Rationalisation

(i)

=

=

(ii)

=

=

(i)

=

=

(ii)

=

=

=

(iii)

=

=

=

(i)

Because

= 121 - 11 = 110

(ii)

= 25 – 7 = 18

(iii)

= 8 – 2 = 6

(iv)

= 9 – 3 = 6

(v)

= 5 – 2 = 3

(i)

Because:

=

=

(ii)

(iii)

=

=

=

=

(i) As there is √5 in the denominator and we know that √5 x √5 = 5

So, multiply numerator and denominator by √5,

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(i) Given that √2 = 1.414, √3=1.732, √5 = 2.236 and √10= 3.162

So we have,

Rationalising factor of denominator is √3

=1.15466667 = 1.54

(ii) we have rationalisation factor of denominator is √10

(iii) we have rationalisation factor of denominator is √2

(iv) we have rationalisation factor of denominator is √2

=

=

(v) We have

=

(vi) We have rationalising factor of denominator is √5

(i) we have rationalizing factor of the denominator is 3-√2

(ii) we have rationalizing factor of the denominator is

=

(iii) we have rationalizing factor of the denominator is

=

=

=

(iv) we have to rationalize factor of

=

=

=

=

(v) we have to rationalize factor of

(vi) we have to rationalize factor of

(vii) we have to rationalize factor of

=

Because;

(viii) we have to rationalize factor of

(ix) we have to rationalize factor of

i)

ii)

iii)

iv)

v)

vi)

i)

ii)

iii) rationalising factors of denominators are

(iv)

Rationalising factor for

For and

For

(v)

Rationalising factors for denominators are,

For

For and

For

(i)

Given,

Rationalising factor for denominator is

On equating rational and irrational parts,

We get a = 2 and b = 1

(ii) rationalising factor for the denominator is

We have

On equating rational and irrational parts we get,

a=3 and b=2

(iii)

Rationalising factor for the denominator is

On equating rational and irrational parts we get,

(iv) given,

Rationalising factor for denominator is

We have

On equating rational and irrational parts we get,

a= -1 and b=1

(v) given,

We have

On equating rational and irrational parts we get

(vi) given,

We have

On equating rational and irrational parts we have,

Given and given to find the value of

We have

We know that ,

By putting we get

The value of is 52.

Given that

And given to find the value of

We have

The rationalising factor for denominator is

=

=

Also,

We know that,

= 36 - 2 = 34

The value of is 34.

Rationalising factor for the denominator is

We have

= 3(3.968)

= 11.904

(i) We have rationalising factor for denominator is

=

We have

(ii) by putting the value of in the equation we get,

Given and given to find the value of

Squaring on both the sides we get,

Now take

The value of is 10.

√10 × √15 =( √5×√2) × ( √5×√3)

= 5 (√6)

(2+√3) (2-√3)

= (2)² - (√3)² [(a+b) (a-b) = a² –b² ]

= 4 – 3 = 1.

⁵√6 × ⁵√6 = (6)^1/5 × (6)^1/5 = (36)^1/5

= ⁵√36

Reciprocal of 5 + √2 = 1/ (5 + √2)

=

Rationalizing factor of 7- 3√5

=

Rationalisation factor of √3 = 1/√3

Rationalisation factor of 2+√3 = 1/2+√3 = 2-√3

Given,

=

So, x= 2 , y = -1

Given . x = √2-1

Given x = √5+2

−2

= √5 + 2 – ( √5 – 2) = 4

Consider,





As we know,

(a+b)² = a² + b² + 2ab

Given

=

So, a = 2 , b = 1.

³√(125×4) = 5׳√4

√(3-2√2) = √ (√2)² + (1)² - 2× √2×1 = √ (√2 – 1 )² = √2- 1.

Given , a = √2 +1

=

=

Simplest rationalizing factor of √3 + √5

1/ (√3+√5) = √3-√5

Simplest rationalizing factor of 2√5 - √3

= 1/(2√5 - √3)

= 2√5 + √3

Given, x = 2+√3

=

=

Rationalizing factor of √5 – 2

=

Given X = 2/(3+√7)

=

= (x -3 )² = ( 3 - √7 -3 )² = √7² = 7

Given x = 3+ 2√2

)²

=

So, √x – 1 /√x = √2+1 – ( √2- 1)

= 1 + 1 = 2.

Given . x = 7+4√3 , xy = 1

Y = 1/x = 1/7+ 4√3 = 7-4√3

Y² = 1/x² = 49 + 48 - 56√3 = 97 - 56√3

Similarly, x = 1/y

= x² = 1/y² = ( 7 +4√3)² = 49 + 48 + 56√3 = 97+ 56√3

So, 1/x² + 1/y² = 97 + 56√3 + 97 – 56√3 = 194

Given x +√15 = 4

X = 4 - √15

1/x = 1/(4 - √15) = (4 + √15) / 16 -15 = 4 + √15

So, x + 1/x = 4 - √15 + 4+√15 = 8

Given x =

= x³ = 2 + √3

Similarly, 1/x³ = 2 - √3

X³ + 1/x³ = 2+ √3 + 2 - √3 = 4.

Given x = √5 +√3 / √5 - √3 , y =√5 - √3 / √5 + √3

X=

Y =

Xy = 4² - √15² = 16 – 15 = 1

So,

X + y+ xy = 4+√15 + 4 - √15 + 1 = 9.

Given x = , y =

X= 5 – 2√6

X² = (5 – 2√6)² = 25 +24 -20√6) = 49 – 20√6

Similarily y = = 5 + 2√6

Y² = (5 + 2√6 )² = 49 + 20√6

Xy = (5- 2√6) ( 5 + 2√6) = 25 – 24 = 1

So, x²+ xy+y² = 49 – 20√6 +1 + 49 + 20√6 = 99.

( try to break the terms in form of (a+b)² or (a – b )² )

√(√2)²+ 1² – 2 ×√2×1) = √(√2-1)² = √2 – 1 .

( try to break the terms in form of (a+b)² or (a – b )² )

√(√2)²+ 1² – 2 ×√2×1) = √(√2-1)² = √2 – 1 .

Given √2 = 1.4142

√(√2-1)/√2+1) = √(√2-1)² = √2 – 1 = 1.4142 – 1 = 0 .4142

Given , √2 = 1.414

√6 - √3
= √2 × √3 - √3
= √3(√2 - 1)
= 1.732 (1.414 – 1)
= 1.732 × 0.414
= 0.707

7 + √48

= 7 + √(16×3) = 7 + 4√3 ( try to break it in form of (a+b)²)

= (2)² + (√3)² + 2×2×√3 = (2+√3)² = (2+√3) (2+√3).

1/ √9-√8

= 1/(√9 - √8) × (√9 + √8) / (√9+√8)

= √9 + √8 = 3 + 2√2

√48 + √32 / √27 + √18

= 4√3 + 4√2 / 3√3 + 3√2 = (4√3+4√2)/(3√3+3√2) × (3√3- 3√2)/ (3√3-3√2)

=( 36 + 12√6 – 12√6 -24) / (27-18) = 12/9 = 4/3

Given x= √6 + √5

X² = 11 + 2√11

1/x² = 11- 2√11

So, x² + 1/x² – 2 = 11+ 2√11 + 11 – 2√11 -2 = 22-2 = 20.

√(13-a√10) =√8 +√5

Squaring both side,..

= 13 – a√10 = 8 + 5 + 2×√8×√5

= 13 – a√10 = 13 + 2√40

= - a√10 = 4√10

= a = -4

5 - √3/ 2+√3 = x + y√3

= ( 5-√3) / (2+ √3) × (2 -√3) / (2 -√3)

= (10 – 5√3 – 2√3 +3)/ (4-3)

= 10 – 7√3 +3

= 13 – 7√3 = x + y√3

So , x= 13 , y = -7