Question

The number obtained on rationalizing the denominator of 1/(√7--2) is

Answer 1

Given that:

1/(√7 - 2)

The denominator is (√7-2).

We know

The rationalising factor of (√a-b) is (√a+b)

Therefore, the rationalising factor of √7-2 is √7+2.

On rationalising the denominator them

⇛[1/(√7-2)]×[(√7+2)/(√7+2)]

⇛[1(√7+2)]/[(√7-2)(√7+2)]

Since, (a-b)(a+b) = a²-b²

Where, a = √7 and b = √2.

⇛[1(√7+2)]/[(√7)²-(2)²]

⇛[1(√7+2)]/[(√7*7)-(2*2)]

⇛[1(√7+2)]/[7-4]

⇛[1(√7+2)]/3

⇛(√7+2)/3

Hence, the denominator is rationalised.

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Answer 2

`(1)/( √(7) -(- 2)) \\ rationalizing \: \: \: the \: \: \: denominator \: \: \: we \: \: \: have \\ = (1)/( √(7) + 2 ) * ( √(7) - 2)/( √(7) - 2) \\ = \frac{ √(7) - 2}{ {( √(7) )}^(2) - {(2)}^(2) } \\ = ( √(7) - 2 )/(7 - 4) \\ = ( √(7) - 2)/(3)`

Hope you could get an idea from here.

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