ch1.4.15-new-method-quartic

Ch 1.4.15 — Of a New Method of Resolving Equations of the Fourth Degree

Summary: Euler’s own method assumes a root of the form $x=\sqrt{p}+\sqrt{q}+\sqrt{r}$ where $p,q,r$ satisfy an auxiliary cubic, deriving all four roots via sign combinations. Also shows how to eliminate the cubic term from a general quartic.

Sources: chapter-1.4.15

Last updated: 2026-05-03

---

The key ansatz (§774)

Suppose $x=\sqrt{p}+\sqrt{q}+\sqrt{r}$ where $p,q,r$ are roots of $z^3-f{z}^2+gz-h=0$, so that:

$p+q+r=f,pq+pr+qr=g,pqr=h$

Squaring: $x^2=f+2\sqrt{pq}+2\sqrt{pr}+2\sqrt{qr}$, so $x^2-f=2(\sqrt{pq}+\sqrt{pr}+\sqrt{qr})$.

Squaring again and using $4(pq+pr+qr)=4g$ and $\sqrt{pqr}=\sqrt{h}$:

$x^4-2f{x}^2-8x\sqrt{h}+f^2-4g=0$

One root of this quartic is $x=\sqrt{p}+\sqrt{q}+\sqrt{r}$.

Matching the general depressed quartic (§775–776)

For $x^4-a{x}^2-bx-c=0$ (no $x^3$ term), compare with the formula:

$2f=a\Rightarrow f=\frac{a}{2},8\sqrt{h}=b\Rightarrow h=\frac{b^2}{64},f^2-4g=-c\Rightarrow g=\frac{a^2}{16}+\frac{c}{4}$

Then solve the auxiliary cubic:

$z^3-\frac{a}{2}{z}^2+\left(\frac{a^2}{16}+\frac{c}{4}\right)z-\frac{b^2}{64}=0$

to obtain $p$, $q$, $r$.

All four roots (§777)

Naively, each $\sqrt{\cdot }$ can be $\pm$, giving 8 combinations. But the constraint $\sqrt{pqr}=\frac{b}{8}$ (which must match the sign of $b$) cuts these to exactly four valid combinations.

If $b>0$ (product $\sqrt{p}\sqrt{q}\sqrt{r}>0$):

$x=\sqrt{p}+\sqrt{q}+\sqrt{r},\sqrt{p}-\sqrt{q}-\sqrt{r},-\sqrt{p}+\sqrt{q}-\sqrt{r},-\sqrt{p}-\sqrt{q}+\sqrt{r}$

If $b<0$ (product $\sqrt{p}\sqrt{q}\sqrt{r}<0$):

$x=\sqrt{p}+\sqrt{q}-\sqrt{r},\sqrt{p}-\sqrt{q}+\sqrt{r},-\sqrt{p}+\sqrt{q}+\sqrt{r},-\sqrt{p}-\sqrt{q}-\sqrt{r}$

Worked example 1 (§778)

$x^4-25x^2+60x-36=0$

Here $a=25$, $b=-60$, $c=36$. The auxiliary cubic (after clearing fractions via $z=u/4$) becomes $u^3-50u^2+769u-3600=0$. Roots: $u=9,16,25$, so $p=\frac{9}{4}$, $q=4$, $r=\frac{25}{4}$; thus $\sqrt{p}=\frac{3}{2}$, $\sqrt{q}=2$, $\sqrt{r}=\frac{5}{2}$.

Since $b<0$, take all three negative or exactly one negative:

$x=1,2,3,-6$

Verification: $(x-1)(x-2)(x-3)(x+6)=0$ expands to the original equation.

Removing the cubic term from a complete quartic (§779)

To apply the method to $y^4+a{y}^3+b{y}^2+cy+d=0$, substitute $y=x-\frac{a}{4}$. This eliminates the $x^3$ term, yielding:

$x^4+\left(b-\frac{3a^2}{8}\right)x^2+\left(\frac{a^3}{8}-\frac{ab}{2}+c\right)x+\left(-\frac{3a^4}{256}+\frac{a^{2}b}{16}-\frac{ac}{4}+d\right)=0$

After finding $x$ roots, recover $y=x-\frac{a}{4}$.

Note on higher degrees (§780)

Euler observes that no general algebraic solution is known for equations of degree 5 or higher. The only progress for those cases is finding rational roots by the divisor method — which reduces the problem to a lower degree.

Worked example 2 — irrational roots (§781–783)

$y^4-8y^3+14y^2+4y-8=0$

Substitute $y-2=x$ to remove the cubic term, getting $x^4-10x^2-4x+8=0$.

The auxiliary cubic is $z^3-5z^2+\frac{17}{4}z-\frac{1}{4}=0$, with roots $p=1$, $q=\frac{4+\sqrt{15}}{2}$, $r=\frac{4-\sqrt{15}}{2}$.

Using the binomial-surd formula $\sqrt{a\pm \sqrt{b}}=\sqrt{(a+c)/2}\pm \sqrt{(a-c)/2}$ where $c=\sqrt{a^2-b}$:

$\sqrt{q}=\frac{\sqrt{5}+\sqrt{3}}{2},\sqrt{r}=\frac{\sqrt{5}-\sqrt{3}}{2}$

The four values of $x$ are $1\pm \sqrt{5}$ and $-1\pm \sqrt{3}$, giving the four roots of $y$:

$y=3+\sqrt{5},3-\sqrt{5},1+\sqrt{3},1-\sqrt{3}$

Related pages

• quartic-equations
• bombelli-rule
• ch1.4.13-quartic-equations
• ch1.4.14-bombelli-rule
• cubic-equations
• cardanos-rule
• square-roots-of-binomials
• imaginary-numbers
• vieta-formulas
• equations