NCERT Solutions for Class 9 Maths Exercise 1.4 Question 5
Understanding the Question 🧐
In this problem, we need to rationalise the denominator of several fractions. This means we have to rewrite each fraction so that there are no irrational numbers (like square roots) left in the denominator. This is a standard technique that makes expressions simpler and easier to work with.
Rationalise the denominators of the following:
(i) &&\frac{1}{\sqrt{7}}&&
(ii) &&\frac{1}{\sqrt{7} – \sqrt{6}}&&
(iii) &&\frac{1}{\sqrt{5} + \sqrt{2}}&&
(iv) &&\frac{1}{\sqrt{7} – 2}&&
Part (i): Rationalise &&\frac{1}{\sqrt{7}}&& 📝
Step 1: Identify the rationalizing factor.
The denominator is &&\sqrt{7}&&. To make it a rational number, we need to multiply it by &&\sqrt{7}&&, since &&\sqrt{7} \times \sqrt{7} = 7&&.
Step 2: Multiply the numerator and denominator by the factor.
To keep the value of the fraction the same, we must multiply both the top and bottom by &&\sqrt{7}&&.
&&\Rightarrow \frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}&&
&&\Rightarrow \frac{1 \times \sqrt{7}}{(\sqrt{7})^2} = \frac{\sqrt{7}}{7}&&
Final Answer: The rationalised form is &&\frac{\sqrt{7}}{7}&&.
Part (ii): Rationalise &&\frac{1}{\sqrt{7} – \sqrt{6}}&& 📝
Step 1: Find the conjugate of the denominator.
The denominator is &&\sqrt{7} – \sqrt{6}&&. Its conjugate is found by changing the sign in the middle. So, the conjugate is &&\sqrt{7} + \sqrt{6}&&. We use the conjugate because of the identity &&(a-b)(a+b) = a^2 – b^2&&, which will eliminate the square roots.
Step 2: Multiply the numerator and denominator by the conjugate.
&&\Rightarrow \frac{1}{\sqrt{7} – \sqrt{6}} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}}&&
&&\Rightarrow \frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 – (\sqrt{6})^2}&&
&&\Rightarrow \frac{\sqrt{7} + \sqrt{6}}{7 – 6} = \frac{\sqrt{7} + \sqrt{6}}{1}&&
Final Answer: The rationalised form is &&\sqrt{7} + \sqrt{6}&&.
Part (iii): Rationalise &&\frac{1}{\sqrt{5} + \sqrt{2}}&& 📝
Step 1: Find the conjugate of the denominator.
The denominator is &&\sqrt{5} + \sqrt{2}&&. Its conjugate is &&\sqrt{5} – \sqrt{2}&&.
Step 2: Multiply the numerator and denominator by the conjugate.
&&\Rightarrow \frac{1}{\sqrt{5} + \sqrt{2}} \times \frac{\sqrt{5} – \sqrt{2}}{\sqrt{5} – \sqrt{2}}&&
&&\Rightarrow \frac{\sqrt{5} – \sqrt{2}}{(\sqrt{5})^2 – (\sqrt{2})^2}&&
&&\Rightarrow \frac{\sqrt{5} – \sqrt{2}}{5 – 2} = \frac{\sqrt{5} – \sqrt{2}}{3}&&
Final Answer: The rationalised form is &&\frac{\sqrt{5} – \sqrt{2}}{3}&&.
Part (iv): Rationalise &&\frac{1}{\sqrt{7} – 2}&& 📝
Step 1: Find the conjugate of the denominator.
The denominator is &&\sqrt{7} – 2&&. Its conjugate is &&\sqrt{7} + 2&&.
Step 2: Multiply the numerator and denominator by the conjugate.
&&\Rightarrow \frac{1}{\sqrt{7} – 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2}&&
&&\Rightarrow \frac{\sqrt{7} + 2}{(\sqrt{7})^2 – (2)^2}&&
&&\Rightarrow \frac{\sqrt{7} + 2}{7 – 4} = \frac{\sqrt{7} + 2}{3}&&
Final Answer: The rationalised form is &&\frac{\sqrt{7} + 2}{3}&&.
Conclusion and Key Points ✅
Rationalizing the denominator is a fundamental skill. The method depends on the form of the denominator:
- For a single term like &&\frac{1}{\sqrt{a}}&&, multiply by &&\frac{\sqrt{a}}{\sqrt{a}}&&.
- For a binomial term like &&\frac{1}{\sqrt{a} \pm \sqrt{b}}&&, multiply by its conjugate &&\frac{\sqrt{a} \mp \sqrt{b}}{\sqrt{a} \mp \sqrt{b}}&& to use the identity &&(x-y)(x+y) = x^2 – y^2&&.
- The goal is to make the denominator a rational number (an integer).
- Always multiply both the numerator and the denominator by the same factor to not change the fraction’s value.
- The identity &&(a – b)(a + b) = a^2 – b^2&& is essential for rationalizing binomial denominators.
FAQ ❓
Q: What does it mean to “rationalize the denominator”?
A: It’s the process of removing any irrational numbers (like square roots) from the denominator of a fraction, turning it into a rational number (usually a whole number).
Q: What is the conjugate of an expression like &&(\sqrt{a} – b)&&?
A: The conjugate is found by simply changing the sign between the two terms. The conjugate of &&(\sqrt{a} – b)&& is &&(\sqrt{a} + b)&&.
Q: What identity is used when rationalizing a binomial denominator?
A: We use the difference of squares identity: &&(x – y)(x + y) = x^2 – y^2&&. This identity is perfect for eliminating square roots, since &&(\sqrt{a})^2 = a&&.
Further Reading 📖
To practice more problems on real numbers, you can refer to the official NCERT textbook for Class 9 Maths. More resources are available on the NCERT website at https://ncert.nic.in/.
