ch1.4.6-mixt-quadratic-equations

Ch 1.4.6 – Of the Resolution of Mixt Equations of the Second Degree

Summary: Solves complete (mixt) quadratic equations by completing the square, derives the quadratic formula, presents an equivalent substitution method, and illustrates with ten worked commercial and geometric problems.

Sources: chapter-1.4.6

Last updated: 2026-05-03

---

Setup

A mixt (complete) quadratic contains all three kinds of terms. Dividing through by the leading coefficient and rearranging, every such equation can be written as (source: chapter-1.4.6, §639):

$x^2\pm px=\pm q$

where $p$ and $q$ are known. The task is to find $x$.

Method 1: Completing the square

Since $x^2+px$ is not a perfect square, we add $\frac{1}{4}{p}^2$ to both sides (§640–641). This works because

$(x+\frac{1}{2}p{)}^2=x^2+px+\frac{1}{4}{p}^2.$

The equation becomes

$x^2+px+\frac{1}{4}{p}^2=q+\frac{1}{4}{p}^2,$

whose left side is the square of $x+\frac{1}{2}p$. Taking the square root of both sides:

$x+\frac{1}{2}p=\pm \sqrt{\frac{1}{4}{p}^2+q}$

$\boxed{x=-\frac{1}{2}p\pm \sqrt{\frac{1}{4}{p}^2+q}}$

This is the quadratic formula in Euler’s notation. See completing-the-square and quadratic-equations.

Cases by type of coefficient (§644)

Coefficient	Equation form	Formula
$p$ even ($p=2m$)	$x^2=2mx+q$	$x=m\pm \sqrt{m^2+q}$
$p$ odd	$x^2=px+q$	$x=\frac{p\pm \sqrt{p^2+4q}}{2}$
Fractional ($a{x}^2=bx+c$)	divide by $a$	$x=\frac{b\pm \sqrt{b^2+4ac}}{2a}$

The last row is the standard quadratic formula.

Method 2: Substitution (§645)

Substitute $x=y+\frac{1}{2}p$ into $x^2=px+q$. The linear terms in $y$ cancel, leaving:

$y^2=\frac{1}{4}{p}^2+q$

which is a pure quadratic solvable by ch1.4.5-pure-quadratic-equations. Since $x=y+\frac{1}{2}p$, the same formula results.

Worked problems (§646–655)

Q	Problem	Key equation	Answer
1	Two numbers differing by 6 with product 91	$x^2=91-6x$	$7$ and $13$ (or $-13$ and $-7$)
2	$x^2-9$ exceeds 100 by as much as $x$ is less than 23	$x^2=-x+132$	$x=11$
3	Half × third + half = 30	$x^2+3x=180$	$x=12$
4	One number is double the other; sum + product = 90	$2x^2+3x=90$	$x=6$
5	Horse cost $x$, sold for 119 crowns, gain = $x\%$	$x^2=-100x+11900$	$x=70$ crowns
6	$x$ pieces of cloth at $2,4,\dots ,2x$ crowns total 110	$x^2+x=110$	$x=10$ pieces
7	180 crowns buys $x$ pieces; 3 more pieces would cost 3 crowns less each	$x^2=-3x+180$	$x=12$ pieces
8	Partnership: 100 lb stock, each takes out 99 lb	$x^2=19800-395x$	First partner: 45 lb
9	100 eggs divided; cross-price conditions	$25x^2=200000-4000x$	First girl: 40 eggs
10	Two silk merchants, combined price 35 crowns	$x^2=20x-75$	Two solutions: 15 & 18, or 5 & 8

Related pages

• quadratic-equations
• completing-the-square
• ch1.4.5-pure-quadratic-equations
• ch1.4.9-nature-quadratic-equations
• equations
• systems-of-linear-equations