Mixing
Find the positive value of $x$ satisfying the given equation
Clash Royale CLAN TAG#URR8PPP
up vote
3
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$$sqrt x^2- 1over x + sqrtx-1over x = x$$
Tried removing roots. Got a degree $6$ equation which I didn't no how to solve. Also tried substituting $x = sec(y)$ but couldn't even come close to the solution.
algebra-precalculus radicals
edited Jul 26 at 21:03
Daniel Buck
2,3041623
asked Jul 26 at 20:25
user579048
161
add a comment |Â
up vote
3
down vote
favorite
$$sqrt x^2- 1over x + sqrtx-1over x = x$$
Tried removing roots. Got a degree $6$ equation which I didn't no how to solve. Also tried substituting $x = sec(y)$ but couldn't even come close to the solution.
algebra-precalculus radicals
edited Jul 26 at 21:03
Daniel Buck
2,3041623
asked Jul 26 at 20:25
user579048
161
Please show us how you obtained a degree 6 equation.
– amWhy
Jul 26 at 20:28
@user579048 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$$sqrt x^2- 1over x + sqrtx-1over x = x$$
Tried removing roots. Got a degree $6$ equation which I didn't no how to solve. Also tried substituting $x = sec(y)$ but couldn't even come close to the solution.
algebra-precalculus radicals
edited Jul 26 at 21:03
Daniel Buck
2,3041623
asked Jul 26 at 20:25
user579048
161
$$sqrt x^2- 1over x + sqrtx-1over x = x$$
Tried removing roots. Got a degree $6$ equation which I didn't no how to solve. Also tried substituting $x = sec(y)$ but couldn't even come close to the solution.
algebra-precalculus radicals
edited Jul 26 at 21:03
Daniel Buck
2,3041623
asked Jul 26 at 20:25
user579048
161
edited Jul 26 at 21:03
Daniel Buck
2,3041623
edited Jul 26 at 21:03
Daniel Buck
2,3041623
edited Jul 26 at 21:03
Daniel Buck
2,3041623
2,3041623
asked Jul 26 at 20:25
user579048
161
asked Jul 26 at 20:25
user579048
161
asked Jul 26 at 20:25
user579048
161
161
Please show us how you obtained a degree 6 equation.
– amWhy
Jul 26 at 20:28
@user579048 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
add a comment |Â
Please show us how you obtained a degree 6 equation.
– amWhy
Jul 26 at 20:28
@user579048 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
Please show us how you obtained a degree 6 equation.
– amWhy
Jul 26 at 20:28
Please show us how you obtained a degree 6 equation.
– amWhy
Jul 26 at 20:28
@user579048 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
@user579048 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
Let's see how insurmountable this sextic equation is.
Start by deriving the equation. First multiply by $sqrtx$, square both sides, and isolate the remaining radical:
$(x^2-1)+2sqrt(x^2-1)(x-1)+(x-1)=x^3$
$2sqrt(x^2-1)(x-1)=x^2-x^2-x+2$
Square again, expand and collect to get our monster:
$x^6-2x^5-x^4+2x^3+x^2=0$
Clearly since $x$ must be positive (certainly not zero, which the fractional radicands in the original equations would forbid), the factor $x^2$ can be divided out. We are down to degree $4$:
$x^4-2x^3-x^2+2x+1=0$
Next observe that this quartic equation has the following property:
(Linear coefficient)/(Cubic coefficient) $=c$
(Constant)/(Quartic coefficient)$=c^2$
When this occurs, the roots of the quartic equation occur in pairs giving the product $c$ and thus we must have a factorization:
$(x^2+ax+c)(x^2+bx+c)$
Here, $c=-1$ and so:
$x^4-2x^3-x^4+2x+1=0=(x^2+ax-1)(x^2+bx-1)$
Expanding the right side and matching like terms leads to two independent equations, thus:
$a+b=-2; b=-2-a$
$ab=1$
Then $a(-2-a)=1, a^2+2a+1=0$ and we have the root $a=-1$. Thus $b=-1$ and our polynomial is now reduced to a single, squared factor:
$(x^2-x-1)^2=0$
And so, from the quadratic formula,
$x=frac1+sqrt52$
Which the reader can verify, checks out! Note that in the checking process we should find $(x^2-1)/x=1$.
answered Jul 26 at 23:13
Oscar Lanzi
9,93911632
add a comment |Â
up vote
2
down vote
HINT
We can try with
$$sqrt x^2- 1over x + sqrtx-1over x = x$$
$$sqrt x+1sqrt x- 1over x + sqrtx-1over x = x$$
$$(sqrt x+1+1)sqrt x- 1over x = x$$
$$sqrt x- 1over x = fracxsqrt x+1+1fracsqrt x+1-1sqrt x+1-1=sqrt x+1-1$$
$$sqrt x- 1 =sqrt x^2+x-sqrt x$$
and from here we can square to eliminate the square roots.
Recall to check at the end the conditions for the existence related to the original equation
$x^2- 1over xge 0$
$x- 1over xge 0$
$xneq 0$
answered Jul 26 at 20:35
gimusi
65k73583
Can I see the radical eliminated equation? It looks like it would be degree only 4 making it more parsimonious than my (and the OP's) development.
– Oscar Lanzi
Jul 27 at 10:00
1
@OscarLanzi squaring 2 times we obtain $x^4-2x^3-x^2+2x+1=(x^2-x-1)^2=0$.
– gimusi
Jul 27 at 10:03
add a comment |Â
up vote
1
down vote
Checking other answers:
Note that $fracx^2-1xge 0;fracx-1xge 0$ and $x>0$ imply $x>1$.
If $x^2=x+1 (1)$, then:
$$ x^2-1=x Rightarrow fracx-1x=frac1x+1,\
sqrt x^2- 1over x + sqrtx-1over x = x (2) Rightarrow \
sqrt (x+1)- 1over x + sqrt1over x+1 = sqrtx+1 Rightarrow \
1+frac1sqrtx+1=sqrtx+1 Rightarrow \
sqrtx+1+1=x+1 Rightarrow \
x+1=x^2.$$
So, $(1)$ and $(2)$ are equivalent.
edited Jul 28 at 1:08
Oscar Lanzi
9,93911632
answered Jul 27 at 13:53
farruhota
13.5k2632
How d you magically get $x^2=x+1$?
– Oscar Lanzi
Jul 27 at 17:32
@OscarLanzi, the point was to support the given two answers and show the golden mean is indeed the root of the given equation.
– farruhota
Jul 27 at 18:35
See my introductory phrase.
– Oscar Lanzi
Jul 28 at 1:09
1
@OscarLanzi, it is fine, thank you.
– farruhota
Jul 28 at 13:52
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Let's see how insurmountable this sextic equation is.
Start by deriving the equation. First multiply by $sqrtx$, square both sides, and isolate the remaining radical:
$(x^2-1)+2sqrt(x^2-1)(x-1)+(x-1)=x^3$
$2sqrt(x^2-1)(x-1)=x^2-x^2-x+2$
Square again, expand and collect to get our monster:
$x^6-2x^5-x^4+2x^3+x^2=0$
Clearly since $x$ must be positive (certainly not zero, which the fractional radicands in the original equations would forbid), the factor $x^2$ can be divided out. We are down to degree $4$:
$x^4-2x^3-x^2+2x+1=0$
Next observe that this quartic equation has the following property:
(Linear coefficient)/(Cubic coefficient) $=c$
(Constant)/(Quartic coefficient)$=c^2$
When this occurs, the roots of the quartic equation occur in pairs giving the product $c$ and thus we must have a factorization:
$(x^2+ax+c)(x^2+bx+c)$
Here, $c=-1$ and so:
$x^4-2x^3-x^4+2x+1=0=(x^2+ax-1)(x^2+bx-1)$
Expanding the right side and matching like terms leads to two independent equations, thus:
$a+b=-2; b=-2-a$
$ab=1$
Then $a(-2-a)=1, a^2+2a+1=0$ and we have the root $a=-1$. Thus $b=-1$ and our polynomial is now reduced to a single, squared factor:
$(x^2-x-1)^2=0$
And so, from the quadratic formula,
$x=frac1+sqrt52$
Which the reader can verify, checks out! Note that in the checking process we should find $(x^2-1)/x=1$.
answered Jul 26 at 23:13
Oscar Lanzi
9,93911632
add a comment |Â
up vote
4
down vote
Let's see how insurmountable this sextic equation is.
Start by deriving the equation. First multiply by $sqrtx$, square both sides, and isolate the remaining radical:
$(x^2-1)+2sqrt(x^2-1)(x-1)+(x-1)=x^3$
$2sqrt(x^2-1)(x-1)=x^2-x^2-x+2$
Square again, expand and collect to get our monster:
$x^6-2x^5-x^4+2x^3+x^2=0$
Clearly since $x$ must be positive (certainly not zero, which the fractional radicands in the original equations would forbid), the factor $x^2$ can be divided out. We are down to degree $4$:
$x^4-2x^3-x^2+2x+1=0$
Next observe that this quartic equation has the following property:
(Linear coefficient)/(Cubic coefficient) $=c$
(Constant)/(Quartic coefficient)$=c^2$
When this occurs, the roots of the quartic equation occur in pairs giving the product $c$ and thus we must have a factorization:
$(x^2+ax+c)(x^2+bx+c)$
Here, $c=-1$ and so:
$x^4-2x^3-x^4+2x+1=0=(x^2+ax-1)(x^2+bx-1)$
Expanding the right side and matching like terms leads to two independent equations, thus:
$a+b=-2; b=-2-a$
$ab=1$
Then $a(-2-a)=1, a^2+2a+1=0$ and we have the root $a=-1$. Thus $b=-1$ and our polynomial is now reduced to a single, squared factor:
$(x^2-x-1)^2=0$
And so, from the quadratic formula,
$x=frac1+sqrt52$
Which the reader can verify, checks out! Note that in the checking process we should find $(x^2-1)/x=1$.
answered Jul 26 at 23:13
Oscar Lanzi
9,93911632
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Let's see how insurmountable this sextic equation is.
Start by deriving the equation. First multiply by $sqrtx$, square both sides, and isolate the remaining radical:
$(x^2-1)+2sqrt(x^2-1)(x-1)+(x-1)=x^3$
$2sqrt(x^2-1)(x-1)=x^2-x^2-x+2$
Square again, expand and collect to get our monster:
$x^6-2x^5-x^4+2x^3+x^2=0$
Clearly since $x$ must be positive (certainly not zero, which the fractional radicands in the original equations would forbid), the factor $x^2$ can be divided out. We are down to degree $4$:
$x^4-2x^3-x^2+2x+1=0$
Next observe that this quartic equation has the following property:
(Linear coefficient)/(Cubic coefficient) $=c$
(Constant)/(Quartic coefficient)$=c^2$
When this occurs, the roots of the quartic equation occur in pairs giving the product $c$ and thus we must have a factorization:
$(x^2+ax+c)(x^2+bx+c)$
Here, $c=-1$ and so:
$x^4-2x^3-x^4+2x+1=0=(x^2+ax-1)(x^2+bx-1)$
Expanding the right side and matching like terms leads to two independent equations, thus:
$a+b=-2; b=-2-a$
$ab=1$
Then $a(-2-a)=1, a^2+2a+1=0$ and we have the root $a=-1$. Thus $b=-1$ and our polynomial is now reduced to a single, squared factor:
$(x^2-x-1)^2=0$
And so, from the quadratic formula,
$x=frac1+sqrt52$
Which the reader can verify, checks out! Note that in the checking process we should find $(x^2-1)/x=1$.
answered Jul 26 at 23:13
Oscar Lanzi
9,93911632
Let's see how insurmountable this sextic equation is.
Start by deriving the equation. First multiply by $sqrtx$, square both sides, and isolate the remaining radical:
$(x^2-1)+2sqrt(x^2-1)(x-1)+(x-1)=x^3$
$2sqrt(x^2-1)(x-1)=x^2-x^2-x+2$
Square again, expand and collect to get our monster:
$x^6-2x^5-x^4+2x^3+x^2=0$
Clearly since $x$ must be positive (certainly not zero, which the fractional radicands in the original equations would forbid), the factor $x^2$ can be divided out. We are down to degree $4$:
$x^4-2x^3-x^2+2x+1=0$
Next observe that this quartic equation has the following property:
(Linear coefficient)/(Cubic coefficient) $=c$
(Constant)/(Quartic coefficient)$=c^2$
When this occurs, the roots of the quartic equation occur in pairs giving the product $c$ and thus we must have a factorization:
$(x^2+ax+c)(x^2+bx+c)$
Here, $c=-1$ and so:
$x^4-2x^3-x^4+2x+1=0=(x^2+ax-1)(x^2+bx-1)$
Expanding the right side and matching like terms leads to two independent equations, thus:
$a+b=-2; b=-2-a$
$ab=1$
Then $a(-2-a)=1, a^2+2a+1=0$ and we have the root $a=-1$. Thus $b=-1$ and our polynomial is now reduced to a single, squared factor:
$(x^2-x-1)^2=0$
And so, from the quadratic formula,
$x=frac1+sqrt52$
Which the reader can verify, checks out! Note that in the checking process we should find $(x^2-1)/x=1$.
answered Jul 26 at 23:13
Oscar Lanzi
9,93911632
edited Jul 27 at 9:54
answered Jul 26 at 23:13
Oscar Lanzi
9,93911632
answered Jul 26 at 23:13
Oscar Lanzi
9,93911632
answered Jul 26 at 23:13
Oscar Lanzi
9,93911632
9,93911632
add a comment |Â
add a comment |Â
up vote
2
down vote
HINT
We can try with
$$sqrt x^2- 1over x + sqrtx-1over x = x$$
$$sqrt x+1sqrt x- 1over x + sqrtx-1over x = x$$
$$(sqrt x+1+1)sqrt x- 1over x = x$$
$$sqrt x- 1over x = fracxsqrt x+1+1fracsqrt x+1-1sqrt x+1-1=sqrt x+1-1$$
$$sqrt x- 1 =sqrt x^2+x-sqrt x$$
and from here we can square to eliminate the square roots.
Recall to check at the end the conditions for the existence related to the original equation
$x^2- 1over xge 0$
$x- 1over xge 0$
$xneq 0$
answered Jul 26 at 20:35
gimusi
65k73583
Can I see the radical eliminated equation? It looks like it would be degree only 4 making it more parsimonious than my (and the OP's) development.
– Oscar Lanzi
Jul 27 at 10:00
1
@OscarLanzi squaring 2 times we obtain $x^4-2x^3-x^2+2x+1=(x^2-x-1)^2=0$.
– gimusi
Jul 27 at 10:03
add a comment |Â
up vote
2
down vote
HINT
We can try with
$$sqrt x^2- 1over x + sqrtx-1over x = x$$
$$sqrt x+1sqrt x- 1over x + sqrtx-1over x = x$$
$$(sqrt x+1+1)sqrt x- 1over x = x$$
$$sqrt x- 1over x = fracxsqrt x+1+1fracsqrt x+1-1sqrt x+1-1=sqrt x+1-1$$
$$sqrt x- 1 =sqrt x^2+x-sqrt x$$
and from here we can square to eliminate the square roots.
Recall to check at the end the conditions for the existence related to the original equation
$x^2- 1over xge 0$
$x- 1over xge 0$
$xneq 0$
answered Jul 26 at 20:35
gimusi
65k73583
Can I see the radical eliminated equation? It looks like it would be degree only 4 making it more parsimonious than my (and the OP's) development.
– Oscar Lanzi
Jul 27 at 10:00
1
@OscarLanzi squaring 2 times we obtain $x^4-2x^3-x^2+2x+1=(x^2-x-1)^2=0$.
– gimusi
Jul 27 at 10:03
add a comment |Â
up vote
2
down vote
up vote
2
down vote
HINT
We can try with
$$sqrt x^2- 1over x + sqrtx-1over x = x$$
$$sqrt x+1sqrt x- 1over x + sqrtx-1over x = x$$
$$(sqrt x+1+1)sqrt x- 1over x = x$$
$$sqrt x- 1over x = fracxsqrt x+1+1fracsqrt x+1-1sqrt x+1-1=sqrt x+1-1$$
$$sqrt x- 1 =sqrt x^2+x-sqrt x$$
and from here we can square to eliminate the square roots.
Recall to check at the end the conditions for the existence related to the original equation
$x^2- 1over xge 0$
$x- 1over xge 0$
$xneq 0$
answered Jul 26 at 20:35
gimusi
65k73583
HINT
We can try with
$$sqrt x^2- 1over x + sqrtx-1over x = x$$
$$sqrt x+1sqrt x- 1over x + sqrtx-1over x = x$$
$$(sqrt x+1+1)sqrt x- 1over x = x$$
$$sqrt x- 1over x = fracxsqrt x+1+1fracsqrt x+1-1sqrt x+1-1=sqrt x+1-1$$
$$sqrt x- 1 =sqrt x^2+x-sqrt x$$
and from here we can square to eliminate the square roots.
Recall to check at the end the conditions for the existence related to the original equation
$x^2- 1over xge 0$
$x- 1over xge 0$
$xneq 0$
answered Jul 26 at 20:35
gimusi
65k73583
edited Jul 26 at 20:40
answered Jul 26 at 20:35
gimusi
65k73583
answered Jul 26 at 20:35
gimusi
65k73583
answered Jul 26 at 20:35
gimusi
65k73583
65k73583
Can I see the radical eliminated equation? It looks like it would be degree only 4 making it more parsimonious than my (and the OP's) development.
– Oscar Lanzi
Jul 27 at 10:00
1
@OscarLanzi squaring 2 times we obtain $x^4-2x^3-x^2+2x+1=(x^2-x-1)^2=0$.
– gimusi
Jul 27 at 10:03
add a comment |Â
Can I see the radical eliminated equation? It looks like it would be degree only 4 making it more parsimonious than my (and the OP's) development.
– Oscar Lanzi
Jul 27 at 10:00
1
@OscarLanzi squaring 2 times we obtain $x^4-2x^3-x^2+2x+1=(x^2-x-1)^2=0$.
– gimusi
Jul 27 at 10:03
Can I see the radical eliminated equation? It looks like it would be degree only 4 making it more parsimonious than my (and the OP's) development.
– Oscar Lanzi
Jul 27 at 10:00
Can I see the radical eliminated equation? It looks like it would be degree only 4 making it more parsimonious than my (and the OP's) development.
– Oscar Lanzi
Jul 27 at 10:00
1
1
@OscarLanzi squaring 2 times we obtain $x^4-2x^3-x^2+2x+1=(x^2-x-1)^2=0$.
– gimusi
Jul 27 at 10:03
@OscarLanzi squaring 2 times we obtain $x^4-2x^3-x^2+2x+1=(x^2-x-1)^2=0$.
– gimusi
Jul 27 at 10:03
add a comment |Â
up vote
1
down vote
Checking other answers:
Note that $fracx^2-1xge 0;fracx-1xge 0$ and $x>0$ imply $x>1$.
If $x^2=x+1 (1)$, then:
$$ x^2-1=x Rightarrow fracx-1x=frac1x+1,\
sqrt x^2- 1over x + sqrtx-1over x = x (2) Rightarrow \
sqrt (x+1)- 1over x + sqrt1over x+1 = sqrtx+1 Rightarrow \
1+frac1sqrtx+1=sqrtx+1 Rightarrow \
sqrtx+1+1=x+1 Rightarrow \
x+1=x^2.$$
So, $(1)$ and $(2)$ are equivalent.
edited Jul 28 at 1:08
Oscar Lanzi
9,93911632
answered Jul 27 at 13:53
farruhota
13.5k2632
How d you magically get $x^2=x+1$?
– Oscar Lanzi
Jul 27 at 17:32
@OscarLanzi, the point was to support the given two answers and show the golden mean is indeed the root of the given equation.
– farruhota
Jul 27 at 18:35
See my introductory phrase.
– Oscar Lanzi
Jul 28 at 1:09
1
@OscarLanzi, it is fine, thank you.
– farruhota
Jul 28 at 13:52
add a comment |Â
up vote
1
down vote
Checking other answers:
Note that $fracx^2-1xge 0;fracx-1xge 0$ and $x>0$ imply $x>1$.
If $x^2=x+1 (1)$, then:
$$ x^2-1=x Rightarrow fracx-1x=frac1x+1,\
sqrt x^2- 1over x + sqrtx-1over x = x (2) Rightarrow \
sqrt (x+1)- 1over x + sqrt1over x+1 = sqrtx+1 Rightarrow \
1+frac1sqrtx+1=sqrtx+1 Rightarrow \
sqrtx+1+1=x+1 Rightarrow \
x+1=x^2.$$
So, $(1)$ and $(2)$ are equivalent.
edited Jul 28 at 1:08
Oscar Lanzi
9,93911632
answered Jul 27 at 13:53
farruhota
13.5k2632
How d you magically get $x^2=x+1$?
– Oscar Lanzi
Jul 27 at 17:32
@OscarLanzi, the point was to support the given two answers and show the golden mean is indeed the root of the given equation.
– farruhota
Jul 27 at 18:35
See my introductory phrase.
– Oscar Lanzi
Jul 28 at 1:09
1
@OscarLanzi, it is fine, thank you.
– farruhota
Jul 28 at 13:52
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Checking other answers:
Note that $fracx^2-1xge 0;fracx-1xge 0$ and $x>0$ imply $x>1$.
If $x^2=x+1 (1)$, then:
$$ x^2-1=x Rightarrow fracx-1x=frac1x+1,\
sqrt x^2- 1over x + sqrtx-1over x = x (2) Rightarrow \
sqrt (x+1)- 1over x + sqrt1over x+1 = sqrtx+1 Rightarrow \
1+frac1sqrtx+1=sqrtx+1 Rightarrow \
sqrtx+1+1=x+1 Rightarrow \
x+1=x^2.$$
So, $(1)$ and $(2)$ are equivalent.
edited Jul 28 at 1:08
Oscar Lanzi
9,93911632
answered Jul 27 at 13:53
farruhota
13.5k2632
Checking other answers:
Note that $fracx^2-1xge 0;fracx-1xge 0$ and $x>0$ imply $x>1$.
If $x^2=x+1 (1)$, then:
$$ x^2-1=x Rightarrow fracx-1x=frac1x+1,\
sqrt x^2- 1over x + sqrtx-1over x = x (2) Rightarrow \
sqrt (x+1)- 1over x + sqrt1over x+1 = sqrtx+1 Rightarrow \
1+frac1sqrtx+1=sqrtx+1 Rightarrow \
sqrtx+1+1=x+1 Rightarrow \
x+1=x^2.$$
So, $(1)$ and $(2)$ are equivalent.
edited Jul 28 at 1:08
Oscar Lanzi
9,93911632
answered Jul 27 at 13:53
farruhota
13.5k2632
edited Jul 28 at 1:08
Oscar Lanzi
9,93911632
edited Jul 28 at 1:08
Oscar Lanzi
9,93911632
edited Jul 28 at 1:08
Oscar Lanzi
9,93911632
9,93911632
answered Jul 27 at 13:53
farruhota
13.5k2632
answered Jul 27 at 13:53
farruhota
13.5k2632
answered Jul 27 at 13:53
farruhota
13.5k2632
13.5k2632
How d you magically get $x^2=x+1$?
– Oscar Lanzi
Jul 27 at 17:32
@OscarLanzi, the point was to support the given two answers and show the golden mean is indeed the root of the given equation.
– farruhota
Jul 27 at 18:35
See my introductory phrase.
– Oscar Lanzi
Jul 28 at 1:09
1
@OscarLanzi, it is fine, thank you.
– farruhota
Jul 28 at 13:52
add a comment |Â
How d you magically get $x^2=x+1$?
– Oscar Lanzi
Jul 27 at 17:32
@OscarLanzi, the point was to support the given two answers and show the golden mean is indeed the root of the given equation.
– farruhota
Jul 27 at 18:35
See my introductory phrase.
– Oscar Lanzi
Jul 28 at 1:09
1
@OscarLanzi, it is fine, thank you.
– farruhota
Jul 28 at 13:52
How d you magically get $x^2=x+1$?
– Oscar Lanzi
Jul 27 at 17:32
How d you magically get $x^2=x+1$?
– Oscar Lanzi
Jul 27 at 17:32
@OscarLanzi, the point was to support the given two answers and show the golden mean is indeed the root of the given equation.
– farruhota
Jul 27 at 18:35
@OscarLanzi, the point was to support the given two answers and show the golden mean is indeed the root of the given equation.
– farruhota
Jul 27 at 18:35
See my introductory phrase.
– Oscar Lanzi
Jul 28 at 1:09
See my introductory phrase.
– Oscar Lanzi
Jul 28 at 1:09
1
1
@OscarLanzi, it is fine, thank you.
– farruhota
Jul 28 at 13:52
@OscarLanzi, it is fine, thank you.
– farruhota
Jul 28 at 13:52
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Please show us how you obtained a degree 6 equation.
– amWhy
Jul 26 at 20:28
@user579048 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday