Q5(ii):

Rationalise the denominators of the following: (ii) $\frac{1}{\sqrt{7} - \sqrt{6}}$

Initial Setup & Mathematical Objective

We are given the following fractional expression featuring an irrational denominator:

$ E = \frac{1}{\sqrt{7} - \sqrt{6}} $

The mathematical objective is to rationalize the denominator. Rationalization is the process of eliminating irrational numbers (such as surds or roots) from the denominator of an algebraic fraction. This is achieved by multiplying the numerator and the denominator by a suitable rationalizing factor, ensuring the fraction's overall value remains unchanged [Per the Multiplicative Identity Property].

Step 1: Identifying the Conjugate Surd

To eliminate the square roots in a binomial denominator of the form $a - b$, we utilize its conjugate. The conjugate of a binomial $a - b$ is $a + b$.

• The given denominator is: $\sqrt{7} - \sqrt{6}$
• The conjugate of this denominator is: $\sqrt{7} + \sqrt{6}$

Multiplying a binomial surd by its conjugate leverages the difference of squares identity, which squares both terms and effectively removes the square root radicals.

Step 2: Multiplying Numerator and Denominator by the Conjugate

We multiply both the numerator and the denominator of the fraction by the conjugate $\sqrt{7} + \sqrt{6}$. This is equivalent to multiplying the entire expression by $1$, thus preserving its fundamental value.

$ E = \frac{1}{\sqrt{7} - \sqrt{6}} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} $

Step 3: Applying Algebraic Identities

Combine the numerators and the denominators:

$ E = \frac{1 \cdot (\sqrt{7} + \sqrt{6})}{(\sqrt{7} - \sqrt{6})(\sqrt{7} + \sqrt{6})} $

For the denominator, we apply the foundational algebraic identity for the difference of two squares [$(a - b)(a + b) = a^2 - b^2$].

• Let $a = \sqrt{7}$
• Let $b = \sqrt{6}$

Substituting these into the identity yields:

$ (\sqrt{7} - \sqrt{6})(\sqrt{7} + \sqrt{6}) = (\sqrt{7})^2 - (\sqrt{6})^2 $

Step 4: Simplifying the Expression

Now, we evaluate the squares in the denominator. By definition, the square of a square root returns the original radicand [$(\sqrt{x})^2 = x$ for $x \ge 0$].

• $(\sqrt{7})^2 = 7$
• $(\sqrt{6})^2 = 6$

Substitute these values back into the denominator:

$ \text{Denominator} = 7 - 6 = 1 $

The numerator remains unchanged as it was multiplied by $1$:

$ \text{Numerator} = \sqrt{7} + \sqrt{6} $

Reassembling the fraction gives:

$ E = \frac{\sqrt{7} + \sqrt{6}}{1} $

Since any number divided by $1$ is the number itself, the expression simplifies completely to a linear binomial surd.

Final Solution: \sqrt{7} + \sqrt{6}