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.**ðŸ‘‰ This post extends our posts on surds ðŸ‘ˆ

**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.

If you got questions, feel free to tell me in the comment box.

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