Solving quadratics homework [closed]
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$3x = 1 - frac{1}{x}$
My daughter could do this if it was $3(x)^{-2}$ using substitution but the $3x$ is a problem!! One friend multiplied both sides by $x$ but she believes that is the wrong process.
quadratics
edited Nov 16 at 5:20
Makina
966113
asked Nov 16 at 4:58
Traci J Cole
1
closed as unclear what you're asking by Did, Tom-Tom, Kelvin Lois, user10354138, amWhy Nov 16 at 18:16
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-3
down vote
favorite
$3x = 1 - frac{1}{x}$
My daughter could do this if it was $3(x)^{-2}$ using substitution but the $3x$ is a problem!! One friend multiplied both sides by $x$ but she believes that is the wrong process.
quadratics
edited Nov 16 at 5:20
Makina
966113
asked Nov 16 at 4:58
Traci J Cole
1
closed as unclear what you're asking by Did, Tom-Tom, Kelvin Lois, user10354138, amWhy Nov 16 at 18:16
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
Nope, multiplying both sides by $x$ is correct. You'll get $3x^2 = x - 1$. From there, you just move all of the $x$ terms to one side, and solve for $x$ using whatever method you choose.
– Eevee Trainer
Nov 16 at 5:00
3
Since $x = 0$ is not a solution, you can multiply by $x$ and not care about it. Usually the "wrong" comes from the fact, that multiplying by $x$ can bring extra solutions, which don't exist in reality.
– Makina
Nov 16 at 5:04
Thank you very much.
– Traci J Cole
Nov 16 at 5:05
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
$3x = 1 - frac{1}{x}$
My daughter could do this if it was $3(x)^{-2}$ using substitution but the $3x$ is a problem!! One friend multiplied both sides by $x$ but she believes that is the wrong process.
quadratics
edited Nov 16 at 5:20
Makina
966113
asked Nov 16 at 4:58
Traci J Cole
1
$3x = 1 - frac{1}{x}$
My daughter could do this if it was $3(x)^{-2}$ using substitution but the $3x$ is a problem!! One friend multiplied both sides by $x$ but she believes that is the wrong process.
quadratics
quadratics
edited Nov 16 at 5:20
Makina
966113
asked Nov 16 at 4:58
Traci J Cole
1
edited Nov 16 at 5:20
Makina
966113
asked Nov 16 at 4:58
Traci J Cole
1
edited Nov 16 at 5:20
Makina
966113
edited Nov 16 at 5:20
Makina
966113
edited Nov 16 at 5:20
Makina
966113
966113
asked Nov 16 at 4:58
Traci J Cole
1
asked Nov 16 at 4:58
Traci J Cole
1
asked Nov 16 at 4:58
Traci J Cole
1
1
closed as unclear what you're asking by Did, Tom-Tom, Kelvin Lois, user10354138, amWhy Nov 16 at 18:16
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Did, Tom-Tom, Kelvin Lois, user10354138, amWhy Nov 16 at 18:16
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
Nope, multiplying both sides by $x$ is correct. You'll get $3x^2 = x - 1$. From there, you just move all of the $x$ terms to one side, and solve for $x$ using whatever method you choose.
– Eevee Trainer
Nov 16 at 5:00
3
Since $x = 0$ is not a solution, you can multiply by $x$ and not care about it. Usually the "wrong" comes from the fact, that multiplying by $x$ can bring extra solutions, which don't exist in reality.
– Makina
Nov 16 at 5:04
Thank you very much.
– Traci J Cole
Nov 16 at 5:05
add a comment |
2
Nope, multiplying both sides by $x$ is correct. You'll get $3x^2 = x - 1$. From there, you just move all of the $x$ terms to one side, and solve for $x$ using whatever method you choose.
– Eevee Trainer
Nov 16 at 5:00
3
Since $x = 0$ is not a solution, you can multiply by $x$ and not care about it. Usually the "wrong" comes from the fact, that multiplying by $x$ can bring extra solutions, which don't exist in reality.
– Makina
Nov 16 at 5:04
Thank you very much.
– Traci J Cole
Nov 16 at 5:05
2
2
Nope, multiplying both sides by $x$ is correct. You'll get $3x^2 = x - 1$. From there, you just move all of the $x$ terms to one side, and solve for $x$ using whatever method you choose.
– Eevee Trainer
Nov 16 at 5:00
Nope, multiplying both sides by $x$ is correct. You'll get $3x^2 = x - 1$. From there, you just move all of the $x$ terms to one side, and solve for $x$ using whatever method you choose.
– Eevee Trainer
Nov 16 at 5:00
3
3
Since $x = 0$ is not a solution, you can multiply by $x$ and not care about it. Usually the "wrong" comes from the fact, that multiplying by $x$ can bring extra solutions, which don't exist in reality.
– Makina
Nov 16 at 5:04
Since $x = 0$ is not a solution, you can multiply by $x$ and not care about it. Usually the "wrong" comes from the fact, that multiplying by $x$ can bring extra solutions, which don't exist in reality.
– Makina
Nov 16 at 5:04
Thank you very much.
– Traci J Cole
Nov 16 at 5:05
Thank you very much.
– Traci J Cole
Nov 16 at 5:05
add a comment |
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2
Nope, multiplying both sides by $x$ is correct. You'll get $3x^2 = x - 1$. From there, you just move all of the $x$ terms to one side, and solve for $x$ using whatever method you choose.
– Eevee Trainer
Nov 16 at 5:00
3
Since $x = 0$ is not a solution, you can multiply by $x$ and not care about it. Usually the "wrong" comes from the fact, that multiplying by $x$ can bring extra solutions, which don't exist in reality.
– Makina
Nov 16 at 5:04
Thank you very much.
– Traci J Cole
Nov 16 at 5:05