Solve the formula for the indicated variable. [closed]
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I can't seem to understand how to solve for 'y' in this equation for my college algebra class. Please help.
Solve the formula for the indicated variable. (Enter your answers as a comma-separated list.)
$$x^2 - 5xy + y^2 = 0$$
algebra-precalculus
closed as off-topic by amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim Sep 5 at 20:27
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim
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up vote
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I can't seem to understand how to solve for 'y' in this equation for my college algebra class. Please help.
Solve the formula for the indicated variable. (Enter your answers as a comma-separated list.)
$$x^2 - 5xy + y^2 = 0$$
algebra-precalculus
closed as off-topic by amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim Sep 5 at 20:27
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim
1
One way would be to treat the above as a quadratic in the variable $y$ and treat $x$ constant, then use the quadratic formula.
â Kevin
Sep 5 at 11:26
$(x-frac5y2)^2=frac21y^24$ so...
â mrs
Sep 5 at 11:27
This is an ordinary quadratic equation in $y$, use the standard formula.
â Yves Daoust
Sep 5 at 12:40
Treat $x$ as constant and solve the quadratic in $y$.
â Narasimham
Sep 5 at 19:30
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I can't seem to understand how to solve for 'y' in this equation for my college algebra class. Please help.
Solve the formula for the indicated variable. (Enter your answers as a comma-separated list.)
$$x^2 - 5xy + y^2 = 0$$
algebra-precalculus
I can't seem to understand how to solve for 'y' in this equation for my college algebra class. Please help.
Solve the formula for the indicated variable. (Enter your answers as a comma-separated list.)
$$x^2 - 5xy + y^2 = 0$$
algebra-precalculus
algebra-precalculus
edited Sep 5 at 11:25
Kevin
5,138722
5,138722
asked Sep 5 at 11:23
Abdullah Mumtaz
111
111
closed as off-topic by amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim Sep 5 at 20:27
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim
closed as off-topic by amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim Sep 5 at 20:27
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim
1
One way would be to treat the above as a quadratic in the variable $y$ and treat $x$ constant, then use the quadratic formula.
â Kevin
Sep 5 at 11:26
$(x-frac5y2)^2=frac21y^24$ so...
â mrs
Sep 5 at 11:27
This is an ordinary quadratic equation in $y$, use the standard formula.
â Yves Daoust
Sep 5 at 12:40
Treat $x$ as constant and solve the quadratic in $y$.
â Narasimham
Sep 5 at 19:30
add a comment |Â
1
One way would be to treat the above as a quadratic in the variable $y$ and treat $x$ constant, then use the quadratic formula.
â Kevin
Sep 5 at 11:26
$(x-frac5y2)^2=frac21y^24$ so...
â mrs
Sep 5 at 11:27
This is an ordinary quadratic equation in $y$, use the standard formula.
â Yves Daoust
Sep 5 at 12:40
Treat $x$ as constant and solve the quadratic in $y$.
â Narasimham
Sep 5 at 19:30
1
1
One way would be to treat the above as a quadratic in the variable $y$ and treat $x$ constant, then use the quadratic formula.
â Kevin
Sep 5 at 11:26
One way would be to treat the above as a quadratic in the variable $y$ and treat $x$ constant, then use the quadratic formula.
â Kevin
Sep 5 at 11:26
$(x-frac5y2)^2=frac21y^24$ so...
â mrs
Sep 5 at 11:27
$(x-frac5y2)^2=frac21y^24$ so...
â mrs
Sep 5 at 11:27
This is an ordinary quadratic equation in $y$, use the standard formula.
â Yves Daoust
Sep 5 at 12:40
This is an ordinary quadratic equation in $y$, use the standard formula.
â Yves Daoust
Sep 5 at 12:40
Treat $x$ as constant and solve the quadratic in $y$.
â Narasimham
Sep 5 at 19:30
Treat $x$ as constant and solve the quadratic in $y$.
â Narasimham
Sep 5 at 19:30
add a comment |Â
2 Answers
2
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up vote
1
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To solve for y.
I thought of the equation as a quadratic, treating $x$ as a constant:
$$x^2 â 5xy + y^2 = 0 ; iff ; y^2 - 5xy + x^2 = 0$$
After that, I plugged it into the quadratic formula to solve for $y$:
ANSWER: $$ frac5x + x sqrt212 , frac5x - x sqrt212 $$.
Thanks to Kevin and Mrs for the help!
â Abdullah Mumtaz
Sep 5 at 11:51
add a comment |Â
up vote
1
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Alternatively, noticing that the equation is homogeneous, divide by $x^2$ to get
$$left(frac yxright)^2-5frac yx+1=0$$
and
$$frac yx=frac5pmsqrt212.$$
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
To solve for y.
I thought of the equation as a quadratic, treating $x$ as a constant:
$$x^2 â 5xy + y^2 = 0 ; iff ; y^2 - 5xy + x^2 = 0$$
After that, I plugged it into the quadratic formula to solve for $y$:
ANSWER: $$ frac5x + x sqrt212 , frac5x - x sqrt212 $$.
Thanks to Kevin and Mrs for the help!
â Abdullah Mumtaz
Sep 5 at 11:51
add a comment |Â
up vote
1
down vote
To solve for y.
I thought of the equation as a quadratic, treating $x$ as a constant:
$$x^2 â 5xy + y^2 = 0 ; iff ; y^2 - 5xy + x^2 = 0$$
After that, I plugged it into the quadratic formula to solve for $y$:
ANSWER: $$ frac5x + x sqrt212 , frac5x - x sqrt212 $$.
Thanks to Kevin and Mrs for the help!
â Abdullah Mumtaz
Sep 5 at 11:51
add a comment |Â
up vote
1
down vote
up vote
1
down vote
To solve for y.
I thought of the equation as a quadratic, treating $x$ as a constant:
$$x^2 â 5xy + y^2 = 0 ; iff ; y^2 - 5xy + x^2 = 0$$
After that, I plugged it into the quadratic formula to solve for $y$:
ANSWER: $$ frac5x + x sqrt212 , frac5x - x sqrt212 $$.
To solve for y.
I thought of the equation as a quadratic, treating $x$ as a constant:
$$x^2 â 5xy + y^2 = 0 ; iff ; y^2 - 5xy + x^2 = 0$$
After that, I plugged it into the quadratic formula to solve for $y$:
ANSWER: $$ frac5x + x sqrt212 , frac5x - x sqrt212 $$.
edited Sep 5 at 12:30
amWhy
190k27221433
190k27221433
answered Sep 5 at 11:50
Abdullah Mumtaz
111
111
Thanks to Kevin and Mrs for the help!
â Abdullah Mumtaz
Sep 5 at 11:51
add a comment |Â
Thanks to Kevin and Mrs for the help!
â Abdullah Mumtaz
Sep 5 at 11:51
Thanks to Kevin and Mrs for the help!
â Abdullah Mumtaz
Sep 5 at 11:51
Thanks to Kevin and Mrs for the help!
â Abdullah Mumtaz
Sep 5 at 11:51
add a comment |Â
up vote
1
down vote
Alternatively, noticing that the equation is homogeneous, divide by $x^2$ to get
$$left(frac yxright)^2-5frac yx+1=0$$
and
$$frac yx=frac5pmsqrt212.$$
add a comment |Â
up vote
1
down vote
Alternatively, noticing that the equation is homogeneous, divide by $x^2$ to get
$$left(frac yxright)^2-5frac yx+1=0$$
and
$$frac yx=frac5pmsqrt212.$$
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Alternatively, noticing that the equation is homogeneous, divide by $x^2$ to get
$$left(frac yxright)^2-5frac yx+1=0$$
and
$$frac yx=frac5pmsqrt212.$$
Alternatively, noticing that the equation is homogeneous, divide by $x^2$ to get
$$left(frac yxright)^2-5frac yx+1=0$$
and
$$frac yx=frac5pmsqrt212.$$
answered Sep 5 at 12:43
Yves Daoust
115k666209
115k666209
add a comment |Â
add a comment |Â
1
One way would be to treat the above as a quadratic in the variable $y$ and treat $x$ constant, then use the quadratic formula.
â Kevin
Sep 5 at 11:26
$(x-frac5y2)^2=frac21y^24$ so...
â mrs
Sep 5 at 11:27
This is an ordinary quadratic equation in $y$, use the standard formula.
â Yves Daoust
Sep 5 at 12:40
Treat $x$ as constant and solve the quadratic in $y$.
â Narasimham
Sep 5 at 19:30