Solving for $x$ in a system of equations with 2 variables
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$$a^2-b^2-c^2=x^2-y^2quad &quad
ab=xy$$
Hi. So I have been trying to solve this equation (I'm looking for $x$, and $x$ and $y$ are the only variables) but I just can't find a solution. Please help. How do I find an expression for $x$?
algebra-precalculus
edited Nov 18 at 13:21
amWhy
191k27223439
asked Nov 18 at 12:55
Maria Mercedes
1
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up vote
0
down vote
favorite
$$a^2-b^2-c^2=x^2-y^2quad &quad
ab=xy$$
Hi. So I have been trying to solve this equation (I'm looking for $x$, and $x$ and $y$ are the only variables) but I just can't find a solution. Please help. How do I find an expression for $x$?
algebra-precalculus
edited Nov 18 at 13:21
amWhy
191k27223439
asked Nov 18 at 12:55
Maria Mercedes
1
You can use the 2nd one to eliminate $y$ from the first to give a quadratic in $x^2$. there may be a neater way.
– user121049
Nov 18 at 13:01
But still, both sides will still have x. :(
– Maria Mercedes
Nov 18 at 14:52
You should end up with $x^4-(a^2-b^2-c^2)x^2-a^2b^2=0$.
– user121049
Nov 18 at 15:32
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$a^2-b^2-c^2=x^2-y^2quad &quad
ab=xy$$
Hi. So I have been trying to solve this equation (I'm looking for $x$, and $x$ and $y$ are the only variables) but I just can't find a solution. Please help. How do I find an expression for $x$?
algebra-precalculus
edited Nov 18 at 13:21
amWhy
191k27223439
asked Nov 18 at 12:55
Maria Mercedes
1
$$a^2-b^2-c^2=x^2-y^2quad &quad
ab=xy$$
Hi. So I have been trying to solve this equation (I'm looking for $x$, and $x$ and $y$ are the only variables) but I just can't find a solution. Please help. How do I find an expression for $x$?
algebra-precalculus
algebra-precalculus
edited Nov 18 at 13:21
amWhy
191k27223439
asked Nov 18 at 12:55
Maria Mercedes
1
edited Nov 18 at 13:21
amWhy
191k27223439
asked Nov 18 at 12:55
Maria Mercedes
1
edited Nov 18 at 13:21
amWhy
191k27223439
edited Nov 18 at 13:21
amWhy
191k27223439
edited Nov 18 at 13:21
amWhy
191k27223439
191k27223439
asked Nov 18 at 12:55
Maria Mercedes
1
asked Nov 18 at 12:55
Maria Mercedes
1
asked Nov 18 at 12:55
Maria Mercedes
1
1
You can use the 2nd one to eliminate $y$ from the first to give a quadratic in $x^2$. there may be a neater way.
– user121049
Nov 18 at 13:01
But still, both sides will still have x. :(
– Maria Mercedes
Nov 18 at 14:52
You should end up with $x^4-(a^2-b^2-c^2)x^2-a^2b^2=0$.
– user121049
Nov 18 at 15:32
add a comment |
You can use the 2nd one to eliminate $y$ from the first to give a quadratic in $x^2$. there may be a neater way.
– user121049
Nov 18 at 13:01
But still, both sides will still have x. :(
– Maria Mercedes
Nov 18 at 14:52
You should end up with $x^4-(a^2-b^2-c^2)x^2-a^2b^2=0$.
– user121049
Nov 18 at 15:32
You can use the 2nd one to eliminate $y$ from the first to give a quadratic in $x^2$. there may be a neater way.
– user121049
Nov 18 at 13:01
You can use the 2nd one to eliminate $y$ from the first to give a quadratic in $x^2$. there may be a neater way.
– user121049
Nov 18 at 13:01
But still, both sides will still have x. :(
– Maria Mercedes
Nov 18 at 14:52
But still, both sides will still have x. :(
– Maria Mercedes
Nov 18 at 14:52
You should end up with $x^4-(a^2-b^2-c^2)x^2-a^2b^2=0$.
– user121049
Nov 18 at 15:32
You should end up with $x^4-(a^2-b^2-c^2)x^2-a^2b^2=0$.
– user121049
Nov 18 at 15:32
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Multiply all terms by $x^2$
$x^2(a^2-b^2-c^2)= x^4-(xy)^2$ then it turns out to be
$(x^2)^2-x^2(a^2-b^2-c^2) -(ab)^2=0$
Use quadratic formula
answered Nov 18 at 23:30
Ameryr
599211
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Multiply all terms by $x^2$
$x^2(a^2-b^2-c^2)= x^4-(xy)^2$ then it turns out to be
$(x^2)^2-x^2(a^2-b^2-c^2) -(ab)^2=0$
Use quadratic formula
answered Nov 18 at 23:30
Ameryr
599211
add a comment |
up vote
0
down vote
Multiply all terms by $x^2$
$x^2(a^2-b^2-c^2)= x^4-(xy)^2$ then it turns out to be
$(x^2)^2-x^2(a^2-b^2-c^2) -(ab)^2=0$
Use quadratic formula
answered Nov 18 at 23:30
Ameryr
599211
add a comment |
up vote
0
down vote
up vote
0
down vote
Multiply all terms by $x^2$
$x^2(a^2-b^2-c^2)= x^4-(xy)^2$ then it turns out to be
$(x^2)^2-x^2(a^2-b^2-c^2) -(ab)^2=0$
Use quadratic formula
answered Nov 18 at 23:30
Ameryr
599211
Multiply all terms by $x^2$
$x^2(a^2-b^2-c^2)= x^4-(xy)^2$ then it turns out to be
$(x^2)^2-x^2(a^2-b^2-c^2) -(ab)^2=0$
Use quadratic formula
answered Nov 18 at 23:30
Ameryr
599211
answered Nov 18 at 23:30
Ameryr
599211
answered Nov 18 at 23:30
Ameryr
599211
answered Nov 18 at 23:30
Ameryr
599211
599211
add a comment |
add a comment |
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You can use the 2nd one to eliminate $y$ from the first to give a quadratic in $x^2$. there may be a neater way.
– user121049
Nov 18 at 13:01
But still, both sides will still have x. :(
– Maria Mercedes
Nov 18 at 14:52
You should end up with $x^4-(a^2-b^2-c^2)x^2-a^2b^2=0$.
– user121049
Nov 18 at 15:32