Extra Questions For Class 9 Maths Chapter 1 | Q1

Number System | Prime Factorize & Rationalise the Denominator

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

NCERT Solution to Class 10 Maths

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

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