Extra Questions For Class 9 Maths Chapter 1 | Q1

Question 1: Rationalize the denominator of \\frac{14}{{\sqrt {108}} - {\sqrt {96}} + {\sqrt {192}} - {\sqrt {54}}}\\)

Video Explanation

Explanatory Answer | Chapter 1 Extra Question 1

Step 1: Prime factorize the numbers under the root in the denominator

This step will ensure that we break the expression into a more manageable one so that rationalisation will be easier.
\\frac{14} {{\sqrt {2^2 \times 3^3}} - {\sqrt {2^5 \times 3}} + {\sqrt {2^6 \times 3}} - {\sqrt {2 \times 3^3}}}\\)

Let us break the powers of prime numbers into product of even powers and whatever is left, wherever possible, and rewrite as follows
\\frac{14} {{\sqrt {2^2 \times 3^2 \times 3}} - {\sqrt {2^4 \times 2 \times 3}} + {\sqrt {2^6 \times 3}} - {\sqrt {2 \times 3^2 \times 3}}}\\)

Step 2: Take all the even prime powers out of the root

= \\frac{14} {6{\sqrt {3}} - 4{\sqrt {6}} + 8{\sqrt{3}} - 3{\sqrt{6}}}\\)
= \\frac{14} {14{\sqrt {3}} - 7{\sqrt {6}}}\\)
= \\frac{2} {2{\sqrt {3}} - {\sqrt {6}}}\\)
This expression is definitely much easier to rationalise than the original one presented to us.

Step 3: Rationalise the denominator

= \\frac{2} {2{\sqrt {3}} - {\sqrt {6}}}\\) x \\frac{2{\sqrt {3}} + {\sqrt {6}}} {2{\sqrt {3}} + {\sqrt {6}}}\\)
= \\frac{4{\sqrt {3}} + 2{\sqrt {6}}} {12 - 6}\\)
= \\frac{4{\sqrt {3}} + 2{\sqrt {6}}} {6}\\)
= \\frac{2{\sqrt {3}} + {\sqrt {6}}} {3}\\)