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Simplify the following: (sqrt(7) + sqrt(6))/(sqrt(7) – sqrt(6)) + (sqrt(7) – sqrt(6))/(sqrt(7) + sqrt(6)) - Mathematics

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Question

Simplify the following:

`(sqrt(7) + sqrt(6))/(sqrt(7) - sqrt(6)) + (sqrt(7) - sqrt(6))/(sqrt(7) + sqrt(6))`

Simplify
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Solution

We are asked to simplify:

`(sqrt(7) + sqrt(6))/(sqrt(7) - sqrt(6)) + (sqrt(7) - sqrt(6))/(sqrt(7) + sqrt(6))`

Step 1: Rationalise both terms

We’ll rationalise the denominators of both fractions by multiplying the numerator and denominator of each fraction by the conjugate of its denominator.

First fraction: `(sqrt(7) + sqrt(6))/(sqrt(7) - sqrt(6))`

Multiply the numerator and denominator by the conjugate of the denominator `(sqrt(7) + sqrt(6))`:

`(sqrt(7) + sqrt(6))/(sqrt(7) - sqrt(6)) xx (sqrt(7) + sqrt(6))/(sqrt(7) + sqrt(6))`

= `(sqrt(7) + sqrt(6))^2/((sqrt(7))^2 - (sqrt(6))^2`

Simplify the denominator:

`(sqrt(7))^2 - (sqrt(6))^2`

= 7 – 6

= 1

Now expand the numerator:

`(sqrt(7) + sqrt(6))^2`

= `(sqrt(7))^2 + 2sqrt(7)sqrt(6) + (sqrt(6))^2`

= `7 + 2sqrt(42) + 6`

= `13 + 2sqrt(42)`

So, the first fraction becomes:

`(13 + 2sqrt(42))/1 = 13 + 2sqrt(42)`

Second fraction: `(sqrt(7) - sqrt(6))/(sqrt(7) + sqrt(6))`

Multiply the numerator and denominator by the conjugate of the denominator `(sqrt(7) - sqrt(6))`:

`(sqrt(7) - sqrt(6))/(sqrt(7) + sqrt(6)) xx (sqrt(7) - sqrt(6))/(sqrt(7) - sqrt(6))`

= `(sqrt(7) - sqrt(6))^2/((sqrt(7))^2 - (sqrt(6))^2`

Simplify the denominator (same as before):

`(sqrt(7))^2 - (sqrt(6))^2`

= 7 – 6

= 1

Now expand the numerator:

`(sqrt(7)) - sqrt(6))^2`

= `(sqrt(7))^2 - 2sqrt(7)sqrt(6) + (sqrt(6))^2`

= `7 - 2sqrt(42) + 6`

= `13 - 2sqrt(42)`

So, the second fraction becomes:

`(13 - 2sqrt(42))/1 = 13 - 2sqrt(42)`

Step 2: Add the two fractions

Now add the two simplified fractions:

`13 + 2sqrt(42) + 13 - 2sqrt(42)`

The `2sqrt(42)` terms cancel out:

13 + 13 = 26

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Chapter 1: Rational and Irrational Numbers - EXERCISE 1C [Page 15]

APPEARS IN

B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 1 Rational and Irrational Numbers
EXERCISE 1C | Q 9. (i) | Page 15
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