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Q5(ii):
Rationalise the denominators of the following:
(ii) $\frac{1}{\sqrt{7} - \sqrt{6}}$
Solution :
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}
More Questions from Class 9 Mathematics Number Systems EXERCISE 1.4
- Q1(i): Classify the following numbers as rational or irrational: (i) $2 - \sqrt{5}$
- Q1(ii): Classify the following numbers as rational or irrational: (ii) $(3 + \sqrt{23}) - \sqrt{23}$
- Q1(iii): Classify the following numbers as rational or irrational: (iii) $\frac{2\sqrt{7}}{7\sqrt{7}}$
- Q1(iv): Classify the following numbers as rational or irrational: (iv) $\frac{1}{\sqrt{2}}$
- Q1(v): Classify the following numbers as rational or irrational: (v) $2\pi$
- Q2(i): Simplify each of the following expressions: (i) $(3 + \sqrt{3})(2 + \sqrt{2})$
- Q2(ii): Simplify each of the following expressions: (ii) $(3 + \sqrt{3})(3 - \sqrt{3})$
- Q2(iii): Simplify each of the following expressions: (iii) $(\sqrt{5} + \sqrt{2})^2$
- Q2(iv): Simplify each of the following expressions: (iv) $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$
- Q3: Recall, $\pi$ is defined as the ratio of the circumference (say $c$) of a circle to its diameter (say $d$). That is, $\pi = \frac{c}{d}$. This seems to contradict the fact that $\pi$ is irrational. How will you resolve this contradiction?
- Q4: Represent $\sqrt{9.3}$ on the number line.
- Q5(i): Rationalise the denominators of the following: (i) $\frac{1}{\sqrt{7}}$
- Q5(iii): Rationalise the denominators of the following: (iii) $\frac{1}{\sqrt{5} + \sqrt{2}}$
- Q5(iv): Rationalise the denominators of the following: (iv) $\frac{1}{\sqrt{7} - 2}$
CBSE Solutions for Class 9 Mathematics Number Systems
Chapters in CBSE - Class 9 Mathematics
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