Surds – Exercise 7.1 – Class X

Surds – Exercise 7.1

1. Simplify the following surds:

(i) √75 + √108 – √192

(ii)  4√12 – √50 – 7√48

(iii) √45 – 3√20 + 3√5

(iv) 2√(2a) + 3√(8a)  – √(2a)

(v) 3x√x + 3√x³ – 2√(9x³)

(vi) √12 + √50 +5√3  – √147  – √32

(vii) 4√7 – 3√252 + 5√343

(viii) ¹/₈ √50 + ¹/₆√75 – ¹/₈ √18  – ¹/₃√3

1. Find sum of the following surds

(i) x√y , 2x√y, 4x√y

(ii) 5∛p, 3∛p, 2∛p

(iii) x√x, y√y, 3√x³, 4√y³

(iv) (√12+20), (3√3 + 2√5), (√45 – √90)

(v) (√3 + √2), (2√2 ₊ 3√3), (4√2 – 3√3)

(vi) (√x +2√y), (2√x – 3√y), (3√x + √y)

III.

1. Subtract 5√x from 9√x ad express the result in index form.
2. Subtract 3√p from 10√p
3. Subtract 3√a from the sum of 4√a and 2√a
4. Subtract 2√x + 3√y from 5√x – √y

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Surds – Exercise 7.1 – Solution:

1. Simplify the following surds:

(i) √75 + √108 – √192

Solution:

√75 + √108 – √192

= √(3x 25) + √(3×36) – √(3×64)

= 5√3 + 6√3 – 8√3

= (5 + 6 – 8) √3

= 3√3

 

(ii)  4√12 – √50 – 7√48

Solution:

4√12 – √50 – 7√48

= 4√(4×3) – √(25×2) – 7√(16×3)

= 4×2√3 – 5√2 – 7×4√3

= 8√3 – 5√2 – 28√3

= (8 – 28)√3 – 5√2

= – 20√3 – 5√2

= -5(4√3 + √2)

 

(iii) √45 – 3√20 + 3√5

Solution:

√45 – 3√20 + 3√5

= √(9×5) – 3√(4×5) + 3√5

= 3√5 – 3×2√5 + 3√5

= 3√5 – 6√5 + 3√5

= 0

 

(iv) 2√(2a) + 3√(8a)  – √(2a)

Solution:

= 2√(2a) + 3√(8a)  – √(2a)

= 2√(2a) + 3√(4x2a)  – √(2a)

= 2√(2a) + 3×2√(2a)  – √(2a)

= (2 + 6 – 1)√(2a)

= 7√(2a)

 

(v) 3x√x + 3√x³ – 2√(9x³)

Solution:

3x√x + 3√x³ – 2√(9x³)

= 3x√x + 3√(x*x²) – 2√(x*9x²)

= 3x√x + 3x√x – 2*3x√x

= 3x√x + 3x√x – 6x√x

= (3x + 3x – 6x)√x

= 0

 

(vi) √12 + √50 +5√3  – √147  – √32

Solution:

√12 + √50 +5√3  – √147  – √32

= √(4×3) + √(25×2) + 5√3 – √(49×3) – √(16×2)

= 2√3 + 5√2 + 5√3 – 7√3 – 4√2

= (2 + 5 – 7)√3 +(5 – 4)√2

= 0. √3 + √2

= √2

 

(vii) 4√7 – 3√252 + 5√343

Solution:

4√7 – 3√252 + 5√343

= 4√7 – 3√(7×36) + 5√(7×49)

= 4√7 – 3×6√7 + 5×7√7

= (4 – 18 + 35) √7

= 21√7

 

(viii) ¹/₈ √50 + ¹/₆√75 – ¹/₈ √18  – ¹/₃√3

Solution:

¹/₈ √50 + ¹/₆√75 – ¹/₈ √18  – ¹/₃√3

= ¹/₈ √(2×25) + ¹/₆√(25×3) – ¹/₈ √(9×2)  – ¹/₃√3

= ¹/₈ x 5√2 + ¹/₆ x 5√3 – ¹/₈ x 3√2  – ¹/₃√3

= (⁵/₈ – ³/₈)√2 + (⁵/₆ – ¹/₃) √3

=¹/₄ √2 + ¹/₂ √3

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2. Find sum of the following surds

(i) x√y , 2x√y, 4x√y

Solution:

x√y + 2x√y + 4x√y = (x + 2x + 4x)√y

= 7x√y

 

(ii) 5∛p, 3∛p, 2∛p

Solution:

5∛p + 3∛p + 2∛p = (5 + 3+ 2) ∛p

= 10∛p

 

(iii) x√x, y√y, 3√x³, 4√y³

Solution:

x√x + y√y + 3√x³ + 4√y³

= x√x + y√y + 3√(x*x²) + 4√(y*y²)

= (x + 3x)√x + (y + 4y) √y

= 4x√x + 5y√y

 

(iv) (√12+√20), (3√3 + 2√5), (√45 – √90)

Solution:

= √12+√20 + 3√3 + 2√5 + √45 – √90

= √(4×3) + √(4×5) + 3√3 + 2√5 + √(9×5) – √(9×10)

= 2√3 + 2√5 + 3√3 + 2√5 + 3√5 – 3√10

= (2+3)√3 + (2+2+3)√5 – 3√10

= 5√3 + 7√5 – 3√10

 

(v) (√3 + √2), (2√2 ₊ 3√3), (4√2 – 3√3)

Solution:

= √3 + √2 + 2√2 ₊ 3√3 + 4√2 – 3√3

= (1+ 3 – 3)√3 + (1+2+4)√2

= √3 + 7√2

 

(vi) (√x +2√y), (2√x – 3√y), (3√x + √y)

Solution:

= √x +2√y + 2√x – 3√y + 3√x + √y

= (1+2 + 3)√x +(2 – 3 + 1)√y

= 6√x + 0. √y

= 6√x

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III.

1. Subtract 5√x from 9√x ad express the result in index form.

Solution:

9√x – 5√x = (9 – 5) √x = 4√x

Therefore, 4√x is the result when 5√x subtracted from 9√x.

Then the index form of the result 4√x = 4x^1/2

 

2. Subtract 3√p from 10√p

Solution:

10√p – 3√p = (10 – 3)√p = 7√p

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3. Subtract 3√a from the sum of 4√a and 2√a

Solution:

We need to find the sum of 4√a and 2√a:

4√a + 2√a = (4+2)√a + 6√a

We have to subtract 3√a from the sum of 4√a and 2√a:

6√a + 3√a  = (6+3)√a = 9√a

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4. Subtract 2√x + 3√y from 5√x – √y

Solution:

(5√x – √y) – (2√x + 3√y)

= 5√x – √y – 2√x – 3√y

= (5 – 2)√x + (-1 – 3)√y

= 7√x – 4√y

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Next exercise – Surds – Exercise 7.2 – Class X

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