ch1.4.8-extraction-square-roots-binomials

Ch 1.4.8 – Of the Extraction of the Square Roots of Binomials

Summary: Develops a rule for extracting the square root of a binomial surd $a\pm \sqrt{b}$ by setting up a system in $x,y$ and solving it; applies the method to real, imaginary, and quartic-equation contexts.

Sources: chapter-1.4.8

Last updated: 2026-05-03

---

What is a binomial (in this context)?

A binomial here means a two-term expression in which at least one term involves a square root, for example $3+\sqrt{5}$, $\sqrt{8}+\sqrt{3}$, or $3-\sqrt{5}$ (source: chapter-1.4.8, §669). Such quantities arise naturally when solving quadratics: e.g. $x^2=6x-4$ gives $x=3+\sqrt{5}$.

Squaring a binomial — motivation

$(a+\sqrt{b}{)}^2=(a^2+b)+2a\sqrt{b}$ $(\sqrt{a}+\sqrt{b}{)}^2=(a+b)+2\sqrt{ab}$

Reading these backwards: to take the square root of $(a^2+b)+2a\sqrt{b}$ is to find $a+\sqrt{b}$, which is far more informative than simply writing $\sqrt{(a^2+b)+2a\sqrt{b}}$ (§671).

The root formula

To find $\sqrt{a+\sqrt{b}}$, suppose the root is $\sqrt{x}+\sqrt{y}$ (§672–674). Squaring:

$(\sqrt{x}+\sqrt{y}{)}^2=(x+y)+2\sqrt{xy}=a+\sqrt{b}$

This gives the system $x+y=a$ and $4xy=b$. Eliminating $y$:

$x-y=\sqrt{a^2-b}$

Therefore:

$x=\frac{a+\sqrt{a^2-b}}{2},y=\frac{a-\sqrt{a^2-b}}{2}$

and the square root of $a+\sqrt{b}$ is:

$\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a-c}{2}},\text{where }c=\sqrt{a^2-b}$

The extraction is conveniently expressible only when $a^2-b$ is a perfect square $c^2$; otherwise we cannot simplify beyond placing x​ in front (§675). For $a-\sqrt{b}$ simply flip the sign of the second term:

$\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+c}{2}}-\sqrt{\frac{a-c}{2}}$

Summary rule (§677): subtract the square of the irrational part from the square of the rational part; if the remainder $c^2$ is a perfect square, the root is $\sqrt{(a+c)/2}\pm \sqrt{(a-c)/2}$.

Examples (§678–679)

Binomial	$a$	$b$	$a^2-b$	$c$	Square root
$5+2\sqrt{6}$	$5$	$24$	$1$	$1$	$\sqrt{3}+\sqrt{2}$
$2+\sqrt{3}$	$2$	$3$	$1$	$1$	$\sqrt{3/2}+\sqrt{1/2}$
$11+6\sqrt{2}$	$11$	$72$	$49$	$7$	$3+\sqrt{2}$
$11+2\sqrt{30}$	$11$	$120$	$1$	$1$	$\sqrt{6}+\sqrt{5}$
$1+4\sqrt{-3}$	$1$	$-48$	$49$	$7$	$2+\sqrt{-3}$
$2\sqrt{-1}$	$0$	$-4$	$4$	$2$	$1+\sqrt{-1}$

The rule works even for imaginary binomials (§679).

Application to quartic equations

The case $x^4=2a{x}^2-c^2$ reduces, via $x^2=y$, to $y^2=2ay-c^2$, giving $y=a\pm \sqrt{a^2-c^2}$. Then $x^2=a\pm \sqrt{a^2-c^2}$ is a binomial square root problem, solved by the formula above (§680–681):

$x=\sqrt{\frac{a+c}{2}}\pm \sqrt{\frac{a-c}{2}}$

Worked problems (§682–688)

Problem	Key step	Answer
$xy=105$, $x^2+y^2=274$	Quartic → $a=137,c=105$	$x=15,y=7$
$xy=m$, $x^2+y^2=n$	Add/subtract $2xy$	$x,y=\frac{1}{2}\sqrt{n+2m}\pm \frac{1}{2}\sqrt{n-2m}$
$xy=35$, $x^2-y^2=24$	Quartic $z^2=24z+1225$	$x=7,y=5$
Sum = product = diff of squares	$y^2=y+1$	$y=\frac{1+\sqrt{5}}{2}$, $x=\frac{3+\sqrt{5}}{2}$
Sum = product = sum of squares	$y^2=3y-3$	$x=\frac{3-\sqrt{-3}}{2}$, $y=\frac{3+\sqrt{-3}}{2}$

The last two examples connect to the golden ratio and complex numbers. Problem 5 is also solved by the $p+q$/$p-q$ substitution trick (§688), which simplifies the algebra significantly.

Related pages

• square-roots-of-binomials
• quadratic-equations
• ch1.4.9-nature-quadratic-equations
• imaginary-numbers
• irrational-numbers
• square-roots-and-irrational-numbers
• algebraic-identities