Is it possible to isolate the variable X in this equation? [closed]
$begingroup$
$P = dfrac{1}{sqrt{x^2+xac+a^2}} + dfrac{1}{sqrt{x^2+xbc+b^2}}$
algebra-precalculus quadratics
edited Dec 29 '18 at 21:57
ÍgjøgnumMeg
2,79511029
asked Dec 29 '18 at 21:42
Gonzalo FerrandoGonzalo Ferrando
111
$endgroup$
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.
add a comment |
$begingroup$
$P = dfrac{1}{sqrt{x^2+xac+a^2}} + dfrac{1}{sqrt{x^2+xbc+b^2}}$
algebra-precalculus quadratics
edited Dec 29 '18 at 21:57
ÍgjøgnumMeg
2,79511029
asked Dec 29 '18 at 21:42
Gonzalo FerrandoGonzalo Ferrando
111
$endgroup$
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
add a comment |
$begingroup$
$P = dfrac{1}{sqrt{x^2+xac+a^2}} + dfrac{1}{sqrt{x^2+xbc+b^2}}$
algebra-precalculus quadratics
edited Dec 29 '18 at 21:57
ÍgjøgnumMeg
2,79511029
asked Dec 29 '18 at 21:42
Gonzalo FerrandoGonzalo Ferrando
111
$endgroup$
$P = dfrac{1}{sqrt{x^2+xac+a^2}} + dfrac{1}{sqrt{x^2+xbc+b^2}}$
algebra-precalculus quadratics
algebra-precalculus quadratics
edited Dec 29 '18 at 21:57
ÍgjøgnumMeg
2,79511029
asked Dec 29 '18 at 21:42
Gonzalo FerrandoGonzalo Ferrando
111
edited Dec 29 '18 at 21:57
ÍgjøgnumMeg
2,79511029
asked Dec 29 '18 at 21:42
Gonzalo FerrandoGonzalo Ferrando
111
edited Dec 29 '18 at 21:57
ÍgjøgnumMeg
2,79511029
edited Dec 29 '18 at 21:57
ÍgjøgnumMeg
2,79511029
edited Dec 29 '18 at 21:57
ÍgjøgnumMeg
2,79511029
2,79511029
asked Dec 29 '18 at 21:42
Gonzalo FerrandoGonzalo Ferrando
111
asked Dec 29 '18 at 21:42
Gonzalo FerrandoGonzalo Ferrando
111
asked Dec 29 '18 at 21:42
Gonzalo FerrandoGonzalo Ferrando
111
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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!
$${P}^{4}{a}^{2}{b}^{2}{c}^{4}{x}^{4}+2,{P}^{4}{a}^{3}{b}^{2}{c}^{3}{x}
^{3}+2,{P}^{4}{a}^{2}{b}^{3}{c}^{3}{x}^{3}+2,{P}^{4}{a}^{2}b{c}^{3}{
x}^{5}+2,{P}^{4}a{b}^{2}{c}^{3}{x}^{5}+{P}^{4}{a}^{4}{b}^{2}{c}^{2}{x
}^{2}+4,{P}^{4}{a}^{3}{b}^{3}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{3}b{c}^{2}
{x}^{4}+{P}^{4}{a}^{2}{b}^{4}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{2}{b}^{2}{c
}^{2}{x}^{4}+{P}^{4}{a}^{2}{c}^{2}{x}^{6}+4,{P}^{4}a{b}^{3}{c}^{2}{x}
^{4}+4,{P}^{4}ab{c}^{2}{x}^{6}+{P}^{4}{b}^{2}{c}^{2}{x}^{6}+2,{P}^{4
}{a}^{4}{b}^{3}cx+2,{P}^{4}{a}^{4}bc{x}^{3}+2,{P}^{4}{a}^{3}{b}^{4}c
x+4,{P}^{4}{a}^{3}{b}^{2}c{x}^{3}+2,{P}^{4}{a}^{3}c{x}^{5}+4,{P}^{4
}{a}^{2}{b}^{3}c{x}^{3}+4,{P}^{4}{a}^{2}bc{x}^{5}+2,{P}^{4}a{b}^{4}c
{x}^{3}+4,{P}^{4}a{b}^{2}c{x}^{5}+2,{P}^{4}ac{x}^{7}+2,{P}^{4}{b}^{
3}c{x}^{5}+2,{P}^{4}bc{x}^{7}+{P}^{4}{a}^{4}{b}^{4}+2,{P}^{4}{a}^{4}
{b}^{2}{x}^{2}+{P}^{4}{a}^{4}{x}^{4}+2,{P}^{4}{a}^{2}{b}^{4}{x}^{2}+4
,{P}^{4}{a}^{2}{b}^{2}{x}^{4}+2,{P}^{4}{a}^{2}{x}^{6}+{P}^{4}{b}^{4}
{x}^{4}+2,{P}^{4}{b}^{2}{x}^{6}+{P}^{4}{x}^{8}-2,{P}^{2}{a}^{2}b{c}^
{3}{x}^{3}-2,{P}^{2}a{b}^{2}{c}^{3}{x}^{3}-4,{P}^{2}{a}^{3}b{c}^{2}{
x}^{2}-4,{P}^{2}{a}^{2}{b}^{2}{c}^{2}{x}^{2}-2,{P}^{2}{a}^{2}{c}^{2}
{x}^{4}-4,{P}^{2}a{b}^{3}{c}^{2}{x}^{2}-8,{P}^{2}ab{c}^{2}{x}^{4}-2
,{P}^{2}{b}^{2}{c}^{2}{x}^{4}-2,{P}^{2}{a}^{4}bcx-4,{P}^{2}{a}^{3}{
b}^{2}cx-4,{P}^{2}{a}^{3}c{x}^{3}-4,{P}^{2}{a}^{2}{b}^{3}cx-8,{P}^{
2}{a}^{2}bc{x}^{3}-2,{P}^{2}a{b}^{4}cx-8,{P}^{2}a{b}^{2}c{x}^{3}-6,
{P}^{2}ac{x}^{5}-4,{P}^{2}{b}^{3}c{x}^{3}-6,{P}^{2}bc{x}^{5}-2,{P}^
{2}{a}^{4}{b}^{2}-2,{P}^{2}{a}^{4}{x}^{2}-2,{P}^{2}{a}^{2}{b}^{4}-8
,{P}^{2}{a}^{2}{b}^{2}{x}^{2}-6,{P}^{2}{a}^{2}{x}^{4}-2,{P}^{2}{b}^
{4}{x}^{2}-6,{P}^{2}{b}^{2}{x}^{4}-4,{P}^{2}{x}^{6}+{a}^{2}{c}^{2}{x
}^{2}-2,ab{c}^{2}{x}^{2}+{b}^{2}{c}^{2}{x}^{2}+2,{a}^{3}cx-2,{a}^{2
}bcx-2,a{b}^{2}cx+2,{b}^{3}cx+{a}^{4}-2,{a}^{2}{b}^{2}+{b}^{4}
=0$$
answered Dec 29 '18 at 21:56
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42864
$endgroup$
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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!
$${P}^{4}{a}^{2}{b}^{2}{c}^{4}{x}^{4}+2,{P}^{4}{a}^{3}{b}^{2}{c}^{3}{x}
^{3}+2,{P}^{4}{a}^{2}{b}^{3}{c}^{3}{x}^{3}+2,{P}^{4}{a}^{2}b{c}^{3}{
x}^{5}+2,{P}^{4}a{b}^{2}{c}^{3}{x}^{5}+{P}^{4}{a}^{4}{b}^{2}{c}^{2}{x
}^{2}+4,{P}^{4}{a}^{3}{b}^{3}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{3}b{c}^{2}
{x}^{4}+{P}^{4}{a}^{2}{b}^{4}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{2}{b}^{2}{c
}^{2}{x}^{4}+{P}^{4}{a}^{2}{c}^{2}{x}^{6}+4,{P}^{4}a{b}^{3}{c}^{2}{x}
^{4}+4,{P}^{4}ab{c}^{2}{x}^{6}+{P}^{4}{b}^{2}{c}^{2}{x}^{6}+2,{P}^{4
}{a}^{4}{b}^{3}cx+2,{P}^{4}{a}^{4}bc{x}^{3}+2,{P}^{4}{a}^{3}{b}^{4}c
x+4,{P}^{4}{a}^{3}{b}^{2}c{x}^{3}+2,{P}^{4}{a}^{3}c{x}^{5}+4,{P}^{4
}{a}^{2}{b}^{3}c{x}^{3}+4,{P}^{4}{a}^{2}bc{x}^{5}+2,{P}^{4}a{b}^{4}c
{x}^{3}+4,{P}^{4}a{b}^{2}c{x}^{5}+2,{P}^{4}ac{x}^{7}+2,{P}^{4}{b}^{
3}c{x}^{5}+2,{P}^{4}bc{x}^{7}+{P}^{4}{a}^{4}{b}^{4}+2,{P}^{4}{a}^{4}
{b}^{2}{x}^{2}+{P}^{4}{a}^{4}{x}^{4}+2,{P}^{4}{a}^{2}{b}^{4}{x}^{2}+4
,{P}^{4}{a}^{2}{b}^{2}{x}^{4}+2,{P}^{4}{a}^{2}{x}^{6}+{P}^{4}{b}^{4}
{x}^{4}+2,{P}^{4}{b}^{2}{x}^{6}+{P}^{4}{x}^{8}-2,{P}^{2}{a}^{2}b{c}^
{3}{x}^{3}-2,{P}^{2}a{b}^{2}{c}^{3}{x}^{3}-4,{P}^{2}{a}^{3}b{c}^{2}{
x}^{2}-4,{P}^{2}{a}^{2}{b}^{2}{c}^{2}{x}^{2}-2,{P}^{2}{a}^{2}{c}^{2}
{x}^{4}-4,{P}^{2}a{b}^{3}{c}^{2}{x}^{2}-8,{P}^{2}ab{c}^{2}{x}^{4}-2
,{P}^{2}{b}^{2}{c}^{2}{x}^{4}-2,{P}^{2}{a}^{4}bcx-4,{P}^{2}{a}^{3}{
b}^{2}cx-4,{P}^{2}{a}^{3}c{x}^{3}-4,{P}^{2}{a}^{2}{b}^{3}cx-8,{P}^{
2}{a}^{2}bc{x}^{3}-2,{P}^{2}a{b}^{4}cx-8,{P}^{2}a{b}^{2}c{x}^{3}-6,
{P}^{2}ac{x}^{5}-4,{P}^{2}{b}^{3}c{x}^{3}-6,{P}^{2}bc{x}^{5}-2,{P}^
{2}{a}^{4}{b}^{2}-2,{P}^{2}{a}^{4}{x}^{2}-2,{P}^{2}{a}^{2}{b}^{4}-8
,{P}^{2}{a}^{2}{b}^{2}{x}^{2}-6,{P}^{2}{a}^{2}{x}^{4}-2,{P}^{2}{b}^
{4}{x}^{2}-6,{P}^{2}{b}^{2}{x}^{4}-4,{P}^{2}{x}^{6}+{a}^{2}{c}^{2}{x
}^{2}-2,ab{c}^{2}{x}^{2}+{b}^{2}{c}^{2}{x}^{2}+2,{a}^{3}cx-2,{a}^{2
}bcx-2,a{b}^{2}cx+2,{b}^{3}cx+{a}^{4}-2,{a}^{2}{b}^{2}+{b}^{4}
=0$$
answered Dec 29 '18 at 21:56
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42864
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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!
$${P}^{4}{a}^{2}{b}^{2}{c}^{4}{x}^{4}+2,{P}^{4}{a}^{3}{b}^{2}{c}^{3}{x}
^{3}+2,{P}^{4}{a}^{2}{b}^{3}{c}^{3}{x}^{3}+2,{P}^{4}{a}^{2}b{c}^{3}{
x}^{5}+2,{P}^{4}a{b}^{2}{c}^{3}{x}^{5}+{P}^{4}{a}^{4}{b}^{2}{c}^{2}{x
}^{2}+4,{P}^{4}{a}^{3}{b}^{3}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{3}b{c}^{2}
{x}^{4}+{P}^{4}{a}^{2}{b}^{4}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{2}{b}^{2}{c
}^{2}{x}^{4}+{P}^{4}{a}^{2}{c}^{2}{x}^{6}+4,{P}^{4}a{b}^{3}{c}^{2}{x}
^{4}+4,{P}^{4}ab{c}^{2}{x}^{6}+{P}^{4}{b}^{2}{c}^{2}{x}^{6}+2,{P}^{4
}{a}^{4}{b}^{3}cx+2,{P}^{4}{a}^{4}bc{x}^{3}+2,{P}^{4}{a}^{3}{b}^{4}c
x+4,{P}^{4}{a}^{3}{b}^{2}c{x}^{3}+2,{P}^{4}{a}^{3}c{x}^{5}+4,{P}^{4
}{a}^{2}{b}^{3}c{x}^{3}+4,{P}^{4}{a}^{2}bc{x}^{5}+2,{P}^{4}a{b}^{4}c
{x}^{3}+4,{P}^{4}a{b}^{2}c{x}^{5}+2,{P}^{4}ac{x}^{7}+2,{P}^{4}{b}^{
3}c{x}^{5}+2,{P}^{4}bc{x}^{7}+{P}^{4}{a}^{4}{b}^{4}+2,{P}^{4}{a}^{4}
{b}^{2}{x}^{2}+{P}^{4}{a}^{4}{x}^{4}+2,{P}^{4}{a}^{2}{b}^{4}{x}^{2}+4
,{P}^{4}{a}^{2}{b}^{2}{x}^{4}+2,{P}^{4}{a}^{2}{x}^{6}+{P}^{4}{b}^{4}
{x}^{4}+2,{P}^{4}{b}^{2}{x}^{6}+{P}^{4}{x}^{8}-2,{P}^{2}{a}^{2}b{c}^
{3}{x}^{3}-2,{P}^{2}a{b}^{2}{c}^{3}{x}^{3}-4,{P}^{2}{a}^{3}b{c}^{2}{
x}^{2}-4,{P}^{2}{a}^{2}{b}^{2}{c}^{2}{x}^{2}-2,{P}^{2}{a}^{2}{c}^{2}
{x}^{4}-4,{P}^{2}a{b}^{3}{c}^{2}{x}^{2}-8,{P}^{2}ab{c}^{2}{x}^{4}-2
,{P}^{2}{b}^{2}{c}^{2}{x}^{4}-2,{P}^{2}{a}^{4}bcx-4,{P}^{2}{a}^{3}{
b}^{2}cx-4,{P}^{2}{a}^{3}c{x}^{3}-4,{P}^{2}{a}^{2}{b}^{3}cx-8,{P}^{
2}{a}^{2}bc{x}^{3}-2,{P}^{2}a{b}^{4}cx-8,{P}^{2}a{b}^{2}c{x}^{3}-6,
{P}^{2}ac{x}^{5}-4,{P}^{2}{b}^{3}c{x}^{3}-6,{P}^{2}bc{x}^{5}-2,{P}^
{2}{a}^{4}{b}^{2}-2,{P}^{2}{a}^{4}{x}^{2}-2,{P}^{2}{a}^{2}{b}^{4}-8
,{P}^{2}{a}^{2}{b}^{2}{x}^{2}-6,{P}^{2}{a}^{2}{x}^{4}-2,{P}^{2}{b}^
{4}{x}^{2}-6,{P}^{2}{b}^{2}{x}^{4}-4,{P}^{2}{x}^{6}+{a}^{2}{c}^{2}{x
}^{2}-2,ab{c}^{2}{x}^{2}+{b}^{2}{c}^{2}{x}^{2}+2,{a}^{3}cx-2,{a}^{2
}bcx-2,a{b}^{2}cx+2,{b}^{3}cx+{a}^{4}-2,{a}^{2}{b}^{2}+{b}^{4}
=0$$
answered Dec 29 '18 at 21:56
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42864
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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!
$${P}^{4}{a}^{2}{b}^{2}{c}^{4}{x}^{4}+2,{P}^{4}{a}^{3}{b}^{2}{c}^{3}{x}
^{3}+2,{P}^{4}{a}^{2}{b}^{3}{c}^{3}{x}^{3}+2,{P}^{4}{a}^{2}b{c}^{3}{
x}^{5}+2,{P}^{4}a{b}^{2}{c}^{3}{x}^{5}+{P}^{4}{a}^{4}{b}^{2}{c}^{2}{x
}^{2}+4,{P}^{4}{a}^{3}{b}^{3}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{3}b{c}^{2}
{x}^{4}+{P}^{4}{a}^{2}{b}^{4}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{2}{b}^{2}{c
}^{2}{x}^{4}+{P}^{4}{a}^{2}{c}^{2}{x}^{6}+4,{P}^{4}a{b}^{3}{c}^{2}{x}
^{4}+4,{P}^{4}ab{c}^{2}{x}^{6}+{P}^{4}{b}^{2}{c}^{2}{x}^{6}+2,{P}^{4
}{a}^{4}{b}^{3}cx+2,{P}^{4}{a}^{4}bc{x}^{3}+2,{P}^{4}{a}^{3}{b}^{4}c
x+4,{P}^{4}{a}^{3}{b}^{2}c{x}^{3}+2,{P}^{4}{a}^{3}c{x}^{5}+4,{P}^{4
}{a}^{2}{b}^{3}c{x}^{3}+4,{P}^{4}{a}^{2}bc{x}^{5}+2,{P}^{4}a{b}^{4}c
{x}^{3}+4,{P}^{4}a{b}^{2}c{x}^{5}+2,{P}^{4}ac{x}^{7}+2,{P}^{4}{b}^{
3}c{x}^{5}+2,{P}^{4}bc{x}^{7}+{P}^{4}{a}^{4}{b}^{4}+2,{P}^{4}{a}^{4}
{b}^{2}{x}^{2}+{P}^{4}{a}^{4}{x}^{4}+2,{P}^{4}{a}^{2}{b}^{4}{x}^{2}+4
,{P}^{4}{a}^{2}{b}^{2}{x}^{4}+2,{P}^{4}{a}^{2}{x}^{6}+{P}^{4}{b}^{4}
{x}^{4}+2,{P}^{4}{b}^{2}{x}^{6}+{P}^{4}{x}^{8}-2,{P}^{2}{a}^{2}b{c}^
{3}{x}^{3}-2,{P}^{2}a{b}^{2}{c}^{3}{x}^{3}-4,{P}^{2}{a}^{3}b{c}^{2}{
x}^{2}-4,{P}^{2}{a}^{2}{b}^{2}{c}^{2}{x}^{2}-2,{P}^{2}{a}^{2}{c}^{2}
{x}^{4}-4,{P}^{2}a{b}^{3}{c}^{2}{x}^{2}-8,{P}^{2}ab{c}^{2}{x}^{4}-2
,{P}^{2}{b}^{2}{c}^{2}{x}^{4}-2,{P}^{2}{a}^{4}bcx-4,{P}^{2}{a}^{3}{
b}^{2}cx-4,{P}^{2}{a}^{3}c{x}^{3}-4,{P}^{2}{a}^{2}{b}^{3}cx-8,{P}^{
2}{a}^{2}bc{x}^{3}-2,{P}^{2}a{b}^{4}cx-8,{P}^{2}a{b}^{2}c{x}^{3}-6,
{P}^{2}ac{x}^{5}-4,{P}^{2}{b}^{3}c{x}^{3}-6,{P}^{2}bc{x}^{5}-2,{P}^
{2}{a}^{4}{b}^{2}-2,{P}^{2}{a}^{4}{x}^{2}-2,{P}^{2}{a}^{2}{b}^{4}-8
,{P}^{2}{a}^{2}{b}^{2}{x}^{2}-6,{P}^{2}{a}^{2}{x}^{4}-2,{P}^{2}{b}^
{4}{x}^{2}-6,{P}^{2}{b}^{2}{x}^{4}-4,{P}^{2}{x}^{6}+{a}^{2}{c}^{2}{x
}^{2}-2,ab{c}^{2}{x}^{2}+{b}^{2}{c}^{2}{x}^{2}+2,{a}^{3}cx-2,{a}^{2
}bcx-2,a{b}^{2}cx+2,{b}^{3}cx+{a}^{4}-2,{a}^{2}{b}^{2}+{b}^{4}
=0$$
answered Dec 29 '18 at 21:56
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42864
$endgroup$
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!
$${P}^{4}{a}^{2}{b}^{2}{c}^{4}{x}^{4}+2,{P}^{4}{a}^{3}{b}^{2}{c}^{3}{x}
^{3}+2,{P}^{4}{a}^{2}{b}^{3}{c}^{3}{x}^{3}+2,{P}^{4}{a}^{2}b{c}^{3}{
x}^{5}+2,{P}^{4}a{b}^{2}{c}^{3}{x}^{5}+{P}^{4}{a}^{4}{b}^{2}{c}^{2}{x
}^{2}+4,{P}^{4}{a}^{3}{b}^{3}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{3}b{c}^{2}
{x}^{4}+{P}^{4}{a}^{2}{b}^{4}{c}^{2}{x}^{2}+4,{P}^{4}{a}^{2}{b}^{2}{c
}^{2}{x}^{4}+{P}^{4}{a}^{2}{c}^{2}{x}^{6}+4,{P}^{4}a{b}^{3}{c}^{2}{x}
^{4}+4,{P}^{4}ab{c}^{2}{x}^{6}+{P}^{4}{b}^{2}{c}^{2}{x}^{6}+2,{P}^{4
}{a}^{4}{b}^{3}cx+2,{P}^{4}{a}^{4}bc{x}^{3}+2,{P}^{4}{a}^{3}{b}^{4}c
x+4,{P}^{4}{a}^{3}{b}^{2}c{x}^{3}+2,{P}^{4}{a}^{3}c{x}^{5}+4,{P}^{4
}{a}^{2}{b}^{3}c{x}^{3}+4,{P}^{4}{a}^{2}bc{x}^{5}+2,{P}^{4}a{b}^{4}c
{x}^{3}+4,{P}^{4}a{b}^{2}c{x}^{5}+2,{P}^{4}ac{x}^{7}+2,{P}^{4}{b}^{
3}c{x}^{5}+2,{P}^{4}bc{x}^{7}+{P}^{4}{a}^{4}{b}^{4}+2,{P}^{4}{a}^{4}
{b}^{2}{x}^{2}+{P}^{4}{a}^{4}{x}^{4}+2,{P}^{4}{a}^{2}{b}^{4}{x}^{2}+4
,{P}^{4}{a}^{2}{b}^{2}{x}^{4}+2,{P}^{4}{a}^{2}{x}^{6}+{P}^{4}{b}^{4}
{x}^{4}+2,{P}^{4}{b}^{2}{x}^{6}+{P}^{4}{x}^{8}-2,{P}^{2}{a}^{2}b{c}^
{3}{x}^{3}-2,{P}^{2}a{b}^{2}{c}^{3}{x}^{3}-4,{P}^{2}{a}^{3}b{c}^{2}{
x}^{2}-4,{P}^{2}{a}^{2}{b}^{2}{c}^{2}{x}^{2}-2,{P}^{2}{a}^{2}{c}^{2}
{x}^{4}-4,{P}^{2}a{b}^{3}{c}^{2}{x}^{2}-8,{P}^{2}ab{c}^{2}{x}^{4}-2
,{P}^{2}{b}^{2}{c}^{2}{x}^{4}-2,{P}^{2}{a}^{4}bcx-4,{P}^{2}{a}^{3}{
b}^{2}cx-4,{P}^{2}{a}^{3}c{x}^{3}-4,{P}^{2}{a}^{2}{b}^{3}cx-8,{P}^{
2}{a}^{2}bc{x}^{3}-2,{P}^{2}a{b}^{4}cx-8,{P}^{2}a{b}^{2}c{x}^{3}-6,
{P}^{2}ac{x}^{5}-4,{P}^{2}{b}^{3}c{x}^{3}-6,{P}^{2}bc{x}^{5}-2,{P}^
{2}{a}^{4}{b}^{2}-2,{P}^{2}{a}^{4}{x}^{2}-2,{P}^{2}{a}^{2}{b}^{4}-8
,{P}^{2}{a}^{2}{b}^{2}{x}^{2}-6,{P}^{2}{a}^{2}{x}^{4}-2,{P}^{2}{b}^
{4}{x}^{2}-6,{P}^{2}{b}^{2}{x}^{4}-4,{P}^{2}{x}^{6}+{a}^{2}{c}^{2}{x
}^{2}-2,ab{c}^{2}{x}^{2}+{b}^{2}{c}^{2}{x}^{2}+2,{a}^{3}cx-2,{a}^{2
}bcx-2,a{b}^{2}cx+2,{b}^{3}cx+{a}^{4}-2,{a}^{2}{b}^{2}+{b}^{4}
=0$$
answered Dec 29 '18 at 21:56
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42864
edited Dec 30 '18 at 10:08
answered Dec 29 '18 at 21:56
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42864
answered Dec 29 '18 at 21:56
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42864
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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