How to solve equations of this type [on hold]
My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:
Q1. If: $x - frac{1}{x} = 3$ then what is $x^2 + frac{1}{x^2}$ equal to?
The answer for this question is 11, but I don't know how, I thought it should be 9, but my answer was wrong.
Q2. If: $frac{x}{x + y} = 5$ then what is $frac{y}{x + y}$ equal to?
The answer for this is -5, I also don't know how.
Q3. If: $x^4 + y^4 = 6 x^2 y^2 land xneq y$ then what is $frac{x^2 + y^2}{x^2 - y^2}$ equal to?
I guess the answer for this was $sqrt{2}$ but I'm not sure.
Any body can explain how to solve these questions, and questions of the same pattern?
exponentiation
edited Dec 22 at 20:12
dantopa
6,41132042
asked Dec 22 at 15:47
Ameer Taweel
234
New contributor
Ameer Taweel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as too broad by José Carlos Santos, user10354138, Holo, Cesareo, Saad Dec 23 at 1:35
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. 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.
add a comment |
My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:
Q1. If: $x - frac{1}{x} = 3$ then what is $x^2 + frac{1}{x^2}$ equal to?
The answer for this question is 11, but I don't know how, I thought it should be 9, but my answer was wrong.
Q2. If: $frac{x}{x + y} = 5$ then what is $frac{y}{x + y}$ equal to?
The answer for this is -5, I also don't know how.
Q3. If: $x^4 + y^4 = 6 x^2 y^2 land xneq y$ then what is $frac{x^2 + y^2}{x^2 - y^2}$ equal to?
I guess the answer for this was $sqrt{2}$ but I'm not sure.
Any body can explain how to solve these questions, and questions of the same pattern?
exponentiation
edited Dec 22 at 20:12
dantopa
6,41132042
asked Dec 22 at 15:47
Ameer Taweel
234
New contributor
Ameer Taweel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as too broad by José Carlos Santos, user10354138, Holo, Cesareo, Saad Dec 23 at 1:35
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. 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
The first one should be $x^2 + 1/x^2$.
– Rebellos
Dec 22 at 15:53
2
$frac x{x+y}+frac y{x+y}=1$.
– Lord Shark the Unknown
Dec 22 at 15:55
Sorry, didn't notice, now it's fixed.
– Ameer Taweel
Dec 22 at 15:55
@LordSharktheUnknown this solves the second one, thanks. But what about the others.
– Ameer Taweel
Dec 22 at 15:56
The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
– Sik Feng Cheong
Dec 22 at 16:21
add a comment |
My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:
Q1. If: $x - frac{1}{x} = 3$ then what is $x^2 + frac{1}{x^2}$ equal to?
The answer for this question is 11, but I don't know how, I thought it should be 9, but my answer was wrong.
Q2. If: $frac{x}{x + y} = 5$ then what is $frac{y}{x + y}$ equal to?
The answer for this is -5, I also don't know how.
Q3. If: $x^4 + y^4 = 6 x^2 y^2 land xneq y$ then what is $frac{x^2 + y^2}{x^2 - y^2}$ equal to?
I guess the answer for this was $sqrt{2}$ but I'm not sure.
Any body can explain how to solve these questions, and questions of the same pattern?
exponentiation
edited Dec 22 at 20:12
dantopa
6,41132042
asked Dec 22 at 15:47
Ameer Taweel
234
New contributor
Ameer Taweel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:
Q1. If: $x - frac{1}{x} = 3$ then what is $x^2 + frac{1}{x^2}$ equal to?
The answer for this question is 11, but I don't know how, I thought it should be 9, but my answer was wrong.
Q2. If: $frac{x}{x + y} = 5$ then what is $frac{y}{x + y}$ equal to?
The answer for this is -5, I also don't know how.
Q3. If: $x^4 + y^4 = 6 x^2 y^2 land xneq y$ then what is $frac{x^2 + y^2}{x^2 - y^2}$ equal to?
I guess the answer for this was $sqrt{2}$ but I'm not sure.
Any body can explain how to solve these questions, and questions of the same pattern?
exponentiation
exponentiation
edited Dec 22 at 20:12
dantopa
6,41132042
asked Dec 22 at 15:47
Ameer Taweel
234
New contributor
Ameer Taweel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 22 at 20:12
dantopa
6,41132042
asked Dec 22 at 15:47
Ameer Taweel
234
New contributor
Ameer Taweel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 22 at 20:12
dantopa
6,41132042
edited Dec 22 at 20:12
dantopa
6,41132042
edited Dec 22 at 20:12
dantopa
6,41132042
6,41132042
asked Dec 22 at 15:47
Ameer Taweel
234
New contributor
Ameer Taweel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 22 at 15:47
Ameer Taweel
234
asked Dec 22 at 15:47
Ameer Taweel
234
234
New contributor
Ameer Taweel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ameer Taweel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ameer Taweel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as too broad by José Carlos Santos, user10354138, Holo, Cesareo, Saad Dec 23 at 1:35
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. 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.
put on hold as too broad by José Carlos Santos, user10354138, Holo, Cesareo, Saad Dec 23 at 1:35
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. 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
The first one should be $x^2 + 1/x^2$.
– Rebellos
Dec 22 at 15:53
2
$frac x{x+y}+frac y{x+y}=1$.
– Lord Shark the Unknown
Dec 22 at 15:55
Sorry, didn't notice, now it's fixed.
– Ameer Taweel
Dec 22 at 15:55
@LordSharktheUnknown this solves the second one, thanks. But what about the others.
– Ameer Taweel
Dec 22 at 15:56
The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
– Sik Feng Cheong
Dec 22 at 16:21
add a comment |
2
The first one should be $x^2 + 1/x^2$.
– Rebellos
Dec 22 at 15:53
2
$frac x{x+y}+frac y{x+y}=1$.
– Lord Shark the Unknown
Dec 22 at 15:55
Sorry, didn't notice, now it's fixed.
– Ameer Taweel
Dec 22 at 15:55
@LordSharktheUnknown this solves the second one, thanks. But what about the others.
– Ameer Taweel
Dec 22 at 15:56
The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
– Sik Feng Cheong
Dec 22 at 16:21
2
2
The first one should be $x^2 + 1/x^2$.
– Rebellos
Dec 22 at 15:53
The first one should be $x^2 + 1/x^2$.
– Rebellos
Dec 22 at 15:53
2
2
$frac x{x+y}+frac y{x+y}=1$.
– Lord Shark the Unknown
Dec 22 at 15:55
$frac x{x+y}+frac y{x+y}=1$.
– Lord Shark the Unknown
Dec 22 at 15:55
Sorry, didn't notice, now it's fixed.
– Ameer Taweel
Dec 22 at 15:55
Sorry, didn't notice, now it's fixed.
– Ameer Taweel
Dec 22 at 15:55
@LordSharktheUnknown this solves the second one, thanks. But what about the others.
– Ameer Taweel
Dec 22 at 15:56
@LordSharktheUnknown this solves the second one, thanks. But what about the others.
– Ameer Taweel
Dec 22 at 15:56
The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
– Sik Feng Cheong
Dec 22 at 16:21
The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
– Sik Feng Cheong
Dec 22 at 16:21
add a comment |
4 Answers
4
active
oldest
votes
For the first one :
$$left(x-frac{1}{x}right)^2 = x^2 - 2 + frac{1}{x^2} Leftrightarrow x^2 + frac{1}{x^2} = left(x-frac{1}{x}right)^2 + 2 implies x^2+ frac{1}{x^2} = 11$$
For the second one, observe that :
$$frac{x}{x+y} + frac{y}{x+y} = 1 Rightarrow 5 + frac{y}{x+y} = 1 Leftrightarrow frac{y}{x+y} = -4 $$
For the third one, a small hint :
$$(x^2-y^2)^2 = x^4 - 2x^2y^2 + y^4$$
Alternativelly, observe that it also is :
$$x^4 + y^4 = 6 x^2 y^2 Leftrightarrow frac{x^2}{y^2} + frac{y^2}{x^2} = 6$$
answered Dec 22 at 15:56
Rebellos
14.3k31244
@ÍgjøgnumMeg Thanks. You may as well edit to correct those.
– Rebellos
Dec 22 at 15:59
Thanks very much, I guess I'll get A+ tomorrow.
– Ameer Taweel
Dec 22 at 16:15
1
@AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
– Rebellos
Dec 22 at 18:25
thanks for the advice.
– Ameer Taweel
Dec 22 at 20:10
add a comment |
Hint:
1: $x - dfrac{1}{x} = 3 implies x^2 + dfrac{1}{x^2} - 2 = 9$
Now use, $big(x + dfrac{1}{x}big)^2 = x^2 + dfrac{1}{x^2} + 2$
2: $dfrac{x}{x+y} = dfrac{1}{1 + dfrac{y}{x}}$
3: $x^4 + y^4 = 6 * x^2 * y^2 implies dfrac{x^2}{y^2} + dfrac{y^2}{x^2} = 6$, solve for y/x or x/y use that to get the value of final expression
answered Dec 22 at 15:57
l''''''''l
2,188726
add a comment |
1
Notice that
$$x^2 + frac{1}{x^2} = left(x -frac{1}{x}right)^2 + 2 = 3^2 + 2 = 11$$
2
Notice that the sum of the two fractions gives $1$ hence it's rather trivial to obtain the second value.
3
You can transform the equation and solve for $frac{y}{x}$ for example, and find everything you need.
answered Dec 22 at 15:58
Von Neumann
16.3k72543
add a comment |
Hints.
$1) x-frac1x=3implies x^2+frac1{x^2}-2=9\2)frac y{x+y}+frac x{x+y}=1\3)x^4+y^4=6x^2y^2\implies x^4+y^4+2x^2y^2=(x^2+y^2)^2=8x^2y^2\implies x^4+y^4-2x^2y^2=(x^2-y^2)^2=4x^2y^2$
Divide the two and take the square root.
answered Dec 22 at 16:08
Shubham Johri
3,859716
Thanks for the detailed explanation of the last one.
– Ameer Taweel
Dec 22 at 16:16
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
For the first one :
$$left(x-frac{1}{x}right)^2 = x^2 - 2 + frac{1}{x^2} Leftrightarrow x^2 + frac{1}{x^2} = left(x-frac{1}{x}right)^2 + 2 implies x^2+ frac{1}{x^2} = 11$$
For the second one, observe that :
$$frac{x}{x+y} + frac{y}{x+y} = 1 Rightarrow 5 + frac{y}{x+y} = 1 Leftrightarrow frac{y}{x+y} = -4 $$
For the third one, a small hint :
$$(x^2-y^2)^2 = x^4 - 2x^2y^2 + y^4$$
Alternativelly, observe that it also is :
$$x^4 + y^4 = 6 x^2 y^2 Leftrightarrow frac{x^2}{y^2} + frac{y^2}{x^2} = 6$$
answered Dec 22 at 15:56
Rebellos
14.3k31244
@ÍgjøgnumMeg Thanks. You may as well edit to correct those.
– Rebellos
Dec 22 at 15:59
Thanks very much, I guess I'll get A+ tomorrow.
– Ameer Taweel
Dec 22 at 16:15
1
@AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
– Rebellos
Dec 22 at 18:25
thanks for the advice.
– Ameer Taweel
Dec 22 at 20:10
add a comment |
For the first one :
$$left(x-frac{1}{x}right)^2 = x^2 - 2 + frac{1}{x^2} Leftrightarrow x^2 + frac{1}{x^2} = left(x-frac{1}{x}right)^2 + 2 implies x^2+ frac{1}{x^2} = 11$$
For the second one, observe that :
$$frac{x}{x+y} + frac{y}{x+y} = 1 Rightarrow 5 + frac{y}{x+y} = 1 Leftrightarrow frac{y}{x+y} = -4 $$
For the third one, a small hint :
$$(x^2-y^2)^2 = x^4 - 2x^2y^2 + y^4$$
Alternativelly, observe that it also is :
$$x^4 + y^4 = 6 x^2 y^2 Leftrightarrow frac{x^2}{y^2} + frac{y^2}{x^2} = 6$$
answered Dec 22 at 15:56
Rebellos
14.3k31244
@ÍgjøgnumMeg Thanks. You may as well edit to correct those.
– Rebellos
Dec 22 at 15:59
Thanks very much, I guess I'll get A+ tomorrow.
– Ameer Taweel
Dec 22 at 16:15
1
@AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
– Rebellos
Dec 22 at 18:25
thanks for the advice.
– Ameer Taweel
Dec 22 at 20:10
add a comment |
For the first one :
$$left(x-frac{1}{x}right)^2 = x^2 - 2 + frac{1}{x^2} Leftrightarrow x^2 + frac{1}{x^2} = left(x-frac{1}{x}right)^2 + 2 implies x^2+ frac{1}{x^2} = 11$$
For the second one, observe that :
$$frac{x}{x+y} + frac{y}{x+y} = 1 Rightarrow 5 + frac{y}{x+y} = 1 Leftrightarrow frac{y}{x+y} = -4 $$
For the third one, a small hint :
$$(x^2-y^2)^2 = x^4 - 2x^2y^2 + y^4$$
Alternativelly, observe that it also is :
$$x^4 + y^4 = 6 x^2 y^2 Leftrightarrow frac{x^2}{y^2} + frac{y^2}{x^2} = 6$$
answered Dec 22 at 15:56
Rebellos
14.3k31244
For the first one :
$$left(x-frac{1}{x}right)^2 = x^2 - 2 + frac{1}{x^2} Leftrightarrow x^2 + frac{1}{x^2} = left(x-frac{1}{x}right)^2 + 2 implies x^2+ frac{1}{x^2} = 11$$
For the second one, observe that :
$$frac{x}{x+y} + frac{y}{x+y} = 1 Rightarrow 5 + frac{y}{x+y} = 1 Leftrightarrow frac{y}{x+y} = -4 $$
For the third one, a small hint :
$$(x^2-y^2)^2 = x^4 - 2x^2y^2 + y^4$$
Alternativelly, observe that it also is :
$$x^4 + y^4 = 6 x^2 y^2 Leftrightarrow frac{x^2}{y^2} + frac{y^2}{x^2} = 6$$
answered Dec 22 at 15:56
Rebellos
14.3k31244
edited Dec 22 at 15:58
answered Dec 22 at 15:56
Rebellos
14.3k31244
answered Dec 22 at 15:56
Rebellos
14.3k31244
answered Dec 22 at 15:56
Rebellos
14.3k31244
14.3k31244
@ÍgjøgnumMeg Thanks. You may as well edit to correct those.
– Rebellos
Dec 22 at 15:59
Thanks very much, I guess I'll get A+ tomorrow.
– Ameer Taweel
Dec 22 at 16:15
1
@AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
– Rebellos
Dec 22 at 18:25
thanks for the advice.
– Ameer Taweel
Dec 22 at 20:10
add a comment |
@ÍgjøgnumMeg Thanks. You may as well edit to correct those.
– Rebellos
Dec 22 at 15:59
Thanks very much, I guess I'll get A+ tomorrow.
– Ameer Taweel
Dec 22 at 16:15
1
@AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
– Rebellos
Dec 22 at 18:25
thanks for the advice.
– Ameer Taweel
Dec 22 at 20:10
@ÍgjøgnumMeg Thanks. You may as well edit to correct those.
– Rebellos
Dec 22 at 15:59
@ÍgjøgnumMeg Thanks. You may as well edit to correct those.
– Rebellos
Dec 22 at 15:59
Thanks very much, I guess I'll get A+ tomorrow.
– Ameer Taweel
Dec 22 at 16:15
Thanks very much, I guess I'll get A+ tomorrow.
– Ameer Taweel
Dec 22 at 16:15
1
1
@AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
– Rebellos
Dec 22 at 18:25
@AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
– Rebellos
Dec 22 at 18:25
thanks for the advice.
– Ameer Taweel
Dec 22 at 20:10
thanks for the advice.
– Ameer Taweel
Dec 22 at 20:10
add a comment |
Hint:
1: $x - dfrac{1}{x} = 3 implies x^2 + dfrac{1}{x^2} - 2 = 9$
Now use, $big(x + dfrac{1}{x}big)^2 = x^2 + dfrac{1}{x^2} + 2$
2: $dfrac{x}{x+y} = dfrac{1}{1 + dfrac{y}{x}}$
3: $x^4 + y^4 = 6 * x^2 * y^2 implies dfrac{x^2}{y^2} + dfrac{y^2}{x^2} = 6$, solve for y/x or x/y use that to get the value of final expression
answered Dec 22 at 15:57
l''''''''l
2,188726
add a comment |
Hint:
1: $x - dfrac{1}{x} = 3 implies x^2 + dfrac{1}{x^2} - 2 = 9$
Now use, $big(x + dfrac{1}{x}big)^2 = x^2 + dfrac{1}{x^2} + 2$
2: $dfrac{x}{x+y} = dfrac{1}{1 + dfrac{y}{x}}$
3: $x^4 + y^4 = 6 * x^2 * y^2 implies dfrac{x^2}{y^2} + dfrac{y^2}{x^2} = 6$, solve for y/x or x/y use that to get the value of final expression
answered Dec 22 at 15:57
l''''''''l
2,188726
add a comment |
Hint:
1: $x - dfrac{1}{x} = 3 implies x^2 + dfrac{1}{x^2} - 2 = 9$
Now use, $big(x + dfrac{1}{x}big)^2 = x^2 + dfrac{1}{x^2} + 2$
2: $dfrac{x}{x+y} = dfrac{1}{1 + dfrac{y}{x}}$
3: $x^4 + y^4 = 6 * x^2 * y^2 implies dfrac{x^2}{y^2} + dfrac{y^2}{x^2} = 6$, solve for y/x or x/y use that to get the value of final expression
answered Dec 22 at 15:57
l''''''''l
2,188726
Hint:
1: $x - dfrac{1}{x} = 3 implies x^2 + dfrac{1}{x^2} - 2 = 9$
Now use, $big(x + dfrac{1}{x}big)^2 = x^2 + dfrac{1}{x^2} + 2$
2: $dfrac{x}{x+y} = dfrac{1}{1 + dfrac{y}{x}}$
3: $x^4 + y^4 = 6 * x^2 * y^2 implies dfrac{x^2}{y^2} + dfrac{y^2}{x^2} = 6$, solve for y/x or x/y use that to get the value of final expression
answered Dec 22 at 15:57
l''''''''l
2,188726
answered Dec 22 at 15:57
l''''''''l
2,188726
answered Dec 22 at 15:57
l''''''''l
2,188726
answered Dec 22 at 15:57
l''''''''l
2,188726
2,188726
add a comment |
add a comment |
1
Notice that
$$x^2 + frac{1}{x^2} = left(x -frac{1}{x}right)^2 + 2 = 3^2 + 2 = 11$$
2
Notice that the sum of the two fractions gives $1$ hence it's rather trivial to obtain the second value.
3
You can transform the equation and solve for $frac{y}{x}$ for example, and find everything you need.
answered Dec 22 at 15:58
Von Neumann
16.3k72543
add a comment |
1
Notice that
$$x^2 + frac{1}{x^2} = left(x -frac{1}{x}right)^2 + 2 = 3^2 + 2 = 11$$
2
Notice that the sum of the two fractions gives $1$ hence it's rather trivial to obtain the second value.
3
You can transform the equation and solve for $frac{y}{x}$ for example, and find everything you need.
answered Dec 22 at 15:58
Von Neumann
16.3k72543
add a comment |
1
Notice that
$$x^2 + frac{1}{x^2} = left(x -frac{1}{x}right)^2 + 2 = 3^2 + 2 = 11$$
2
Notice that the sum of the two fractions gives $1$ hence it's rather trivial to obtain the second value.
3
You can transform the equation and solve for $frac{y}{x}$ for example, and find everything you need.
answered Dec 22 at 15:58
Von Neumann
16.3k72543
1
Notice that
$$x^2 + frac{1}{x^2} = left(x -frac{1}{x}right)^2 + 2 = 3^2 + 2 = 11$$
2
Notice that the sum of the two fractions gives $1$ hence it's rather trivial to obtain the second value.
3
You can transform the equation and solve for $frac{y}{x}$ for example, and find everything you need.
answered Dec 22 at 15:58
Von Neumann
16.3k72543
answered Dec 22 at 15:58
Von Neumann
16.3k72543
answered Dec 22 at 15:58
Von Neumann
16.3k72543
answered Dec 22 at 15:58
Von Neumann
16.3k72543
16.3k72543
add a comment |
add a comment |
Hints.
$1) x-frac1x=3implies x^2+frac1{x^2}-2=9\2)frac y{x+y}+frac x{x+y}=1\3)x^4+y^4=6x^2y^2\implies x^4+y^4+2x^2y^2=(x^2+y^2)^2=8x^2y^2\implies x^4+y^4-2x^2y^2=(x^2-y^2)^2=4x^2y^2$
Divide the two and take the square root.
answered Dec 22 at 16:08
Shubham Johri
3,859716
Thanks for the detailed explanation of the last one.
– Ameer Taweel
Dec 22 at 16:16
add a comment |
Hints.
$1) x-frac1x=3implies x^2+frac1{x^2}-2=9\2)frac y{x+y}+frac x{x+y}=1\3)x^4+y^4=6x^2y^2\implies x^4+y^4+2x^2y^2=(x^2+y^2)^2=8x^2y^2\implies x^4+y^4-2x^2y^2=(x^2-y^2)^2=4x^2y^2$
Divide the two and take the square root.
answered Dec 22 at 16:08
Shubham Johri
3,859716
Thanks for the detailed explanation of the last one.
– Ameer Taweel
Dec 22 at 16:16
add a comment |
Hints.
$1) x-frac1x=3implies x^2+frac1{x^2}-2=9\2)frac y{x+y}+frac x{x+y}=1\3)x^4+y^4=6x^2y^2\implies x^4+y^4+2x^2y^2=(x^2+y^2)^2=8x^2y^2\implies x^4+y^4-2x^2y^2=(x^2-y^2)^2=4x^2y^2$
Divide the two and take the square root.
answered Dec 22 at 16:08
Shubham Johri
3,859716
Hints.
$1) x-frac1x=3implies x^2+frac1{x^2}-2=9\2)frac y{x+y}+frac x{x+y}=1\3)x^4+y^4=6x^2y^2\implies x^4+y^4+2x^2y^2=(x^2+y^2)^2=8x^2y^2\implies x^4+y^4-2x^2y^2=(x^2-y^2)^2=4x^2y^2$
Divide the two and take the square root.
answered Dec 22 at 16:08
Shubham Johri
3,859716
answered Dec 22 at 16:08
Shubham Johri
3,859716
answered Dec 22 at 16:08
Shubham Johri
3,859716
answered Dec 22 at 16:08
Shubham Johri
3,859716
3,859716
Thanks for the detailed explanation of the last one.
– Ameer Taweel
Dec 22 at 16:16
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Thanks for the detailed explanation of the last one.
– Ameer Taweel
Dec 22 at 16:16
Thanks for the detailed explanation of the last one.
– Ameer Taweel
Dec 22 at 16:16
Thanks for the detailed explanation of the last one.
– Ameer Taweel
Dec 22 at 16:16
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2
The first one should be $x^2 + 1/x^2$.
– Rebellos
Dec 22 at 15:53
2
$frac x{x+y}+frac y{x+y}=1$.
– Lord Shark the Unknown
Dec 22 at 15:55
Sorry, didn't notice, now it's fixed.
– Ameer Taweel
Dec 22 at 15:55
@LordSharktheUnknown this solves the second one, thanks. But what about the others.
– Ameer Taweel
Dec 22 at 15:56
The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
– Sik Feng Cheong
Dec 22 at 16:21