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Rationalise the denominator of the following: 3⁢√5 − 4⁢√2/3⁢√5 + 4⁢√2 - Mathematics

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Question

Rationalise the denominator of the following:

`(3sqrt(5) - 4sqrt(2))/(3sqrt(5) + 4sqrt(2))`

Sum
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Solution

To rationalise the denominator of the expression:

`(3sqrt(5) - 4sqrt(2))/(3sqrt(5) + 4sqrt(2))`

Step 1: Multiply numerator and denominator by the conjugate of the denominator

The conjugate of `3sqrt(5) + 4sqrt(2)` is `3sqrt(5) - 4sqrt(2)`.

Multiplying by the conjugate eliminates the square roots in the denominator.

`(3sqrt(5) - 4sqrt(2))/(3sqrt(5) + 4sqrt(2)) xx (3sqrt(5) - 4sqrt(2))/(3sqrt(5) - 4sqrt(2))`

= `(3sqrt(5) - 4sqrt(2))^2/((3sqrt(5))^2 - (4sqrt(2))^2`

Step 2: Simplify the denominator using the difference of squares

The denominator becomes:

`(3sqrt(5))^2 - (4sqrt(2))^2`

= 9 × 5 – 16 × 2

= 45 – 32

= 13

Step 3: Expand and simplify the numerator

The numerator is:

`(3sqrt(5) - 4sqrt(2))^2`

= `(3sqrt(5))^2 - 2 xx 3sqrt(5) xx 4sqrt(2) + (4sqrt(2))^2`

Calculating each term:

`(3sqrt(5))^2 = 9 xx 5 = 45`

`(4sqrt(2))^2 = 16 xx 2 = 32`

`-2 xx 3sqrt(5) xx 4sqrt(2) = -24sqrt(10)`

So, the numerator becomes:

`45 - 24sqrt(10) + 32 = 77 - 24sqrt(10)`

Step 4: Combine the results

The rationalized expression is:

`(77 - 24sqrt(10))/13`

This is the simplified form of the original expression with a rationalized denominator.

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Chapter 1: Rational and Irrational Numbers - Exercise 1E [Page 32]

APPEARS IN

Nootan Mathematics [English] Class 9 ICSE
Chapter 1 Rational and Irrational Numbers
Exercise 1E | Q 2. (ii) | Page 32
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