UPSKILL MATH PLUS

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Let $$a > 0$$ be a real number and $$n$$ be a positive integer. Then $\sqrt[n]{a}=b$, if ${b}^{n}=a$ and $$b > 0$$.
Important!
The symbol ‘$\sqrt{}$ ’ used is called the radical sign.
Let $$a$$ and $$b$$ be positive real numbers.

$\begin{array}{l}i\right)\phantom{\rule{0.147em}{0ex}}\sqrt{\mathit{ab}}=\sqrt{a}×\sqrt{b}\\ \\ \mathit{ii}\right)\phantom{\rule{0.147em}{0ex}}\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\\ \\ \mathit{iii}\right)\phantom{\rule{0.147em}{0ex}}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b\\ \\ \mathit{iv}\right)\phantom{\rule{0.147em}{0ex}}\left(a-\sqrt{b}\right)\left(a+\sqrt{b}\right)={a}^{2}-b\\ \\ v\right)\phantom{\rule{0.147em}{0ex}}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{c}+\sqrt{d}\right)=\sqrt{\mathit{ac}}+\sqrt{\mathit{ad}}+\sqrt{\mathit{bc}}+\sqrt{\mathit{bd}}\\ \\ \mathit{vi}\right){\phantom{\rule{0.147em}{0ex}}\left(\sqrt{a}+\sqrt{b}\right)}^{2}=a+2\sqrt{\mathit{ab}}+b\end{array}$
The process of converting the expression in the denominator contains a term with a square root (or a number under a radical sign), to an equivalent expression whose denominator is a rational number is called rationalising the denominator.
To be in a simpler form, the denominator of the expression should not have an irrational number.
To rationalise the denominator containing single term with only one irrational number, multiply and divide by the same irrational number and simplify it.
Example:
Let us rationalise $\frac{1}{\sqrt{2}}$.

Here the denominator containing single irrational number $\sqrt{2}$.

Multiply and divide by the irrational number $\sqrt{2}$.

$=\frac{1}{\sqrt{2}}×\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$.
To rationalise the denominator containing two terms:

Step 1: Find the conjugate of the denominator. Need to change the sign between the two terms.
Step 2: Multiply both the numerator and the denominator by the conjugate of the denominator.
Step 3: Multiply both the numerator and the denominator.
Step 4: Combine like term and simplify the radicals.
Step 5: Reduce the fraction as much as possible.
Important!
To rationalise the denominator of $\frac{1}{a+\sqrt{b}}$, multiply and divide this by $a-\sqrt{b}$. That is, $\frac{1}{a+\sqrt{b}}×\frac{a-\sqrt{b}}{a-\sqrt{b}}=\frac{a-\sqrt{b}}{{a}^{2}-b}$, where $$a$$ and $$b$$ are integers.
Example:
Consider the expression $\frac{2}{\sqrt{7}-4}$.
Here the denominator containing two terms. Let us follow the procedure to rationalise the expression.

Step 1: The conjugate of $\sqrt{7}-4$ is $\sqrt{7}+4$.

Step 2: Multiply both the numerator and the denominator by the conjugate of the denominator.

$\frac{2}{\sqrt{7}-4}×\frac{\sqrt{7}+4}{\sqrt{7}+4}$

Step 3: Multiply both the numerator and the denominator.

$\frac{2}{\sqrt{7}-4}×\frac{\sqrt{7}+4}{\sqrt{7}+4}=\frac{2\sqrt{7}+8}{7-{4}^{2}}$

Step 4: Combining like term and simplify the radicals.

$\frac{2\sqrt{7}+8}{7-{4}^{2}}=-\frac{2\left(\sqrt{7}+4\right)}{9}$

Step 5: As it is already in reduced form, we cannot reduce further.

Thus, the required rationalised form is $-\frac{2\left(\sqrt{7}+4\right)}{9}$.