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Q5(iii):
Rationalise the denominators of the following:
(iii) $\frac{1}{\sqrt{5} + \sqrt{2}}$
Solution :
Initial Setup & Objective
We are given the following fractional expression containing irrational terms in the denominator:
$ \frac{1}{\sqrt{5} + \sqrt{2}} $
The mathematical objective is to rationalise the denominator. Rationalisation is the process of eliminating irrational numbers (such as surds or square roots) from the denominator of an algebraic fraction. This is achieved by multiplying the numerator and the denominator by a carefully chosen multiplier known as the conjugate.
Step 1: Identifying the Conjugate (Rationalising Factor)
To eliminate the square roots in a binomial denominator of the form $\sqrt{a} + \sqrt{b}$, we utilize its conjugate pair, which is $\sqrt{a} - \sqrt{b}$. [Per the fundamental properties of surds, multiplying a binomial by its conjugate leverages the difference of squares identity to yield a rational integer].
For our specific denominator, $\sqrt{5} + \sqrt{2}$, the conjugate is:
$ \sqrt{5} - \sqrt{2} $
| Original Binomial Denominator | Conjugate (Rationalising Factor) | Algebraic Product (Rational Result) |
|---|---|---|
| $\sqrt{a} + \sqrt{b}$ | $\sqrt{a} - \sqrt{b}$ | $(\sqrt{a})^2 - (\sqrt{b})^2 = a - b$ |
| $\sqrt{a} - \sqrt{b}$ | $\sqrt{a} + \sqrt{b}$ | $(\sqrt{a})^2 - (\sqrt{b})^2 = a - b$ |
| $a + \sqrt{b}$ | $a - \sqrt{b}$ | $a^2 - (\sqrt{b})^2 = a^2 - b$ |
Step 2: Multiplying Numerator and Denominator
To maintain the equivalence of the fraction [Per the Multiplicative Identity Property, multiplying by $1$ does not change the value], we multiply both the numerator and the denominator by the conjugate $\sqrt{5} - \sqrt{2}$:
$ \frac{1}{\sqrt{5} + \sqrt{2}} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} $
Step 3: Applying the Difference of Squares Identity
We now expand the numerator and the denominator. The numerator is simply multiplied by $1$:
$ \text{Numerator} = 1 \times (\sqrt{5} - \sqrt{2}) = \sqrt{5} - \sqrt{2} $
For the denominator, we apply the algebraic identity for the difference of squares: $(x + y)(x - y) = x^2 - y^2$.
Let $x = \sqrt{5}$ and $y = \sqrt{2}$:
$ \text{Denominator} = (\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) = (\sqrt{5})^2 - (\sqrt{2})^2 $
Step 4: Evaluating the Squares and Final Simplification
Evaluating the squares in the denominator [Per the definition of a square root, $(\sqrt{n})^2 = n$ for any non-negative real number $n$]:
- $(\sqrt{5})^2 = 5$
- $(\sqrt{2})^2 = 2$
Substitute these values back into the denominator's expression:
$ \text{Denominator} = 5 - 2 = 3 $
Now, recombine the evaluated numerator and the rationalised denominator to form the final simplified fraction:
$ \frac{\sqrt{5} - \sqrt{2}}{3} $
The denominator is now $3$, which is a rational integer. The rationalisation process is complete.
Final Solution: $ \frac{\sqrt{5} - \sqrt{2}}{3} $
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(ii): Rationalise the denominators of the following: (ii) $\frac{1}{\sqrt{7} - \sqrt{6}}$
- 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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