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Q5(iv):
Rationalise the denominators of the following:
(iv) $\frac{1}{\sqrt{7} - 2}$
Solution :
Initial Setup & Given Expression
We are tasked with rationalizing the denominator of the following irrational fraction:
$E = \frac{1}{\sqrt{7} - 2}$
Objective: To rationalize the denominator means to transform the expression such that the denominator contains only rational numbers, without altering the overall mathematical value of the expression.
Step 1: Identifying the Rationalizing Factor (The Conjugate)
The denominator is a binomial of the form $a - b$, specifically $\sqrt{7} - 2$. To eliminate the square root, we utilize the algebraic identity for the difference of squares:
$(a - b)(a + b) = a^2 - b^2$
[Per the Difference of Squares Theorem], multiplying the binomial by its conjugate will square both terms, thereby neutralizing the square root. The conjugate of $\sqrt{7} - 2$ is obtained by changing the sign between the terms, resulting in $\sqrt{7} + 2$.
Step 2: Multiplying by the Multiplicative Identity
To maintain the equivalence of the fraction, we must multiply both the numerator and the denominator by the conjugate. This is equivalent to multiplying the entire expression by $1$ [Per the Multiplicative Identity Property, $\frac{x}{x} = 1$].
$E = \frac{1}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2}$
Step 3: Expanding the Numerator and Denominator
Now, we perform the algebraic multiplication for both the top and bottom of the fraction.
- Numerator: $1 \cdot (\sqrt{7} + 2) = \sqrt{7} + 2$
- Denominator: $(\sqrt{7} - 2)(\sqrt{7} + 2)$
Applying the difference of squares identity $(a-b)(a+b) = a^2 - b^2$ to the denominator, where $a = \sqrt{7}$ and $b = 2$:
$\text{Denominator} = (\sqrt{7})^2 - (2)^2$
Step 4: Simplifying the Expression
Evaluate the squares in the denominator:
- $(\sqrt{7})^2 = 7$ [Since squaring a square root yields the original radicand]
- $(2)^2 = 4$
Subtract the squared values:
$\text{Denominator} = 7 - 4 = 3$
Substitute the simplified denominator back into the fraction:
$E = \frac{\sqrt{7} + 2}{3}$
The denominator is now $3$, which is a rational integer. The rationalization process is complete.
Final Solution: The rationalized form of the expression is $\frac{\sqrt{7} + 2}{3}$.
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(iii): Rationalise the denominators of the following: (iii) $\frac{1}{\sqrt{5} + \sqrt{2}}$
CBSE Solutions for Class 9 Mathematics Number Systems
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