratus, Eevee Trainer, user91500

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    Move one of the fractions to the other side to get $P= 1/texts[stuff}$ on the left. Square both sides. That will leave one radical on the left. Isolate it and square again. Then you will have a fourth degree equation in $x$ you may be able to work with.
    $endgroup$
    – Ethan Bolker
    Dec 29 '18 at 21:48










  • $begingroup$
    I think the tag abstract-algebra does not apply in this case. I am assuming that $+$ refers to addition and this question has nothing to do with groups and rings.
    $endgroup$
    – Bob
    Dec 29 '18 at 21:56










  • $begingroup$
    @user376343 Yes. Should be $P -$ instead of $P =$. Moot since there's an answer along these lines below.
    $endgroup$
    – Ethan Bolker
    Dec 30 '18 at 1:10


















1












$begingroup$


$P = dfrac{1}{sqrt{x^2+xac+a^2}} + dfrac{1}{sqrt{x^2+xbc+b^2}}$






algebra-precalculus quadratics







share|cite|improve this question











$endgroup$



closed as off-topic by José Carlos Santos, Cesareo, clathratus, Eevee Trainer, user91500 Dec 30 '18 at 6:59


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." – José Carlos Santos, Cesareo, clathratus, Eevee Trainer, user91500

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    Move one of the fractions to the other side to get $P= 1/texts[stuff}$ on the left. Square both sides. That will leave one radical on the left. Isolate it and square again. Then you will have a fourth degree equation in $x$ you may be able to work with.
    $endgroup$
    – Ethan Bolker
    Dec 29 '18 at 21:48










  • $begingroup$
    I think the tag abstract-algebra does not apply in this case. I am assuming that $+$ refers to addition and this question has nothing to do with groups and rings.
    $endgroup$
    – Bob
    Dec 29 '18 at 21:56










  • $begingroup$
    @user376343 Yes. Should be $P -$ instead of $P =$. Moot since there's an answer along these lines below.
    $endgroup$
    – Ethan Bolker
    Dec 30 '18 at 1:10
















1












1








1


0



$begingroup$


$P = dfrac{1}{sqrt{x^2+xac+a^2}} + dfrac{1}{sqrt{x^2+xbc+b^2}}$






algebra-precalculus quadratics







share|cite|improve this question











$endgroup$




$P = dfrac{1}{sqrt{x^2+xac+a^2}} + dfrac{1}{sqrt{x^2+xbc+b^2}}$






algebra-precalculus quadratics




algebra-precalculus quadratics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 21:57









ÍgjøgnumMeg

2,79511029




2,79511029










asked Dec 29 '18 at 21:42









Gonzalo FerrandoGonzalo Ferrando

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111




closed as off-topic by José Carlos Santos, Cesareo, clathratus, Eevee Trainer, user91500 Dec 30 '18 at 6:59


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." – José Carlos Santos, Cesareo, clathratus, Eevee Trainer, user91500

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by José Carlos Santos, Cesareo, clathratus, Eevee Trainer, user91500 Dec 30 '18 at 6:59


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." – José Carlos Santos, Cesareo, clathratus, Eevee Trainer, user91500

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Move one of the fractions to the other side to get $P= 1/texts[stuff}$ on the left. Square both sides. That will leave one radical on the left. Isolate it and square again. Then you will have a fourth degree equation in $x$ you may be able to work with.
    $endgroup$
    – Ethan Bolker
    Dec 29 '18 at 21:48










  • $begingroup$
    I think the tag abstract-algebra does not apply in this case. I am assuming that $+$ refers to addition and this question has nothing to do with groups and rings.
    $endgroup$
    – Bob
    Dec 29 '18 at 21:56










  • $begingroup$
    @user376343 Yes. Should be $P -$ instead of $P =$. Moot since there's an answer along these lines below.
    $endgroup$
    – Ethan Bolker
    Dec 30 '18 at 1:10




















  • $begingroup$
    Move one of the fractions to the other side to get $P= 1/texts[stuff}$ on the left. Square both sides. That will leave one radical on the left. Isolate it and square again. Then you will have a fourth degree equation in $x$ you may be able to work with.
    $endgroup$
    – Ethan Bolker
    Dec 29 '18 at 21:48










  • $begingroup$
    I think the tag abstract-algebra does not apply in this case. I am assuming that $+$ refers to addition and this question has nothing to do with groups and rings.
    $endgroup$
    – Bob
    Dec 29 '18 at 21:56










  • $begingroup$
    @user376343 Yes. Should be $P -$ instead of $P =$. Moot since there's an answer along these lines below.
    $endgroup$
    – Ethan Bolker
    Dec 30 '18 at 1:10


















$begingroup$
Move one of the fractions to the other side to get $P= 1/texts[stuff}$ on the left. Square both sides. That will leave one radical on the left. Isolate it and square again. Then you will have a fourth degree equation in $x$ you may be able to work with.
$endgroup$
– Ethan Bolker
Dec 29 '18 at 21:48




$begingroup$
Move one of the fractions to the other side to get $P= 1/texts[stuff}$ on the left. Square both sides. That will leave one radical on the left. Isolate it and square again. Then you will have a fourth degree equation in $x$ you may be able to work with.
$endgroup$
– Ethan Bolker
Dec 29 '18 at 21:48












$begingroup$
I think the tag abstract-algebra does not apply in this case. I am assuming that $+$ refers to addition and this question has nothing to do with groups and rings.
$endgroup$
– Bob
Dec 29 '18 at 21:56




$begingroup$
I think the tag abstract-algebra does not apply in this case. I am assuming that $+$ refers to addition and this question has nothing to do with groups and rings.
$endgroup$
– Bob
Dec 29 '18 at 21:56












$begingroup$
@user376343 Yes. Should be $P -$ instead of $P =$. Moot since there's an answer along these lines below.
$endgroup$
– Ethan Bolker
Dec 30 '18 at 1:10






$begingroup$
@user376343 Yes. Should be $P -$ instead of $P =$. Moot since there's an answer along these lines below.
$endgroup$
– Ethan Bolker
Dec 30 '18 at 1:10












1 Answer
1






active

oldest

votes


















2












$begingroup$

Hint: Squaring your $P$ you will get
$$P^2=frac{1}{x^2+acx+a^2}+frac{1}{x^2+bcx+b^2}+frac{2}{sqrt{x^2+acx+a^2}sqrt{x^2+bcx+b^2}}$$. Now we can write
$$P^2-frac{1}{x^2+acx+a^2}-frac{1}{x^2+bcx+b^2}=frac{2}{sqrt{x^2+acx+a^2}sqrt{x^2+bcx+b^2}}$$
Now you must square again.
Good luck!
$${P}^{4}{a}^{2}{b}^{2}{c}^{4}{x}^{4}+2,{P}^{4}{a}^{3}{b}^{2}{c}^{3}{x}
^{3}+2,{P}^{4}{a}^{2}{b}^{3}{c}^{3}{x}^{3}+2,{P}^{4}{a}^{2}b{c}^{3}{
x}^{5}+2,{P}^{4}a{b}^{2}{c}^{3}{x}^{5}+{P}^{4}{a}^{4}{b}^{2}{c}^{2}{x
}^{2}+4,{P}^{4}{a}^{3}{b}^{3}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{3}b{c}^{2}
{x}^{4}+{P}^{4}{a}^{2}{b}^{4}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{2}{b}^{2}{c
}^{2}{x}^{4}+{P}^{4}{a}^{2}{c}^{2}{x}^{6}+4,{P}^{4}a{b}^{3}{c}^{2}{x}
^{4}+4,{P}^{4}ab{c}^{2}{x}^{6}+{P}^{4}{b}^{2}{c}^{2}{x}^{6}+2,{P}^{4
}{a}^{4}{b}^{3}cx+2,{P}^{4}{a}^{4}bc{x}^{3}+2,{P}^{4}{a}^{3}{b}^{4}c
x+4,{P}^{4}{a}^{3}{b}^{2}c{x}^{3}+2,{P}^{4}{a}^{3}c{x}^{5}+4,{P}^{4
}{a}^{2}{b}^{3}c{x}^{3}+4,{P}^{4}{a}^{2}bc{x}^{5}+2,{P}^{4}a{b}^{4}c
{x}^{3}+4,{P}^{4}a{b}^{2}c{x}^{5}+2,{P}^{4}ac{x}^{7}+2,{P}^{4}{b}^{
3}c{x}^{5}+2,{P}^{4}bc{x}^{7}+{P}^{4}{a}^{4}{b}^{4}+2,{P}^{4}{a}^{4}
{b}^{2}{x}^{2}+{P}^{4}{a}^{4}{x}^{4}+2,{P}^{4}{a}^{2}{b}^{4}{x}^{2}+4
,{P}^{4}{a}^{2}{b}^{2}{x}^{4}+2,{P}^{4}{a}^{2}{x}^{6}+{P}^{4}{b}^{4}
{x}^{4}+2,{P}^{4}{b}^{2}{x}^{6}+{P}^{4}{x}^{8}-2,{P}^{2}{a}^{2}b{c}^
{3}{x}^{3}-2,{P}^{2}a{b}^{2}{c}^{3}{x}^{3}-4,{P}^{2}{a}^{3}b{c}^{2}{
x}^{2}-4,{P}^{2}{a}^{2}{b}^{2}{c}^{2}{x}^{2}-2,{P}^{2}{a}^{2}{c}^{2}
{x}^{4}-4,{P}^{2}a{b}^{3}{c}^{2}{x}^{2}-8,{P}^{2}ab{c}^{2}{x}^{4}-2
,{P}^{2}{b}^{2}{c}^{2}{x}^{4}-2,{P}^{2}{a}^{4}bcx-4,{P}^{2}{a}^{3}{
b}^{2}cx-4,{P}^{2}{a}^{3}c{x}^{3}-4,{P}^{2}{a}^{2}{b}^{3}cx-8,{P}^{
2}{a}^{2}bc{x}^{3}-2,{P}^{2}a{b}^{4}cx-8,{P}^{2}a{b}^{2}c{x}^{3}-6,
{P}^{2}ac{x}^{5}-4,{P}^{2}{b}^{3}c{x}^{3}-6,{P}^{2}bc{x}^{5}-2,{P}^
{2}{a}^{4}{b}^{2}-2,{P}^{2}{a}^{4}{x}^{2}-2,{P}^{2}{a}^{2}{b}^{4}-8
,{P}^{2}{a}^{2}{b}^{2}{x}^{2}-6,{P}^{2}{a}^{2}{x}^{4}-2,{P}^{2}{b}^
{4}{x}^{2}-6,{P}^{2}{b}^{2}{x}^{4}-4,{P}^{2}{x}^{6}+{a}^{2}{c}^{2}{x
}^{2}-2,ab{c}^{2}{x}^{2}+{b}^{2}{c}^{2}{x}^{2}+2,{a}^{3}cx-2,{a}^{2
}bcx-2,a{b}^{2}cx+2,{b}^{3}cx+{a}^{4}-2,{a}^{2}{b}^{2}+{b}^{4}
=0$$






share|cite|improve this answer











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  • $begingroup$
    If you square the given equation again you will get a polynomial in degree six, to solve this you will need a numerical method.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 30 '18 at 9:58


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Hint: Squaring your $P$ you will get
$$P^2=frac{1}{x^2+acx+a^2}+frac{1}{x^2+bcx+b^2}+frac{2}{sqrt{x^2+acx+a^2}sqrt{x^2+bcx+b^2}}$$. Now we can write
$$P^2-frac{1}{x^2+acx+a^2}-frac{1}{x^2+bcx+b^2}=frac{2}{sqrt{x^2+acx+a^2}sqrt{x^2+bcx+b^2}}$$
Now you must square again.
Good luck!
$${P}^{4}{a}^{2}{b}^{2}{c}^{4}{x}^{4}+2,{P}^{4}{a}^{3}{b}^{2}{c}^{3}{x}
^{3}+2,{P}^{4}{a}^{2}{b}^{3}{c}^{3}{x}^{3}+2,{P}^{4}{a}^{2}b{c}^{3}{
x}^{5}+2,{P}^{4}a{b}^{2}{c}^{3}{x}^{5}+{P}^{4}{a}^{4}{b}^{2}{c}^{2}{x
}^{2}+4,{P}^{4}{a}^{3}{b}^{3}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{3}b{c}^{2}
{x}^{4}+{P}^{4}{a}^{2}{b}^{4}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{2}{b}^{2}{c
}^{2}{x}^{4}+{P}^{4}{a}^{2}{c}^{2}{x}^{6}+4,{P}^{4}a{b}^{3}{c}^{2}{x}
^{4}+4,{P}^{4}ab{c}^{2}{x}^{6}+{P}^{4}{b}^{2}{c}^{2}{x}^{6}+2,{P}^{4
}{a}^{4}{b}^{3}cx+2,{P}^{4}{a}^{4}bc{x}^{3}+2,{P}^{4}{a}^{3}{b}^{4}c
x+4,{P}^{4}{a}^{3}{b}^{2}c{x}^{3}+2,{P}^{4}{a}^{3}c{x}^{5}+4,{P}^{4
}{a}^{2}{b}^{3}c{x}^{3}+4,{P}^{4}{a}^{2}bc{x}^{5}+2,{P}^{4}a{b}^{4}c
{x}^{3}+4,{P}^{4}a{b}^{2}c{x}^{5}+2,{P}^{4}ac{x}^{7}+2,{P}^{4}{b}^{
3}c{x}^{5}+2,{P}^{4}bc{x}^{7}+{P}^{4}{a}^{4}{b}^{4}+2,{P}^{4}{a}^{4}
{b}^{2}{x}^{2}+{P}^{4}{a}^{4}{x}^{4}+2,{P}^{4}{a}^{2}{b}^{4}{x}^{2}+4
,{P}^{4}{a}^{2}{b}^{2}{x}^{4}+2,{P}^{4}{a}^{2}{x}^{6}+{P}^{4}{b}^{4}
{x}^{4}+2,{P}^{4}{b}^{2}{x}^{6}+{P}^{4}{x}^{8}-2,{P}^{2}{a}^{2}b{c}^
{3}{x}^{3}-2,{P}^{2}a{b}^{2}{c}^{3}{x}^{3}-4,{P}^{2}{a}^{3}b{c}^{2}{
x}^{2}-4,{P}^{2}{a}^{2}{b}^{2}{c}^{2}{x}^{2}-2,{P}^{2}{a}^{2}{c}^{2}
{x}^{4}-4,{P}^{2}a{b}^{3}{c}^{2}{x}^{2}-8,{P}^{2}ab{c}^{2}{x}^{4}-2
,{P}^{2}{b}^{2}{c}^{2}{x}^{4}-2,{P}^{2}{a}^{4}bcx-4,{P}^{2}{a}^{3}{
b}^{2}cx-4,{P}^{2}{a}^{3}c{x}^{3}-4,{P}^{2}{a}^{2}{b}^{3}cx-8,{P}^{
2}{a}^{2}bc{x}^{3}-2,{P}^{2}a{b}^{4}cx-8,{P}^{2}a{b}^{2}c{x}^{3}-6,
{P}^{2}ac{x}^{5}-4,{P}^{2}{b}^{3}c{x}^{3}-6,{P}^{2}bc{x}^{5}-2,{P}^
{2}{a}^{4}{b}^{2}-2,{P}^{2}{a}^{4}{x}^{2}-2,{P}^{2}{a}^{2}{b}^{4}-8
,{P}^{2}{a}^{2}{b}^{2}{x}^{2}-6,{P}^{2}{a}^{2}{x}^{4}-2,{P}^{2}{b}^
{4}{x}^{2}-6,{P}^{2}{b}^{2}{x}^{4}-4,{P}^{2}{x}^{6}+{a}^{2}{c}^{2}{x
}^{2}-2,ab{c}^{2}{x}^{2}+{b}^{2}{c}^{2}{x}^{2}+2,{a}^{3}cx-2,{a}^{2
}bcx-2,a{b}^{2}cx+2,{b}^{3}cx+{a}^{4}-2,{a}^{2}{b}^{2}+{b}^{4}
=0$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    If you square the given equation again you will get a polynomial in degree six, to solve this you will need a numerical method.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 30 '18 at 9:58
















2












$begingroup$

Hint: Squaring your $P$ you will get
$$P^2=frac{1}{x^2+acx+a^2}+frac{1}{x^2+bcx+b^2}+frac{2}{sqrt{x^2+acx+a^2}sqrt{x^2+bcx+b^2}}$$. Now we can write
$$P^2-frac{1}{x^2+acx+a^2}-frac{1}{x^2+bcx+b^2}=frac{2}{sqrt{x^2+acx+a^2}sqrt{x^2+bcx+b^2}}$$
Now you must square again.
Good luck!
$${P}^{4}{a}^{2}{b}^{2}{c}^{4}{x}^{4}+2,{P}^{4}{a}^{3}{b}^{2}{c}^{3}{x}
^{3}+2,{P}^{4}{a}^{2}{b}^{3}{c}^{3}{x}^{3}+2,{P}^{4}{a}^{2}b{c}^{3}{
x}^{5}+2,{P}^{4}a{b}^{2}{c}^{3}{x}^{5}+{P}^{4}{a}^{4}{b}^{2}{c}^{2}{x
}^{2}+4,{P}^{4}{a}^{3}{b}^{3}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{3}b{c}^{2}
{x}^{4}+{P}^{4}{a}^{2}{b}^{4}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{2}{b}^{2}{c
}^{2}{x}^{4}+{P}^{4}{a}^{2}{c}^{2}{x}^{6}+4,{P}^{4}a{b}^{3}{c}^{2}{x}
^{4}+4,{P}^{4}ab{c}^{2}{x}^{6}+{P}^{4}{b}^{2}{c}^{2}{x}^{6}+2,{P}^{4
}{a}^{4}{b}^{3}cx+2,{P}^{4}{a}^{4}bc{x}^{3}+2,{P}^{4}{a}^{3}{b}^{4}c
x+4,{P}^{4}{a}^{3}{b}^{2}c{x}^{3}+2,{P}^{4}{a}^{3}c{x}^{5}+4,{P}^{4
}{a}^{2}{b}^{3}c{x}^{3}+4,{P}^{4}{a}^{2}bc{x}^{5}+2,{P}^{4}a{b}^{4}c
{x}^{3}+4,{P}^{4}a{b}^{2}c{x}^{5}+2,{P}^{4}ac{x}^{7}+2,{P}^{4}{b}^{
3}c{x}^{5}+2,{P}^{4}bc{x}^{7}+{P}^{4}{a}^{4}{b}^{4}+2,{P}^{4}{a}^{4}
{b}^{2}{x}^{2}+{P}^{4}{a}^{4}{x}^{4}+2,{P}^{4}{a}^{2}{b}^{4}{x}^{2}+4
,{P}^{4}{a}^{2}{b}^{2}{x}^{4}+2,{P}^{4}{a}^{2}{x}^{6}+{P}^{4}{b}^{4}
{x}^{4}+2,{P}^{4}{b}^{2}{x}^{6}+{P}^{4}{x}^{8}-2,{P}^{2}{a}^{2}b{c}^
{3}{x}^{3}-2,{P}^{2}a{b}^{2}{c}^{3}{x}^{3}-4,{P}^{2}{a}^{3}b{c}^{2}{
x}^{2}-4,{P}^{2}{a}^{2}{b}^{2}{c}^{2}{x}^{2}-2,{P}^{2}{a}^{2}{c}^{2}
{x}^{4}-4,{P}^{2}a{b}^{3}{c}^{2}{x}^{2}-8,{P}^{2}ab{c}^{2}{x}^{4}-2
,{P}^{2}{b}^{2}{c}^{2}{x}^{4}-2,{P}^{2}{a}^{4}bcx-4,{P}^{2}{a}^{3}{
b}^{2}cx-4,{P}^{2}{a}^{3}c{x}^{3}-4,{P}^{2}{a}^{2}{b}^{3}cx-8,{P}^{
2}{a}^{2}bc{x}^{3}-2,{P}^{2}a{b}^{4}cx-8,{P}^{2}a{b}^{2}c{x}^{3}-6,
{P}^{2}ac{x}^{5}-4,{P}^{2}{b}^{3}c{x}^{3}-6,{P}^{2}bc{x}^{5}-2,{P}^
{2}{a}^{4}{b}^{2}-2,{P}^{2}{a}^{4}{x}^{2}-2,{P}^{2}{a}^{2}{b}^{4}-8
,{P}^{2}{a}^{2}{b}^{2}{x}^{2}-6,{P}^{2}{a}^{2}{x}^{4}-2,{P}^{2}{b}^
{4}{x}^{2}-6,{P}^{2}{b}^{2}{x}^{4}-4,{P}^{2}{x}^{6}+{a}^{2}{c}^{2}{x
}^{2}-2,ab{c}^{2}{x}^{2}+{b}^{2}{c}^{2}{x}^{2}+2,{a}^{3}cx-2,{a}^{2
}bcx-2,a{b}^{2}cx+2,{b}^{3}cx+{a}^{4}-2,{a}^{2}{b}^{2}+{b}^{4}
=0$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    If you square the given equation again you will get a polynomial in degree six, to solve this you will need a numerical method.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 30 '18 at 9:58














2












2








2





$begingroup$

Hint: Squaring your $P$ you will get
$$P^2=frac{1}{x^2+acx+a^2}+frac{1}{x^2+bcx+b^2}+frac{2}{sqrt{x^2+acx+a^2}sqrt{x^2+bcx+b^2}}$$. Now we can write
$$P^2-frac{1}{x^2+acx+a^2}-frac{1}{x^2+bcx+b^2}=frac{2}{sqrt{x^2+acx+a^2}sqrt{x^2+bcx+b^2}}$$
Now you must square again.
Good luck!
$${P}^{4}{a}^{2}{b}^{2}{c}^{4}{x}^{4}+2,{P}^{4}{a}^{3}{b}^{2}{c}^{3}{x}
^{3}+2,{P}^{4}{a}^{2}{b}^{3}{c}^{3}{x}^{3}+2,{P}^{4}{a}^{2}b{c}^{3}{
x}^{5}+2,{P}^{4}a{b}^{2}{c}^{3}{x}^{5}+{P}^{4}{a}^{4}{b}^{2}{c}^{2}{x
}^{2}+4,{P}^{4}{a}^{3}{b}^{3}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{3}b{c}^{2}
{x}^{4}+{P}^{4}{a}^{2}{b}^{4}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{2}{b}^{2}{c
}^{2}{x}^{4}+{P}^{4}{a}^{2}{c}^{2}{x}^{6}+4,{P}^{4}a{b}^{3}{c}^{2}{x}
^{4}+4,{P}^{4}ab{c}^{2}{x}^{6}+{P}^{4}{b}^{2}{c}^{2}{x}^{6}+2,{P}^{4
}{a}^{4}{b}^{3}cx+2,{P}^{4}{a}^{4}bc{x}^{3}+2,{P}^{4}{a}^{3}{b}^{4}c
x+4,{P}^{4}{a}^{3}{b}^{2}c{x}^{3}+2,{P}^{4}{a}^{3}c{x}^{5}+4,{P}^{4
}{a}^{2}{b}^{3}c{x}^{3}+4,{P}^{4}{a}^{2}bc{x}^{5}+2,{P}^{4}a{b}^{4}c
{x}^{3}+4,{P}^{4}a{b}^{2}c{x}^{5}+2,{P}^{4}ac{x}^{7}+2,{P}^{4}{b}^{
3}c{x}^{5}+2,{P}^{4}bc{x}^{7}+{P}^{4}{a}^{4}{b}^{4}+2,{P}^{4}{a}^{4}
{b}^{2}{x}^{2}+{P}^{4}{a}^{4}{x}^{4}+2,{P}^{4}{a}^{2}{b}^{4}{x}^{2}+4
,{P}^{4}{a}^{2}{b}^{2}{x}^{4}+2,{P}^{4}{a}^{2}{x}^{6}+{P}^{4}{b}^{4}
{x}^{4}+2,{P}^{4}{b}^{2}{x}^{6}+{P}^{4}{x}^{8}-2,{P}^{2}{a}^{2}b{c}^
{3}{x}^{3}-2,{P}^{2}a{b}^{2}{c}^{3}{x}^{3}-4,{P}^{2}{a}^{3}b{c}^{2}{
x}^{2}-4,{P}^{2}{a}^{2}{b}^{2}{c}^{2}{x}^{2}-2,{P}^{2}{a}^{2}{c}^{2}
{x}^{4}-4,{P}^{2}a{b}^{3}{c}^{2}{x}^{2}-8,{P}^{2}ab{c}^{2}{x}^{4}-2
,{P}^{2}{b}^{2}{c}^{2}{x}^{4}-2,{P}^{2}{a}^{4}bcx-4,{P}^{2}{a}^{3}{
b}^{2}cx-4,{P}^{2}{a}^{3}c{x}^{3}-4,{P}^{2}{a}^{2}{b}^{3}cx-8,{P}^{
2}{a}^{2}bc{x}^{3}-2,{P}^{2}a{b}^{4}cx-8,{P}^{2}a{b}^{2}c{x}^{3}-6,
{P}^{2}ac{x}^{5}-4,{P}^{2}{b}^{3}c{x}^{3}-6,{P}^{2}bc{x}^{5}-2,{P}^
{2}{a}^{4}{b}^{2}-2,{P}^{2}{a}^{4}{x}^{2}-2,{P}^{2}{a}^{2}{b}^{4}-8
,{P}^{2}{a}^{2}{b}^{2}{x}^{2}-6,{P}^{2}{a}^{2}{x}^{4}-2,{P}^{2}{b}^
{4}{x}^{2}-6,{P}^{2}{b}^{2}{x}^{4}-4,{P}^{2}{x}^{6}+{a}^{2}{c}^{2}{x
}^{2}-2,ab{c}^{2}{x}^{2}+{b}^{2}{c}^{2}{x}^{2}+2,{a}^{3}cx-2,{a}^{2
}bcx-2,a{b}^{2}cx+2,{b}^{3}cx+{a}^{4}-2,{a}^{2}{b}^{2}+{b}^{4}
=0$$






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$endgroup$



Hint: Squaring your $P$ you will get
$$P^2=frac{1}{x^2+acx+a^2}+frac{1}{x^2+bcx+b^2}+frac{2}{sqrt{x^2+acx+a^2}sqrt{x^2+bcx+b^2}}$$. Now we can write
$$P^2-frac{1}{x^2+acx+a^2}-frac{1}{x^2+bcx+b^2}=frac{2}{sqrt{x^2+acx+a^2}sqrt{x^2+bcx+b^2}}$$
Now you must square again.
Good luck!
$${P}^{4}{a}^{2}{b}^{2}{c}^{4}{x}^{4}+2,{P}^{4}{a}^{3}{b}^{2}{c}^{3}{x}
^{3}+2,{P}^{4}{a}^{2}{b}^{3}{c}^{3}{x}^{3}+2,{P}^{4}{a}^{2}b{c}^{3}{
x}^{5}+2,{P}^{4}a{b}^{2}{c}^{3}{x}^{5}+{P}^{4}{a}^{4}{b}^{2}{c}^{2}{x
}^{2}+4,{P}^{4}{a}^{3}{b}^{3}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{3}b{c}^{2}
{x}^{4}+{P}^{4}{a}^{2}{b}^{4}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{2}{b}^{2}{c
}^{2}{x}^{4}+{P}^{4}{a}^{2}{c}^{2}{x}^{6}+4,{P}^{4}a{b}^{3}{c}^{2}{x}
^{4}+4,{P}^{4}ab{c}^{2}{x}^{6}+{P}^{4}{b}^{2}{c}^{2}{x}^{6}+2,{P}^{4
}{a}^{4}{b}^{3}cx+2,{P}^{4}{a}^{4}bc{x}^{3}+2,{P}^{4}{a}^{3}{b}^{4}c
x+4,{P}^{4}{a}^{3}{b}^{2}c{x}^{3}+2,{P}^{4}{a}^{3}c{x}^{5}+4,{P}^{4
}{a}^{2}{b}^{3}c{x}^{3}+4,{P}^{4}{a}^{2}bc{x}^{5}+2,{P}^{4}a{b}^{4}c
{x}^{3}+4,{P}^{4}a{b}^{2}c{x}^{5}+2,{P}^{4}ac{x}^{7}+2,{P}^{4}{b}^{
3}c{x}^{5}+2,{P}^{4}bc{x}^{7}+{P}^{4}{a}^{4}{b}^{4}+2,{P}^{4}{a}^{4}
{b}^{2}{x}^{2}+{P}^{4}{a}^{4}{x}^{4}+2,{P}^{4}{a}^{2}{b}^{4}{x}^{2}+4
,{P}^{4}{a}^{2}{b}^{2}{x}^{4}+2,{P}^{4}{a}^{2}{x}^{6}+{P}^{4}{b}^{4}
{x}^{4}+2,{P}^{4}{b}^{2}{x}^{6}+{P}^{4}{x}^{8}-2,{P}^{2}{a}^{2}b{c}^
{3}{x}^{3}-2,{P}^{2}a{b}^{2}{c}^{3}{x}^{3}-4,{P}^{2}{a}^{3}b{c}^{2}{
x}^{2}-4,{P}^{2}{a}^{2}{b}^{2}{c}^{2}{x}^{2}-2,{P}^{2}{a}^{2}{c}^{2}
{x}^{4}-4,{P}^{2}a{b}^{3}{c}^{2}{x}^{2}-8,{P}^{2}ab{c}^{2}{x}^{4}-2
,{P}^{2}{b}^{2}{c}^{2}{x}^{4}-2,{P}^{2}{a}^{4}bcx-4,{P}^{2}{a}^{3}{
b}^{2}cx-4,{P}^{2}{a}^{3}c{x}^{3}-4,{P}^{2}{a}^{2}{b}^{3}cx-8,{P}^{
2}{a}^{2}bc{x}^{3}-2,{P}^{2}a{b}^{4}cx-8,{P}^{2}a{b}^{2}c{x}^{3}-6,
{P}^{2}ac{x}^{5}-4,{P}^{2}{b}^{3}c{x}^{3}-6,{P}^{2}bc{x}^{5}-2,{P}^
{2}{a}^{4}{b}^{2}-2,{P}^{2}{a}^{4}{x}^{2}-2,{P}^{2}{a}^{2}{b}^{4}-8
,{P}^{2}{a}^{2}{b}^{2}{x}^{2}-6,{P}^{2}{a}^{2}{x}^{4}-2,{P}^{2}{b}^
{4}{x}^{2}-6,{P}^{2}{b}^{2}{x}^{4}-4,{P}^{2}{x}^{6}+{a}^{2}{c}^{2}{x
}^{2}-2,ab{c}^{2}{x}^{2}+{b}^{2}{c}^{2}{x}^{2}+2,{a}^{3}cx-2,{a}^{2
}bcx-2,a{b}^{2}cx+2,{b}^{3}cx+{a}^{4}-2,{a}^{2}{b}^{2}+{b}^{4}
=0$$







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share|cite|improve this answer



share|cite|improve this answer








edited Dec 30 '18 at 10:08

























answered Dec 29 '18 at 21:56









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

73.5k42864




73.5k42864












  • $begingroup$
    If you square the given equation again you will get a polynomial in degree six, to solve this you will need a numerical method.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 30 '18 at 9:58


















  • $begingroup$
    If you square the given equation again you will get a polynomial in degree six, to solve this you will need a numerical method.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 30 '18 at 9:58
















$begingroup$
If you square the given equation again you will get a polynomial in degree six, to solve this you will need a numerical method.
$endgroup$
– Dr. Sonnhard Graubner
Dec 30 '18 at 9:58




$begingroup$
If you square the given equation again you will get a polynomial in degree six, to solve this you will need a numerical method.
$endgroup$
– Dr. Sonnhard Graubner
Dec 30 '18 at 9:58



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