Quadratic Surd

Given:

√175

It can be written as:

√(5×5×7)

= 5√7

Hence, √175 can be written in the form of product of rational and irrational numbers as : 5√7

Where 5 is a rational number and √7 is an irrational number.

Given:

√112

It can be written as:

√(2×2×2×2×7)

= 4√7

Hence, √112 can be written in the form of product of rational and irrational numbers as : 4√7

Where 4 is a rational number and √7 is an irrational number.

Given:

√108

It can be written as:

√(3×3×3×2×2)

= 6√3

Hence, √108 can be written in the form of product of rational and irrational numbers as : 6√3

Where 6 is a rational number and √3 is an irrational number.

Given:

√125

It can be written as:

√(5×5×5)

= 5√5

Hence, √125 can be written in the form of product of rational and irrational numbers as : 5√5

Where 5 is a rational number and √5 is an irrational number.

Given:

5√119

It can be written as:

5√(7×17)

= 5√119

Hence, 5√119 can be written in the form of product of rational and irrational numbers as : 5√119

Where 5 is a rational number and √119 is an irrational number.

Given:

√108-√75 = √3

Take L.H.S. = √108-√75

It can be written as:

√(3×3×3×2×2)-√(3×5×5)

= 6√3-5√3

= √3 = R.H.S.

Hence, L.H.S. = R.H.S.

Given:

√98 + √8-2√32 = √2

Take L.H.S. = √98 + √8-2√32

It can be written as:

√(2×7×7) + √(2×2×2)-2√(2×2×2×2×2)

= 7√2 + 2√2-8√2

= 9√2-8√2

= √2 = R.H.S.

Hence, L.H.S. = R.H.S.

Given:

3√48-4√75 + √192 = 0

Take L.H.S. = 3√48-4√75 + √192

It can be written as:

3√(2×2×2×2×3)-4√(3×5×5) + √(2×2×2×2×2×2×3)

= 12√3-20√3 + 8√3

= -8√3 + 8√3

= 0 = R.H.S.

Hence, L.H.S. = R.H.S.

Given:

√12 + √18 + √27-√32

It can be written as:

√(2×2×3) + √(3×3×2) + √(3×3×3)-√(2×2×2×2×2)

= 2√3 + 3√2 + 3√3-4√2

= 5√3-1√2

Given:

√5 + √3

Let the term added to √5 + √3 is x :

It can be written as:

√5 + √3 + x = 2√5

⇒x = 2√5-√5-√3

⇒x = √5-√3

Hence, we need to add √5-√3 to get 2√5

Given:

7-√3

Let the term subtracted from 7-√3 is x :

It can be written as:

7-√3-x = 3 + √3

⇒x = 7-√3-3-√3

⇒x = 4-2√3

Hence, we need to subtract 4-2√3 to get 3 + √3

Given:

2 + √3, √3 + √5,2 + √7

Addition of all is:

It can be written as:

2 + √3 + √3 + √5 + 2 + √7

= 4 + 2√3 + √5 + √7

Hence, result is 4 + 2√3 + √5 + √7

Given:

10-√11

To subtracted -5 + 3√11 is :

It can be written as:

10-√11-(-5 + 3√11)

= 10-√11 + 5-3√11

= 15-4√11

Hence, the result is 15-4√11

Given: 5 + √2 + √7, - 5 + √7 and √7 + √2

Sum of – 5 + √7 and √7 + √2 is:

- 5 + √7 + √7 + √2

= - 5 + 2√7 + √2 ….(1)

subtract 5 + √2 + √7 from (1)

We get,

- 5 + 2√7 + √2-(5 + √2 + √7 )

= - 5 + 2√7 + √2-5-√2-√7

= - 10-√7

Given: two quadratic surds

A quadratic surd is an expression containing square roots, such that number under square root is a rational number and is not a perfect square.

Let we take quadratic surds as:

7-√3 and 7 + √3

Sum of these surds will be:

7-√3 + 7 + √3

= 14

Where 14 is a rational number.

Given:

It can be written as:

Product of given values :

Hence, the product of

Is .

Given:

Let the term multiplied to is x :

It can be written as:

Hence, we need to multiply .

Given:

It can be written as:

Product of given values :

Hence, the product of

Is .

Given:

It can be written as:

⇒ x = 3.

Hence, the value of x = 3.

Given:

It can be written as:

Hence, the value of x = .

Given:

It can be written as:

Product of given values :

Hence, the product of is .

Given:

It can be written as:

Product of given values :

= 2 × 2× 3

= 12

Hence, the product of
is 12.

Given:

It can be written as:

Product of given values :

= 5× 3

= 15

Hence, the product of
is 15.

Given:

It can be written as:

Product of given values :

Hence, the product of
is 15.

Given:

It can be written as:

= 2 – 3

= -1

Hence, the product of
is -1.

Given:

It can be written as:

Hence, the product of

is .

Given:

It can be written as:

= 2× 6

= 12

Hence, the product of is 12.

Given:

As we know, rationalization factor is the factor which make the given irrational number as a rational number.

Hence, smallest rational factor of .

As,

Which is a rational number.

Given:

Hence, the value of

Given:

It can be written as:

So, to make it rationalize we must multiply the denominator by .

Hence, we get,

Given:

Its conjugate surd will be :

Which is also its rationalising factor.

Given:

It can be written as:

Given:

Conjugate surd for the given expression will be (√3 + 2), So we have to find a factor other than this

Rationalize with

We get,

Given:

Its conjugate surd will be :

(i) Given:

Its conjugate surds will be :

(ii) Given:

Its conjugate surds will be :

(iii) Given:

Its conjugate surds will be :

(iv) Given:

Its conjugate surds will be :

Given:

Rationalize with

We get,

Given:

Rationalize with

We get,

Given:

Rationalize with

We get,

Given:

Rationalize with

We get,

Given:

Rationalize with

We get,

Given:

Rationalize with

We get,

Given:

Acc. To condition:

Rationalize with

We get,

Given:

Acc. To condition:

Rationalize with

We get,

Given:

Acc. To condition:

Rationalize with We get,

Given:

After rationalization we get,

We get,

Given:

After rationalization we get,

We get,

Formula used.

(a + b)² = a² + b² + 2ab

(a + b)³ = a³ + b³ + 3ab(a + b)

Let a = m and b =

(a + b)² = a² + b² + 2ab

(m + )² = m² + ² + 2 × m ×

(√3)² = m² + + 2 × 1

3 = m² + + 2

m² + = 3 – 2 = 1

(a + b)³ = a³ + b³ + 3ab(a + b)

(m + )³ = m³ + ³ + 2 × m × × (m + )

(√3)³ = m³ + + 2 × 1 × (√3)

3√3 = m³ + + 2√3

m³ + = 3√3 – 2√3 = √3[3 – 2]

= √ 3

= 2√15

]

]

]

=

=

Simplifying part 1

= √35 – √14

Simplifying part 2

= √35 – √10

Simplifying part 3

= √14 – √10

Putting values we get;

[√35 – √14] – [√35 – √10] + [√14 – √10]

√35 – √14 – √35 + √10 + √14 – √10

= 0

Simplifying 1^st part by rationalizing the expression by multiplying and dividing by 2 + √2

= = = 4√3 + 2√6

Simplifying 2^nd part by rationalizing the expression by multiplying and dividing by 4√3 + √18

=

Simplifying 3^rd part by rationalizing the expression by multiplying and dividing by 3 + √12

= = -

Putting all values we get;

4√3 + 2√6 – (4√3 – √18) + (3√2 + √24)

4√3 + √6 × 2² – (4√3 – √18) + (√2 × 3² + √24)

4√3 + √24 – (4√3 – √18) + (√18 + √24)

= 2√24 = 4√6

Simplifying 1^st part

= -√6 + √12

Simplifying 2^nd part

= √18-√6

Simplifying 3^rd part

= -√12 + √18

Putting all values we get;

(-√6 + √12) – (√18 - √6) + (-√12 + √18)

-√6 + √12 – √18 + √6 – √12 + √18

= 0

Putting value x = 2, y = 3, z = 6;

Simplifying 1^st part

= -√6 + √12

Simplifying 2^nd part

= √18-√6

Simplifying 3^rd part

= -√12 + √18

Putting all values we get;

(-√6 + √12) – (√18 - √6) + (-√12 + √18)

-√6 + √12 – √18 + √6 – √12 + √18

= 0

x = √7 + √6

then;

=

By simplifying

=

=

=

=

= √7 – √6

Hence;

x – = (√7 + √6) – (√7 – √6)

= 2√6

x = √7 + √6

then;

=

By simplifying

=

=

=

=

= √7 – √6

Hence;

x + = (√7 + √6) + (√7 – √6)

= 2√7

Formula used.

(a + b)² = a² + b² + 2ab

x + = 2√7

(a + b)² = a² + b² + 2ab

Put a = x and b =

(x + )² = x² + ² + 2 × x ×

(2√7)² = x² + + 2

4 × 7 = x² + + 2

28 = x² + + 2

x² + = 28 – 2 = 26

Formula used.

(a + b)³ = a³ + b³ + 3ab(a + b)

If

x + = 2√7

(a + b)³ = a³ + b³ + 3ab(a + b)

Put a = x and b =

(x + )³ = x³ + ³ + 3 × x × × (x + )

(2√7)³ = x³ + + 3 × (2√7)

56√7 = x³ + + 6√7

x³ + = 56√7 – 6√7

x³ + = √7 [56 – 6]

= 50√7

4x² – 2

If 4x² – 2 = 14

4x² = 14 + 2 = 16

x² = = 4

x = √4 = ±2

a + b =

a + b = = = = 3

a-b =

a-b = = = = √5

ab = = 1

= = =

Putting values we get;

=

a + b =

a + b = = = = 3

a-b =

a-b = = = = √5

= =

a + b =

a + b = = = = 3

a-b =

a-b = = = = √5

ab = = 1

= = =

Putting values we get;

=

a + b =

a + b = = = = 3

a-b =

a-b = = = = √5

ab = = 1

(a + b)³ = a³ + b³ + 3ab(a + b)

a³ + b³ = (a + b)³ - 3ab(a + b)

= (3)³ – 3 × 1 × 3

= 27 – 9 = 18

(a-b)³ = a³-b³-3ab(a-b)

a³-b³ = (a-b)³ + 3ab(a-b)

= (√5)³ + 3 × 1 × (√5)

= 5√5 + 3√5

= √5 [5 + 3]

= 8√5

If x = 2 + √3

Then;

Simplifying it we get;

= = 2-√3

x – = 2 + √3 – [2 - √3]

= 2√3

If y = 2-√3

Then;

Simplifying it we get;

= = 2 + √3

y + = 2 – √3 + [2 + √3] = 4

(y + )² = y² + []² + 2 × y ×

(4)² = y² + []² + 2

y² + []² = 16 – 2 = 14

Formula used.

(a-b)³ = a³-b³-3ab(a-b)

If x = 2 + √3

Then;

Simplifying it we get;

= = 2-√3

x – = 2 + √3 – [2 - √3]

= 2√3

(x – )³ = x³- ³-3 × x × × (x – )

(x – )³ = x³- -3 × 1 × (x – )

(2√3)³ = x³- -3 × (2√3)

x³ - = 24√3 + 6√3

x³ - = 30√3

x = 2 + √3

y = 2 – √3

xy = (2 + √3) × ( 2-√3)

xy = (2)² – (√3)²

= 4 – 3

= 1

= = 1 + 1 = 2

Formula used.

(a – b)² = a² – 2ab + b²

3x² – 5xy + 3y²

Add and subtract xy to the equation.

3x² – 5xy + 3y² [ + xy – xy]

3x² – 6xy + 3y² + xy

3[x² – 2xy + y²] + xy

3[x-y]² + xy

x = 2 + √3

y = 2 – √3

xy = (2 + √3) × ( 2-√3)

xy = (2)² – (√3)²

= 4 – 3

= 1

x – y = 2 + √3 – [2-√3]

x – y = 2√3

Putting the values we get;

3[2√3]² + 1

3[12] + 1

36 + 1 = 37

Formula used.

(a – b)² = a² – 2ab + b²

x =

xy = 1

y =

y = =

x + y = +

=

=

= = 5

x-y = –

=

=

= = √21

Add and Subtract xy both on numerator and denominator

=

Putting values we get;

=

Hence proved.

Formula used.

(a – b)² = a² – 2ab + b²

1^st value is √7 + 1

Its square is

(√7 + 1)² = (√7)² + 1² + 2 × 1 × √7 = 7 + 1 + 2√7 = 8 + 2√7

2^nd value is √5 + √3

Its square is

(√5 + √3)² = (√5)² + (√3)² + 2 × √3 × √5 = 5 + 3 + 2√15 = 8 + 2√15

If 7<15

Then √7 < √15

Then 8 + √7 < 8 + √15

Then √(8 + √7) < √(8 + √15)

∴ √7 + 1 < √5 + √3

If x = 2 + √3

Then;

Simplifying it we get;

= = 2-√3

x + = 2 + √3 + [2 - √3]

= 4

p + q = √13

p = √13 – q

p–q = √5

(√13 – q) – q = √5

2q = √13 - √5

q =

p = √13 – q = √13 – =

pq = × = = = 2

a + b = √5

a = √5 – b

a–b = √3

(√5 – b) – b = √3

2b = √5 - √3

b =

a = √5 – b = √5 – =

ab = × = =

(a + b)² = a² + b² + 2ab

(√5)² = a² + b² + 2 ×

a² + b² = 5 – 1 = 4

√125 = √ (5 × 5 × 5) = 5√5

√ 125 – √ 5

= 5√5 – √5

= √5 [5-1]

= 4√5

= √((4 × 4) × 5)

= √80

(5-√3)(√3-1)(5 + √3)(√3 + 1)

(5-√3)(5 + √3)(√3-1)(√3 + 1)

(5² – (√3)²)((√3)² – 1²)

(25 – 3)(3 – 1)

22 × 2 = 44

(i) True.

√75 = √(5 × 5 × 3) = 5√3

√147 = √(7 × 7 × 3) = 7√3

√3 is common on both surds

(ii) False

π itself is an irrational number

hence;

Square root of π is not a surd.

(a) Irrational

As √11 is irrational number

Multiplying it with 5

Also get irrational number

(b) √3 + 5

Conjugate surds are surds which are having same terms but having different symbol( + to- and – to + ) in between both the terms.

(c) Conjugate Surds

While product of conjugate surds

They get square to become rational number

While sum of conjugate surds

They get cancel to become rational number

If x = 3 + 2√2

Then;

Simplifying it we get;

= = 3-2√2

x + = 3 + 2√2 + [3 – 2√2]

= 6

1^st value is √15 + √3

Its square is

(√15 + √3)² = (√15)² + (√3)² + 2 × √15 × √3 = 15 + 3 + 2√45 = 18 + 2√45

2^nd value is √10 + √8

Its square is

(√10 + √8)² = (√10)² + (√8)² + 2 × √10 × √8 = 10 + 8 + 2√80 = 18 + 2√80

If 45<80

Then √45 < √80

Then 18 + √45 < 18 + √80

Then √(18 + √45) < √(18 + √80)

∴ √15 + √3 < √10 + √8

Two mixed surds are

5√6 and 7√6

Multiplying both

5√6 × 7√6

35 × 6 = 210

Which is a rational number.

Let the number be x

√72 – x = √32

6√2 – x = 4√2

x = 6√2 – 4√2

= √2 [6 – 4]

= 2√2

Simplifying part 1

= √2-1

Simplifying part 2

= √3-√2

Simplifying part 3

= √4-√3

Adding all we get;

√2-1 + √3-√2 + √4-√3

= √4 – 1

= 2 – 1 = 1