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### solve result contain variable which to solved

Hi!

I do: (a,l,x) = var('a,l,x') solve(1/(a(x-1))==(sqrt(x^2+1^2)/2/l),x) --> [x == (sqrt(x^2 + 1)a + 2l)/(sqrt(x^2 + 1)a)]

Why x result contain x itself!? It's not ture for any x.

If I set a,l params (a,l,x) = var('a,l,x') a=24 l=1.28 solve(1/(a(x-1))==(sqrt(x^2+1^2)/2/l),x) --> [x == 1/75(75*sqrt(x^2 + 1) + 8)/sqrt(x^2 + 1)]

So result still contain x itself... I'm confused about it.

Thanks for help.

### solve result contain variable which to solved

Hi!

I do: (a,l,x) = var('a,l,x') solve(1/(a(x-1))==(sqrt(x^2+1^2)/2/l),x) --> solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == (sqrt(x^2 + 1)a + 2l)/(sqrt(x^2 + 1)a)]1)*a)]

Why x result contain x itself!? It's not ture for any x.

If I set a,l params (a,l,x) = var('a,l,x') a=24 l=1.28 solve(1/(a(x-1))==(sqrt(x^2+1^2)/2/l),x) --> solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == 1/75(75(75*sqrt(x^2 sqrt(x^2 + 1) + 8)/sqrt(x^2 + 1)]

So result still contain x itself... I'm confused about it.

Remark: Already I found that if I reorder the eqution by hand into the form below: solve(sqrt(x^4-2x^3+2x^2-2x+1)-(2a/l),x) the solve() function give me the roots well.

Thanks for help.

### solve result contain variable which to solved

Hi!

I do: (a,l,x) = var('a,l,x') solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == (sqrt(x^2 + 1)a + 2l)/(sqrt(x^2 + 1)*a)]

Why x result contain x itself!? It's not ture for any x.

If I set a,l params (a,l,x) = var('a,l,x') a=24 l=1.28 solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == 1/75(75sqrt(x^2 + 1) + 8)/sqrt(x^2 + 1)]

So result still contain x itself... I'm confused about it.

Remark: Already I found that if I reorder the eqution by hand into the form below: solve(sqrt(x^4-2x^3+2x^2-2x+1)-(2a/l),x) the solve() function give me the roots well.

Thanks for help.

### solve result contain variable which to solved

Hi!

I do: (a,l,x) = var('a,l,x') solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == (sqrt(x^2 + 1)a + 2l)/(sqrt(x^2 + 1)*a)]

Why x result contain x itself!? It's not ture for any x.

If I set a,l params (a,l,x) = var('a,l,x') a=24 l=1.28 solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == 1/75(75sqrt(x^2 + 1) + 8)/sqrt(x^2 + 1)]

So result still contain x itself... I'm confused about it.

Remark: Already I found that if I reorder the eqution by hand into the form below: solve(sqrt(x^4-2x^3+2x^2-2x+1)-(2a/l),x) the solve() function give me the roots well.

Thanks for help.

### solve result contain variable which to solved

Hi!

I do: (a,l,x) = var('a,l,x') solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == (sqrt(x^2 + 1)a + 2l)/(sqrt(x^2 + 1)*a)]

Why x result contain x itself!? It's not ture for any x.

If I set a,l params (a,l,x) = var('a,l,x') a=24 l=1.28 solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == 1/75(75sqrt(x^2 + 1) + 8)/sqrt(x^2 + 1)]

So result still contain x itself... I'm confused about it.

Remark: Already I found that if I reorder the eqution by hand into the form below: solve(sqrt(x^4-2x^3+2x^2-2x+1)-(2a/l),x) the solve() function give me the roots well.

Thanks for help.

### solve result contain variable which to solved

Hi!

I do: (a,l,x) = var('a,l,x') solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == (sqrt(x^2 + 1)a + 2l)/(sqrt(x^2 + 1)*a)]

Why x result contain x itself!? It's not ture for any x.

If I set a,l params (a,l,x) = var('a,l,x') a=24 l=1.28 solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == 1/75(75sqrt(x^2 + 1) + 8)/sqrt(x^2 + 1)]

So result still contain x itself... I'm confused about it.

Remark: Already I found that if I reorder the eqution by hand into the form below: solve(sqrt(x^4-2x^3+2x^2-2x+1)-(2a/l),x) the solve() function give me the roots well.

Thanks for help.

### solve result contain variable which to solved

Hi!

I do: do:

(a,l,x) = var('a,l,x') var('a,l,x')

solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == (sqrt(x^2 + 1)a + 2l)/(sqrt(x^2 + 1)*a)]

Why x result contain x itself!? It's not ture for any x.

If I set a,l params

(a,l,x) = var('a,l,x') a=24 l=1.28

a=24

l=1.28

solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == 1/75(75sqrt(x^2 + 1) + 8)/sqrt(x^2 + 1)]

So result still contain x itself... I'm confused about it.

Remark: Already I found that if I reorder the eqution by hand into the form below: solve(sqrt(x^4-2x^3+2x^2-2x+1)-(2a/l),x) the solve() function give me the roots well.

Thanks for help.

### solve result contain variable which to solved

Hi!

I do:

(a,l,x) = var('a,l,x')

solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == (sqrt(x^2 + 1)a + 2l)/(sqrt(x^2 + 1)*a)]

Why x result contain x itself!? It's not ture for any x.

If I set a,l params

a=24

l=1.28

solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == 1/75(75sqrt(x^2 + 1) + 8)/sqrt(x^2 + 1)]

So result still contain x itself... I'm confused about it.

Remark: Already I found that if I reorder the eqution equation by hand into the form below: solve(sqrt(x^4-2x^3+2x^2-2x+1)-(2a/l),x) the solve() function give me the roots well.

Thanks for help.

### solve result contain variable which to solved

Hi!

I do:

(a,l,x) = var('a,l,x')

solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == (sqrt(x^2 + 1)a + 2l)/(sqrt(x^2 + 1)*a)]

Why x result contain x itself!? It's not ture for any x.

If I set a,l params

a=24

l=1.28

solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == 1/75(75sqrt(x^2 + 1) + 8)/sqrt(x^2 + 1)]

So result still contain x itself... I'm confused about it.

Remark: Already I found that if I reorder the equation by hand into the form below: solve(sqrt(x^4-2x^3+2x^2-2x+1)-(2a/l),x) x+1)-(2*a/l),x)

the solve() function give me the roots well.well.*

Thanks for help.

### solve result contain variable which to solved

Hi!

I do:

(a,l,x) = var('a,l,x')

solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == (sqrt(x^2 + 1)a + 2l)/(sqrt(x^2 + 1)*a)]

Why x result contain x itself!? It's not ture for any x.

If I set a,l params

a=24

l=1.28

solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == 1/75(75sqrt(x^2 + 1) + 8)/sqrt(x^2 + 1)]

So result still contain x itself... I'm confused about it.

Remark: *Remark: Already I found that if I reorder the equation by hand into the form below: below:

solve(sqrt(x^4-2x^3+2x^3+2x^2-2x^2-2x+1)-(2x+1)-(2*a/l),x)a/l),x)

the solve() function give me the roots well.*

Thanks for help.

### solve result contain variable which to solved

Hi!

I do:

(a,l,x) = var('a,l,x')

solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == (sqrt(x^2 + 1)a + 2l)/(sqrt(x^2 + 1)*a)]

Why x result contain x itself!? It's not ture for any x.

If I set a,l params

a=24

l=1.28

solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == 1/75(75sqrt(x^2 + 1) + 8)/sqrt(x^2 + 1)]

So result still contain x itself... I'm confused about it.

*Remark: Remark: Already I found that if I reorder the equation by hand into the form below:

solve(sqrt(x^4-2x^3+2x^3+2x^2-2x^2-2x+1)-(2x+1)-(2a/l),x)a/l),x)

the solve() function give me the roots well.*well.

Thanks for help.

### solve result contain variable which to solved

Hi!

I do:

(a,l,x) = var('a,l,x')

solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == (sqrt(x^2 + 1)a + 2l)/(sqrt(x^2 + 1)*a)]

Why x result contain x itself!? It's not ture for any x.

If I set a,l params

a=24

l=1.28

solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == 1/75(75sqrt(x^2 + 1) + 8)/sqrt(x^2 + 1)]

So result still contain x itself... I'm confused about it.

Remark: Already I found that if I reorder the equation by hand into the form below:

solve(sqrt(x^4-2x^3+2x^2-2x+1)-(2a/l),x)

the solve() function give me the roots well.

But I still not understand the problem with the first form.

Thanks for help.

### solve result contain variable which to solved

Hi!

I do:

(a,l,x) = var('a,l,x')

solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == (sqrt(x^2 + 1)a + 2l)/(sqrt(x^2 + 1)*a)]

Why x result contain x itself!? It's not ture for any x.

If I set a,l params

a=24

l=1.28

solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)

--> [x == 1/75(75sqrt(x^2 + 1) + 8)/sqrt(x^2 + 1)]

So result still contain x itself... I'm confused about it.

Remark: Already I found that if I reorder the equation by hand into the form below:

solve(sqrt(x^4-2x^3+2x^2-2x+1)-(2a/l),x)

the solve() function give me the roots well.

But I still not understand the problem with the first form.

Thanks for help.

 14 None tmonteil 24353 ●27 ●176 ●449 http://wiki.sagemath.o...

### solve result contain variable which to solved

Hi!

I do:

(a,l,x) = var('a,l,x')var('a,l,x')
solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)
solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x) --> [x == (sqrt(x^2 + 1)a 1)*a + 2l)/(sqrt(x^2 2*l)/(sqrt(x^2 + 1)*a)]1)*a)]


Why x result contain x itself!? It's not ture for any x.

If I set a,l params

a=24

a=24
l=1.28
l=1.28solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x)
solve(1/(a*(x-1))==(sqrt(x^2+1^2)/2/l),x) --> [x == 1/75(75sqrt(x^2 1/75*(75*sqrt(x^2 + 1) + 8)/sqrt(x^2 + 1)]1)]


So result still contain x itself... I'm confused about it.

Thanks for help.