Square Roots of Binomials

Summary: A binomial surd $a\pm \sqrt{b}$ has a simplified square root $\sqrt{(a+c)/2}\pm \sqrt{(a-c)/2}$ (where $c=\sqrt{a^2-b}$) whenever $a^2-b$ is a perfect square; otherwise the extraction cannot be simplified further.

Sources: chapter-1.4.8

Last updated: 2026-05-03

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Setup

A binomial in Euler’s sense is a two-term expression in which at least one term contains a square root, e.g. $3+\sqrt{5}$ or $\sqrt{8}-\sqrt{3}$ (source: chapter-1.4.8, §669). Such binomials arise as solutions to quadratic equations and appear throughout the theory of equations.

The root formula

To find $\sqrt{a+\sqrt{b}}$, write the root as $\sqrt{x}+\sqrt{y}$ and square:

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

This gives $x+y=a$ and $4xy=b$. Subtracting four times the product equation from the square of the sum equation:

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

Hence:

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

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

Condition: the simplification is possible only when $a^2-b=c^2$ is a perfect square. Otherwise, $\sqrt{a+\sqrt{b}}$ cannot be written more simply. (§675–676)

Practical rule (§677)

Subtract the square of the irrational part from the square of the rational part. If the remainder is a perfect square $c^2$, the square root of $a\pm \sqrt{b}$ is $\sqrt{(a+c)/2}\pm \sqrt{(a-c)/2}$.

Key examples

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

Extension to imaginary binomials (§679)

The formula works unchanged when $b<0$ (imaginary surds). For instance, $\sqrt{2\sqrt{-1}}$ has $a=0$, $b=-4$, $a^2-b=4$, $c=2$, giving root $1+\sqrt{-1}$ — verified by squaring: $(1+\sqrt{-1}{)}^2=2\sqrt{-1}$.

Connection to quartic equations (§681)

The equation $x^4=2a{x}^2-c^2$ substitutes $y=x^2$ to get $y=a\pm \sqrt{a^2-c^2}$, then requires extracting $\sqrt{a\pm \sqrt{a^2-c^2}}$ — precisely this binomial-root problem. When $a^2-(a^2-c^2)=c^2$ is a perfect square, the answer is:

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

Related pages

• ch1.4.8-extraction-square-roots-binomials
• quadratic-equations
• irrational-numbers
• imaginary-numbers
• square-roots-and-irrational-numbers
• algebraic-identities