ch1.4.5-pure-quadratic-equations

Ch 1.4.5 – Of the Resolution of Pure Quadratic Equations

Summary: Defines equations of the second degree, introduces pure (incomplete) quadratics lacking the linear term, solves them by square-root extraction, and illustrates with five worked problems.

Sources: chapter-1.4.5

Last updated: 2026-05-03

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Structure of a quadratic equation

A second-degree equation contains the square $x^2$ of the unknown but no higher power. After reducing like terms and moving everything to one side, every quadratic takes the general form (source: chapter-1.4.5, §626):

$a{x}^2\pm bx\pm c=0$

There are exactly three kinds of terms:

1. Constant terms of the form $a$ (known quantities only).
2. Linear terms of the form $bx$ (first power of the unknown).
3. Quadratic terms of the form $c{x}^2$ (second power of the unknown).

Complete vs. pure quadratics

An equation containing all three kinds of terms is called complete (or mixt); its resolution is treated in ch1.4.6-mixt-quadratic-equations.

When the linear term is absent, the equation reduces to:

$a{x}^2\pm c=0⟹x^2=\frac{c}{a}$

Euler calls this a pure quadratic. It is resolved simply by taking the square root of both sides (§629):

$x=\pm \sqrt{\frac{c}{a}}$

The $\pm$ sign is essential: every pure quadratic has two solutions, one positive and one negative.

Three cases by the sign of $c/a$

Case	Condition	Nature of $x$
1	$c/a$ is a perfect square	$x$ is rational (integer or fraction)
2	$c/a$ is positive but not a square	$x$ is irrational (a surd); approximated by the methods of square-roots-and-irrational-numbers
3	$c/a$ is negative	$x$ is imaginary; the original problem is inherently impossible

(source: chapter-1.4.5, §630–631)

Double solution

The two solutions are always $+\sqrt{c/a}$ and $-\sqrt{c/a}$. Even in the special case $a{x}^2=bx$ (constant term absent), dividing by $x$ gives only $x=b/a$, hiding the second solution $x=0$ (§632). This illustrates that equations of the second degree in general admit two solutions, whereas simple equations admit only one.

Worked examples (§633–637)

Q	Problem	Equation	Answer
1	Number whose half × third = 24	$\frac{1}{6}{x}^2=24$	$x=\pm 12$
2	$(x+5)(x-5)=96$	$x^2=121$	$x=11$
3	$(10+x)(10-x)=51$	$x^2=49$	$x=7$
4	Three players; pairwise products sum to $3830\frac{2}{3}$	$\frac{507}{833}{x}^2=3830\frac{2}{3}$	$x=79\frac{1}{3}$, second $34$, third $10$
5	Company of merchants; $\frac{1}{225}{x}^3=x$	$x^2=225$	$x=15$ partners

Question 5 is nominally cubic but factors to a pure quadratic upon dividing by $x$ (§637).

Related pages

• quadratic-equations
• ch1.4.6-mixt-quadratic-equations
• ch1.4.9-nature-quadratic-equations
• square-roots-and-irrational-numbers
• imaginary-numbers
• equations