Rationalizing Substitutions

Summary: A catalog of substitutions from §47–§51 that express $y=f(z)$ (containing a radical) as a pair $(z(x),y(x))$ of rational — or at least radical-free — functions of a new variable $x$.

Sources: chapter3

Last updated: 2026-04-23

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The pattern

Each case starts from a specific form of $y(z)$ involving a radical, and gives a substitution that eliminates the radical by expressing both $z$ and $y$ as functions of a new variable $x$.

§47 — Simple linear radical

$y=\sqrt{a+bz}.$

Let $y=bx$. Then $a+bz=b^2{x}^2$, so

$z=b{x}^2-\frac{a}{b},y=bx.$

Both are polynomials in $x$ (source: chapter3, §47).

§48 — Rational power of a linear expression

$y=(a+bz{)}^{m/n}.$

Let $y=x^m$, so $(a+bz{)}^{1/n}=x$, giving

$z=\frac{x^n-a}{b},y=x^m$

(source: chapter3, §48). Neither $y$ nor $z$ is expressible without a radical in terms of the other, but both are polynomials in $x$.

§49 — Rational power of a linear-over-linear

$y={\left(\frac{a+bz}{f+gz}\right)}^{m/n}.$

Let $y=x^m$, so $\frac{a+bz}{f+gz}=x^n$, giving

$z=\frac{a-f{x}^n}{g{x}^n-b},y=x^m$

(source: chapter3, §49).

Euler also notes a symmetric generalization: if ${\left(\frac{\alpha +\beta y}{\gamma +\delta y}\right)}^n={\left(\frac{a+bz}{f+gz}\right)}^m$, setting both sides equal to $x^{mn}$ gives $y$ and $z$ as linear-fractional functions of $x^m$ and $x^n$ respectively.

§50 — Radical of two linear factors

$y=\sqrt{(a+bz)(c+dz)}.$

Let $\sqrt{(a+bz)(c+dz)}=(a+bz)x$. Squaring and cancelling $(a+bz)$ gives a linear equation for $z$:

$z=\frac{c-a{x}^2}{b{x}^2-d},y=\frac{(bc-ad)x}{b{x}^2-d}$

(source: chapter3, §50). The celebrated special case $b=1$, $c=a$, $d=-1$ gives $y=\sqrt{a^2-z^2}$, the circle. See rational-parametrization-of-the-circle.

Euler remarks: whenever there are two linear real factors under a radical sign, the radical can be removed by this method.

§51 — Radical of a general quadratic

$y=\sqrt{p+qz+r{z}^2}.$

The treatment branches on the signs of $p$ and $r$.

Case I. $p=a^2>0$. Let $\sqrt{a^2+bz+c{z}^2}=a+xz$. Then

$z=\frac{b-2ax}{x^2-c},y=\frac{bx-a{x}^2-ac}{x^2-c}.$

Case II. $r=a^2>0$. Let $\sqrt{a^2{z}^2+bz+c}=az+x$. Then

$z=\frac{x^2-c}{b-2ax},y=\frac{-ac+bx-a{x}^2}{b-2ax}.$

Case III. $p,r<0$. If $q^2>4pr$, the quadratic factors into two real linear factors and the problem reduces to §50. Otherwise $p+qz+r{z}^2$ is always negative and $y$ is imaginary for real $z$.

Euler’s worked example: $y=\sqrt{-1+3z-z^2}=\sqrt{1-(1-z)(2-z)}$. Let $y=1-(1-z)x$; then

$z=\frac{2-2x+x^2}{1+x^2},y=\frac{1+x-x^2}{1+x^2}$

(source: chapter3, §51).

Scope and limits

These are the cases where algebraic substitution alone suffices. Euler notes explicitly that “other cases, which are not discussed in this treatise, cannot be reduced to a form without radicals by a substitution without radicals” (source: chapter3, §51). In modern terms: rational parametrizability is special, and not every algebraic curve admits one. It will later be known that this corresponds to the curve having genus zero.

Related pages

• substitution
• chapter-3-on-the-transformation-of-functions-by-substitution
• rational-parametrization-of-the-circle
• homogeneous-substitution
• factoring-polynomials — Case III of §51 uses the real factorization of a quadratic.