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I found that taking a general nonlinear equation with powers of .0 and .5

e.g. $$x^2 + x^{1.5} - x + x^{0.5} + 2 = 0$$

They can be transformed into a polynomial equation by multiplying by the same equation, switching the signs of the coefficients of variables with non integer powers.

e.g. with the above equations, multiply by $$x^2 - x^{1.5} - x - x^{0.5} + 2 = 0$$

I was wondering if this is already a known, and also (if anyone has the time to check) if this would have any practical applications.

Curtis Brown
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    The substitution $u = x^{1/2}$ reduces this to a quartic equation. Your transformation has something in common with taking the complex conjugate. –  Mar 04 '18 at 21:21
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    What do you mean by a power of .0? – Théophile Mar 04 '18 at 21:21
  • It's not hard to prove that method will work. It is hard for me to imagine a use for it. I don't know whether it's known: if anyone needed it they would just prove it and use it. – Ethan Bolker Mar 04 '18 at 21:22
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    This will always work, but it will result in a polynomial whose roots are not roots of the original equation e.g. your second equation has a root at $x=1$ which will be a root of your resultant Quartic but it is not a root of your original equation. – WW1 Mar 04 '18 at 21:54

2 Answers2

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HINT

Let consider $x=y^2\ge 0$

$$x^2 + x^{1.5} - x + x^{0.5} + 2 = 0 \iff y^4+y^3-y^2+y+2=0\iff (y+1)(y^3-y+2)=0$$

user
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Is there a general method for transforming equations with powers of  ...  0.5 into polynomials?

Such "algebraization" is always possible for equations with rational powers, using polynomial resultants. In the given case for example, substituting $\,\sqrt{x}=y\,$ then eliminating $\,y\,$ between the resulting equation and $\,y^2-x=0\,$ gives resultant[ y^4 + y^3 - y^2 + y + 2, y^2 - x, y ] $\,=x^4 - 3 x^3 + 3 x^2 - 5 x + 4\,$. This is the same polynomial obtained via multiplication by the "conjugate" as suggested in the original post or, alternatively, obtained by isolating the square roots on one side then squaring the equation.

The technique works for arbitrary rational exponents, not just square roots. Consider for example the more complicated equation:

$$ x+\sqrt{x}+\sqrt[3]{x}-1=0 \tag{1} $$

The substitution $\,\sqrt[6]{x}=y\,$ rewrites the equation as $\,y^6+y^3+y^2-1=0\,$, then eliminating $\,y\,$ between the latter and $\,y^6-x=0\,$ via resultant[ y^6 + y^3 + y^2 - 1, y^6 - x, y ] gives a polynomial with integer coefficients which has all solutions of $(1)$ as roots:

$$ x^6 - 9 x^5 + 32 x^4 - 45 x^3 + 31 x^2 - 11 x + 1 \tag{2} $$

It should be noted that in all cases the resulting equation may (and in general will) have extraneous roots introduced along the way, which are not roots of the original equation. For example, $(2)$ has the root $x=1$ which does not satisfy equation $(1)$.

dxiv
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