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Re: Separating square roots

  • To: mathgroup at smc.vnet.net
  • Subject: [mg127236] Re: Separating square roots
  • From: Bob Hanlon <hanlonr357 at gmail.com>
  • Date: Wed, 11 Jul 2012 02:15:42 -0400 (EDT)
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  • References: <20120710044231.0F717683F@smc.vnet.net>

The proposed approach will introduce some extraneous roots

A = 3 + 5 x^2 + 7 x + Sqrt[5 + x] + 7 x Sqrt[5 + x];

nosqrt = Select[A, FreeQ[#, Power[_, 1/2]] &]

3 + 7*x + 5*x^2

withsqrt = A - nosqrt

Sqrt[5 + x] + 7*x*Sqrt[5 + x]

sol1 = Solve[nosqrt^2 == withsqrt^2, x] // Simplify;

sol1 // N // Chop

{{x -> 0.0891586}, {x -> 2.37923}, {x -> -3.06206}, {x -> -0.246324}}

However, the first two of these values are not roots of the original equation.

A /. sol1 // N // Chop

{7.32771, 95.9166, 0, 0}

Direct use of Solve provides the desired results.

sol2 = Solve[A == 0, x] // Simplify;

sol2 // N // Chop

{{x -> -3.06206}, {x -> -0.246324}}

A /. sol2 // N // Chop

{0, 0}


Bob Hanlon


On Tue, Jul 10, 2012 at 12:42 AM,  <rhartley.anu at gmail.com> wrote:
> I have an expression which is a sum of terms, some monomials
> and others involving square roots, such as
> A = 3  + 5 x^2 + 7 x + Sqrt[5+x] + 7 x Sqrt[5+x] (but much more complicated).
> I want to solve this, which involves separating terms with the square root
> from those that do not have a square root, then squaring each.
>
> I can not work out how to separate out those terms that have the square
> root.  I do not want to hard code, such as nosqrt = A[[1]] + A[[2]] , etc.
> Can anyone help?
>
> r
>



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