# BINOMIAL SURD

A binomial surd is the sum of two roots of rational numbers, one of which is irrational.

For example, $1+\sqrt{3}$, $2+\sqrt{5}$ are binomial surd because they contain one irrational number and one rational number

To rationalize a binomial surd, we use the idea of the difference of two squares.

The difference of squares of two surds is the product of their sum and their difference. This is illustrated below

$(\sqrt{x}+y)(\sqrt{x}-y)=(\sqrt{x})^2-(y)^2$

So this means to rationalize $\sqrt{x}+y$, we multiply it by $\sqrt{x}-y$ so that we get $(\sqrt{x})^2-(y)^2$ which is a rational number.

Therefore $\sqrt{x}+y$ and $\sqrt{x}-y$ are said to be conjugates of one another.

To simplify a fraction that has a binomial sure in the denominator, rationalize the denominator by multiplying it by its conjugate.

👉 This post extends our posts on surds 👈

Example 1
Write $\frac{12}{2\sqrt{2}-1}$ in the lowest possible form.

Solution
Multiplying the numerator and the denominator by the conjugate of the denominator
$\frac{12}{2\sqrt{2}-1}\times\frac{2\sqrt{2}+1}{2\sqrt{2}+1}$

By expansion
$\frac{12(2\sqrt{2}+1)}{(2\sqrt{2})^2-(1)^2}$
Note that the denominator is a difference of two squares.
$\frac{24\sqrt{2}+12}{8-1}$
$\frac{24\sqrt{2}+12}{7}$

Example 2
Simplify $\frac{1+\sqrt{7}}{3-\sqrt{7}}$

Solution
Multiplying $3+\sqrt{7}$
$\frac{1+\sqrt{7}}{2-\sqrt{7}}\times\frac{3+\sqrt{7}}{3+\sqrt{7}}$

By expansion
$\frac{1(3+\sqrt{7})+\sqrt{7}(3+\sqrt{7})}{(3)^2-(\sqrt{7})^2}$
$\frac{3+\sqrt{7}+3\sqrt{7}+7}{9-7}$
$\frac{3+4\sqrt{7}+7}{2}$
$\frac{10+4\sqrt{7}}{2}$

$2$ is common in the numerator and denominator.
$\frac{2(5+2\sqrt{7})}{2}$
$2$ divided by $2$ is $1$, hence
$5+2\sqrt{7}$

Example 3
Write $\frac{3-\sqrt{3}}{3+\sqrt{3}}$

Solution
Rationalizing the denominator
$\frac{3-\sqrt{3}}{3+\sqrt{3}}\times\frac{3-\sqrt{3}}{3-\sqrt{3}}$

By expansion
$\frac{3(3-\sqrt{3})-\sqrt{3}(3-\sqrt{3})}{(3)^2-(\sqrt{3})^2)}$
$\frac{9-3\sqrt{3}-3\sqrt{3}+\sqrt{9}}{9-3}$
$\frac{9-3\sqrt{3}-3\sqrt{3}+3}{9-3}$
$\frac{12-6\sqrt{3}}{6}$

$6$ is common, so we factorize
$\frac{6(2-\sqrt{3}}{6}$

$6$ divided by $6$ is one, therefore,
$2-\sqrt{3}$

Example 4
Write $\frac{\sqrt{2}}{1+\sqrt{2}}$ in the simple form possible.

Solution
By rationalizing the denominator
$\frac{\sqrt{2}}{1+\sqrt{2}}\times\frac{1-\sqrt{2}}{1-\sqrt{2}}$

By expansion
$\sqrt{2}(1-\sqrt{2})((1)^2-(\sqrt{2})^2)$
$\frac{\sqrt{2}-\sqrt{4}}{1-2}$
$\frac{\sqrt{2}-2}{-1}$
$-\sqrt{2}+2$
$2-\sqrt{2}$

Related post
Example 5
Write $\frac{98}{(3+\sqrt{2})^2}$ in the simplest form possible.

Solution:
$\frac{98}{(3+\sqrt{2})(3+\sqrt{2})}$

Expanding the denominator
$\frac{98}{3(3+\sqrt{2})+\sqrt{2}(3+\sqrt{2})}$
$\frac{98}{9+3\sqrt{2}+3\sqrt{2}+2}$

Rationalizing the denominator
$\frac{98}{9+3\sqrt{2}+3\sqrt{2}+2}\times\frac{11-6\sqrt{2}}{11-6\sqrt{2}}$

By expansion
$\frac{98(11-6\sqrt{2})}{(11)^2-(6\sqrt{2})^2}$
$\frac{1078-588\sqrt{2})}{121-36(2)}$

Note, square cancel square root.
$\frac{1078-588\sqrt{2})}{121-72}$
$\frac{1078-588\sqrt{2})}{49}$
$\frac{49(22-12\sqrt{2})}{49}$
$22-12\sqrt{2}$

We would conclude our series on surds when we address some hards questions on surds.

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