Surds – Exercise 7.4 – Class X

Previous exercise – Surds –  Exercise 7.3 – Class X

Surds – Exercise 7.4

I.Rationalise the denominator and simplify

1. 8/√3
2. 3/2√x
3. √(5/2y)
4. 1/2√(2a/5)
5. 3√5/√6

B.

1. 2/√3 + √2
2. x/√x – √y
3. √10/√5 + √3
4. 3√5/√6 -√3
5. √ab/√a – √b

C.

1. √3 + √2/√3 – √2
2. 5√2 – √3/5√2 – √3
3. 4√3 + √2/4√3 – √2
4. 3 + √6/√3 + 6

II. Simplify each of the following:

1. √2/√3 – √2 + √5/√3 + √2
2. √5/√5 – √3√3/√5 + √3
3. √6/√5 + 2/√5 + √2
4. 7√3/√6 – √32√5/√8 + √2
5. √21/√3 + √7 + 2√5/√21 + √5
6. If x = 2√6 + 5 find x + 1/x

III. Solve for x

(i) 3x-4/√3x + 2 = 2 + √3x – 2/2

(ii) x – 1/√x + 1  = 4 + √x – 1/2

Surds – Exercise 7.4 – Solutions:

I. Rationalise the denominator and simplify

1. 8/√3

Solution:

Rational factor of  denominator of 8/√3  is √3

= 8/√3 x √3/√3

= 8x√3/3

= 8√3/3

1. 3/2√x

Solution:

Rational factor of  denominator of 3/2√x is √x

=3/2√x x √x/√x

= 3x√x/2x

= 3√x/2x

1. √(5/2y)

Solution:

Rational factor of  denominator of √(5/2y) is √2y

= √(5/2y) x √(2y)/√(2y)

= √5x√(2y)/√2(y) x √(2y)

= √(10y)/2y

1. 1/2√(2a/5)

Solution:

Rational factor of  denominator of 1/2√(2a/5)is √5

= 1/2 x√(2a)/√5)x √5/√5

=1/2 x √(10a)/5

= √(10a)/10

1. 3√5/√6

Solution:

Rational factor of  denominator of 3√5/√6 is √6

= 3√5/√6 x √6/√6

= 3√5x√6/√6x√6

= 3√30/6

= √30/2

B.

1. 2/√3 + √2

Solution:

In 2/√3 + √2 denominator is √3 + √2.

Rational factor of √3 + √2 is √3 – √2

2/√3 + √2 = 2/√3 + √2 x √3 – √2/√3 – √2

= 2(√3 – √2)/(√3 – √2)(√3 – √2)

= 2(√3 – √2)/3 – 2

= 2(√3 – √2)/1

= 2(√3 – √2)

1. x/√x – √y

Solution:

In x/√x – √y denominator is √x – √y .

Rational factor of √x – √y is √x + √y.

x/√x – √y = x/√x – √y x √x + √y/√x + √y

= x(√x + √y)/(√x – √y)( √x + √y)

= x(√x + √y)/x – y

1. √10/√5 + √3

Solution:

In √10/√5 + √3 denominator is √5 + √3.

Rational factor of √5 + √3 is √5 – √3.

√10/√5 + √3 = √10/√5 + √3 x √5 – √3/√5 – √3

= √10(√5 – √3)/(√5 + √3)(√5 – √3)

= √10(√5 – √3)/5 – 3

= √10(√5 – √3)/2

1. 3√5/√6 -√3

Solution:

In 3√5/√6 -√3 denominator is √6 -√3.

Rational factor of √6 -√3 is √6 +√3.

3√5/√6 -√3 = 3√5/√6 -√3 x √6 +√3/√6 +√3

= 3√5(√6 +√3)/6 -3

= 3√5(√6 +√3)/3

= √5(√6 +√3)

1. √(ab)/√a – √b

Solution:

In √(ab)/√a – √b denominator is √a – √b

Rational factor of √a – √b is √a + √b.

√(ab)/√a – √b = √(ab)/√a – √b x √a + √b/√a + √b

= √(ab)(√a + √b)/a – b

C.

1. √3 + √2/√3 – √2

Solution:

In  √3 + √2/√3 – √2 denominator is √3 – √2

Rational factor of √3 – √2 is √3 + √2

√3 + √2/√3 – √2 = √3 + √2/√3 – √2 x √3 + √2/√3 + √2

= (√3 + √2)(√3 + √2)/3 – 2

= (√3 + √2) (√3 + √2)

=(√3 + √2)2

= 3 + 2√2√3 + 2

= 5 + 2√6

1. 5√2 – √3/3√2 – √5

Solution:

In  5√2 – √3/3√2 – √5 denominator is 3√2 – √5

Rational factor of 3√2 – √5 is 3√2 + √5

5√2 – √3/3√2 – √5 = 5√2 – √3/3√2 – √5 x 3√2 + √3/3√2 + √5

= (5√2 – √3)( 3√2 + √5)/9×2 – 5

=(5√2 – √3)( 3√2 + √5)/18 – 5

=(5√2 – √3)(3√2 + √5)/13

= (5√2(3√2 + √5) – √3)(3√2 + √5)/13

= 15×2 + 5x√10 – 3√6 – √15/13

1. 4√3 + √2/√3 + √2

Solution:

In  4√3 + √2/√3 + √2 denominator is √3 + √2

Rational factor of √3 + √2  is √3 – √2

4√3 + √2/√3 + √2 = 4√3 + √2/√3 – √2 x √3 – √2/√3 – √2

= (4√3 + √2)(√3 – √2)/3 – 2

=(4√3 + √2)(√3 – √2)

= 4√3(√3 – √2) + √2(√3 – √2)

= 4×3 – 4√6 + √6 – 2

= 12 – 2 +(-4 + 1) √6

= 10 – 3√6

1. 3 + √6/√3 + 6

Solution:

In  3 + √6/√3 + 6 denominator is √3 + 6

Rational factor of √3 + 6 is √3 – 6

3 + √6/√3 + 6 = 3 + √6/√3 + 6 x (√3 – 6)/( √3 – 6)

= (3 + √6) (√3 – 6)/3 – 36

= (3 + √6) (√3 – 6)/-33

= 3(√3 – 6) + √6(√3 – 6)/-33

= 3√3 – 18 + √18 – 6√6/-33

= -(√3 – 6 + √2 – 2√6/11)

II. Simplify each of the following:

1. √2/√3 – √2 + √5/√3 + √2

Solution:

Rational factor of √3 – √2 is √3 + √2 and rational factor of √3 + √2 is √3 – √2

√2/√3 – √2 + √5/√3 + √2  = √2/√3 – √2 x √3 + √2 /√3 + √2  + √5/√3 + √2 x √3 – √2/√3 – √2

= √2(√3 + √2)/3 – 2 + √5(√3 – √2)/3 – 2

=[√2(√3 + √2]+[ √5(√3 – √2)]

= √6 + 2 + √15 – √10

1. √5/√5 – √3√3/√5 + √3

Solution:

Rational factor of √5 – √3 is √5 + √3 and rational factor of √5 + √3 is √5 – √3

√5/√5 – √3√3/√5 + √3 = √5/√5 – √3 x √5 + √3/√5 + √3√3/√5 + √3 x √5 – √3/√5 – √3

= √5(√5 + √3)/5 – 3√3(√5 – √3)/5 – 3

= 5 + √15/2√15 – 3/2

= 5 + √15 – √15 + 3/2

= 8/2

= 4

1. √6/√5 + 2/√5 + √2

Solution:

Rational factor of √5 is √5 and rational factor of √5 + √2 is √5 – √2

√6/√5 + 2/√5 + √2 = √6/√5 x √5/√5 + 2/√5 + √2 x√5 – √2/√5 – √2

= √30/5 + 2/√5 + √2 x √5 – √2/√5 – √2

= √30/5 + 2(√5 – √2)/5 – 2

= √30/5 + 2√5 – 2√2/3

= 3√30 + 10√5 – 10√2/15

= 3√30 + 10(√5 – √2)/15

1. 7√3/√6 – √32√5/√8 + √2

Solution:

Rational factor of √6 – √3 is √6 + √3 and rational factor of √8 + √2 is √8 – √2

7√3/√6 – √32√5/√8 + √27√3/√6 – √3 x √6 + √3/√6 + √32√5/√8 + √2 x √8 – √2/√8 – √2

= 7√3(√6 + √3)/6-32√5(√8 – √2)/8-2

= 7√18 + 7√9/32√40 – 2√10/6

= 7√18 + 7√9/3√40 – √10/3

= 7√18 + 7√9 – √40 + √10/3

= 7×3√2 + 7×3 – 2√10 + √10/3

= 21√2 + 21 – 2√10 + √10/3

= 21(√2 + 1) – 2√10 + √10/3

= 21(√2 + 1) – √10/3

1. √21/√3 + √7 + 2√5/√21 + √5

Solution:

Rational factor of √3 + √7 is √3 – √7 and rational factor of √21 + √5 is √21 – √5

√21/√3 + √7 + 2√5/√21 + √5 = √21/√3 + √7 x √3 – √7/√3 – √7+ 2√5/√21 + √5 x √21 – √5/√21 – √5

= √21 x (√3 – √7)/3 – 7 + 2√5(√21 – √5)/21 – 5

= √63 – √147/-4 + √105 – 5/8

=  -2(√63 – √147) + √105 – 5/8

=  -2√63 + 2√147 + √105 – 5/8

=  -2√(7×9) + 2√(49×3) + √105 – 5/8

=  -6√7 + 14√3 + √105 – 5/8

1. If x = 2√6 + 5 find x + 1/x

Solution:

We have, x = 2√6 + 5.

We have to find, x + 1/x = (2√6 + 5) + 1/2√6 + 5

Rational factor of 2√6 + 5 is 2√6 – 5

(2√6 + 5) + 1/2√6 + 5 = (2√6 + 5) + 1/2√6 + 5 x 2√6 – 5/2√6 – 5

= (2√6 + 5) + 2√6 – 5/4×6 – 25

= (2√6 + 5) + 2√6 – 5/-1

= (2√6 + 5) + 2√6 – 5/-1

=[ – 2√6 – 5 + 2√6 – 5]/-1

= -10/-1

= 10

III. Solve for x

(i) 3x-4/√(3x) + 2 = 2 + √(3x) – 2/2

Solution:

3x-4/√(3x) + 2 = 2 + √(3x) – 2/2

3x-4/√(3x) + 2 = 4+ √(3x) – 2/2

(3x – 4)2 = (4+ √(3x) – 2)( √(3x) + 2)

6x – 8 = √(3x)(4+ √(3x) – 2)+2(4+ √(3x) – 2)

6x – 8 =  4√(3x) + 3x – 2√(3x) + 8 + 2√(3x) – 4

6x – 8 = (4 – 2 + 2)√(3x) + 3x + 8 – 4

6x – 8 = 4√(3x) + 3x + 4

4√(3x) + 3x – 6x + 4 + 8 = 0

4√(3x) – 3x + 12 = 0

3x – 4√(3x) – 12 = 0

[√(3x)]2 –  6√(3x) + 2√(3x) – 12 = 0

√(3x)[ √(3x) – 6)]+2[√(3x) – 6] = 0

(√(3x) – 6)(√(3x) + 2) = 0

√(3x) = 6 or √(3x) = -2

If √(3x) = 6 , taking square  on both the sides,

3x = 36

x = 36/3 = 12

Therefore, x = 12

(ii) x – 1/√x + 1  = 4 + √x – 1/2

Solution:

x – 1/√x + 1  = 8+ √x – 1/2

2(x – 1) = (8 + √x – 1)( √x + 1)

2x – 2 = (√x + 7)( √x + 1)

2x – 2 = √x (√x + 1)+ 7(√x + 1)

2x – 2 = x + √x + 7√x + 7

2x – 2 = 8√x + x + 7

2x – x – 8√x – 2  – 7 = 0

x – 8√x – 9 = 0

(√x)2 – 8√x – 9 = 0

(√x)2 – 9√x + √x  – 9 = 0

√x(√x – 9) +1(√x – 9) = 0

(√x + 1)( √x – 9) = 0

√x = -1 or √x = 9

For √x = 9 then, x = 81

Therefore x = 81.