Order: 3

Order: 2

Order: 4

Order: 2

Order: 3

Solution:

\(\displaystyle \sqrt[ 3 ]{ 51 }\) is an irrational root of a positive rational number.
∴ This is a surd.

Solution:

\(\displaystyle \sqrt[ 4 ]{ 16 }\) = \(\displaystyle \sqrt[ 4 ]{ 2 \times 2 \times 2 \times 2 }\) = 2
But, 2 is a rational number.
∴ This is not a surd.

Solution:

\(\displaystyle \sqrt[ 5 ]{ 81 }\) is an irrational root of a positive rational number.
∴ This is a surd.

Solution:

\(\displaystyle \sqrt{ 256 }\) = 16
But, 16 is a rational number.
∴ This is not a surd.

Solution:

\(\displaystyle \sqrt[ 3 ]{ 64 }\) = \(\displaystyle \sqrt[ 3 ]{ 4 \times 4 \times 4}\) = 4
But, 4 is a rational number.
∴ This is not a surd.

Solution:

\(\displaystyle \sqrt{\frac {22}{7}}\) is an irrational root of a positive rational number.
∴ This is a surd.

Solution:

Let’s simplify the first surd.

 \(\displaystyle \sqrt{ 52 }\) = \(\displaystyle \sqrt{ 4 \times 13 }\) = \(\displaystyle 2 \sqrt{ 13 }\) ... (i)

And the second surd is \(\displaystyle 5 \sqrt{ 13 }\) ... (ii)

Here, the radicands and orders are same.

∴ These are like surds.

Solution:

Let’s simplify the first surd.

 \(\displaystyle \sqrt{ 68 }\) = \(\displaystyle \sqrt{ 4 \times 17 }\) = \(\displaystyle 2 \sqrt{ 17 }\) ... (i)

And the other surd is \(\displaystyle 5 \sqrt{ 3 }\) ... (ii)

Here, the orders are same but the radicands are different.

∴ These are unlike surds.

Solution:

Let’s simplify the first surd.

 \(\displaystyle 4\sqrt{ 18 }\) = \(\displaystyle 4\sqrt{ 9 \times 2 }\) = \(\displaystyle 4 \times 3 \sqrt{ 2 }\) = \(\displaystyle 12 \sqrt{ 2 }\) ... (i)

And the other surd is \(\displaystyle 7 \sqrt{ 2 }\) ... (ii)

Here, the orders and radicands are same.

∴ These are like surds.

Solution:

Let’s simplify the first surd.

 \(\displaystyle 19\sqrt{ 12 }\) = \(\displaystyle 19\sqrt{ 4 \times 3 }\) = \(\displaystyle 19 \times 2 \sqrt{ 3 }\) = \(\displaystyle 38 \sqrt{ 3 }\) ... (i)

And the other surd is \(\displaystyle 6 \sqrt{ 3 }\) ... (ii)

Here, the orders and radicands are same.

∴ These are like surds.

Solution:

Here, the orders are same but the radicands are different.

∴ These are unlike surds.

Solution:

Let’s simplify the second surd.

 \(\displaystyle \sqrt{ 75 }\) = \(\displaystyle \sqrt{ 25 \times 3 }\) = \(\displaystyle 5 \sqrt{ 3 }\) ... (i)

And the first surd is \(\displaystyle 5 \sqrt{ 5 }\) ... (ii)

Here, the orders are same but the radicands are different.

∴ These are unlike surds.

Solution:

 \(\displaystyle \sqrt{ 27 }\)

= \(\displaystyle \sqrt{ 9 \times 3 }\)

= \(\displaystyle 3 \sqrt{ 3 }\)

Solution:

 \(\displaystyle \sqrt{ 50 }\)

= \(\displaystyle \sqrt{ 25 \times 2 }\)

= \(\displaystyle 5 \sqrt{ 2 }\)

Solution:

 \(\displaystyle \sqrt{ 250 }\)

= \(\displaystyle \sqrt{ 25 \times 10 }\)

= \(\displaystyle 5 \sqrt{ 10 }\)

Solution:

 \(\displaystyle \sqrt{ 112 }\)

= \(\displaystyle \sqrt{ 16 \times 7 }\)

= \(\displaystyle 4 \sqrt{ 7 }\)

Solution:

 \(\displaystyle \sqrt{ 168 }\)

= \(\displaystyle \sqrt{ 4 \times 42 }\)

= \(\displaystyle 2 \sqrt{ 42 }\)

Solution:

 \(\displaystyle 7 \sqrt{ 2 }\) = \(\displaystyle \sqrt{ 49 \times 2 }\) = \(\displaystyle \sqrt{ 98 }\) ... (i)

And, \(\displaystyle 5 \sqrt{ 3 }\) = \(\displaystyle \sqrt{ 25 \times 3 }\) = \(\displaystyle \sqrt{ 75 }\) ... (ii)

Now, 98 > 75

∴ \(\displaystyle \sqrt{ 98 }\) > \(\displaystyle \sqrt{ 75 }\)

∴ \(\displaystyle 7 \sqrt{ 2 }\) > \(\displaystyle 5 \sqrt{ 3 }\)

Solution:

Here, 247 < 274

∴ \(\displaystyle \sqrt{ 247 }\) < \(\displaystyle \sqrt{ 274 }\)

Solution:

 \(\displaystyle 2 \sqrt{ 7 }\) = \(\displaystyle \sqrt{ 4 \times 7 }\) = \(\displaystyle \sqrt{ 28 }\) ... (i)

And, \(\displaystyle \sqrt{ 28 }\) = \(\displaystyle \sqrt{ 28 }\) ... (ii)

Now, 28 = 28

∴ \(\displaystyle \sqrt{ 28 }\) = \(\displaystyle \sqrt{ 28 }\)

∴ \(\displaystyle 2 \sqrt{ 7 }\) = \(\displaystyle \sqrt{ 28 }\)

Solution:

 \(\displaystyle 5 \sqrt{ 5 }\) = \(\displaystyle \sqrt{ 25 \times 5 }\) = \(\displaystyle \sqrt{ 125 }\) ... (i)

And, \(\displaystyle 7 \sqrt{ 2 }\) = \(\displaystyle \sqrt{ 49 \times 2 }\) = \(\displaystyle \sqrt{ 98 }\) ... (ii)

Now, 125 > 98

∴ \(\displaystyle \sqrt{ 125 }\) > \(\displaystyle \sqrt{ 98 }\)

∴ \(\displaystyle 5 \sqrt{ 5 }\) > \(\displaystyle 7 \sqrt{ 2 }\)

Solution:

 \(\displaystyle 4 \sqrt{ 42 }\) = \(\displaystyle \sqrt{ 16 \times 42 }\) = \(\displaystyle \sqrt{ 672 }\) ... (i)

And, \(\displaystyle 9 \sqrt{ 2 }\) = \(\displaystyle \sqrt{ 81 \times 2 }\) = \(\displaystyle \sqrt{ 162 }\) ... (ii)

Now, 672 > 162

∴ \(\displaystyle \sqrt{ 672 }\) > \(\displaystyle \sqrt{ 162 }\)

∴ \(\displaystyle 4 \sqrt{ 42 }\) > \(\displaystyle 9 \sqrt{ 2 }\)

Solution:

 \(\displaystyle 5 \sqrt{ 3 }\) = \(\displaystyle \sqrt{ 25 \times 3 }\) = \(\displaystyle \sqrt{ 75 }\) ... (i)

And, \(\displaystyle 9 \) = \(\displaystyle \sqrt{ 81 }\) ... (ii)

Now, 75 < 81

∴ \(\displaystyle \sqrt{ 75 }\) < \(\displaystyle \sqrt{ 81 }\)

∴ \(\displaystyle 5 \sqrt{ 3 }\) < \(\displaystyle 9\)

Solution:

  \(\displaystyle 7 \) = \(\displaystyle \sqrt{ 49 }\) ... (i)

And, \(\displaystyle 2 \sqrt{ 5 }\) = \(\displaystyle \sqrt{ 4 \times 5 }\) = \(\displaystyle \sqrt{ 20 }\) ... (ii)

Now, 49 > 20

∴ \(\displaystyle \sqrt{ 49 }\) > \(\displaystyle \sqrt{ 20 }\)

∴ \(\displaystyle 7\) > \(\displaystyle 2 \sqrt{ 5 }\)

Solution:

 \(\displaystyle 5 \sqrt{ 3 }\) + \(\displaystyle 8 \sqrt{ 3 }\)

= \(\displaystyle (5 + 8) \sqrt{ 3 }\)

= \(\displaystyle 13 \sqrt{ 3 }\)

Solution:

 \(\displaystyle 9 \sqrt{ 5 }\) − \(\displaystyle 4 \sqrt{ 5 }\) + \(\displaystyle \sqrt{ 125 }\)

= \(\displaystyle 9 \sqrt{ 5 }\) − \(\displaystyle 4 \sqrt{ 5 }\) + \(\displaystyle \sqrt{ 25 \times 5 }\)

= \(\displaystyle 9 \sqrt{ 5 }\) − \(\displaystyle 4 \sqrt{ 5 }\) + \(\displaystyle 5 \sqrt{ 5 }\)

= \(\displaystyle (9 - 4 + 5) \sqrt{ 5 }\)

= \(\displaystyle 10 \sqrt{ 5 }\)

Solution:

 \(\displaystyle 7 \sqrt{ 48 }\) − \(\displaystyle \sqrt{ 27 }\) − \(\displaystyle \sqrt{ 3 }\)

= \(\displaystyle 7 \sqrt{ 16 \times 3 }\) − \(\displaystyle \sqrt{ 9 \times 3 }\) − \(\displaystyle \sqrt{ 3 }\)

= \(\displaystyle 7 \times 4 \sqrt{ 3 }\) − \(\displaystyle 3 \sqrt{ 3 }\) − \(\displaystyle \sqrt{ 3 }\)

= \(\displaystyle 28 \sqrt{ 3 }\) − \(\displaystyle 3 \sqrt{ 3 }\) − \(\displaystyle \sqrt{ 3 }\)

= \(\displaystyle (28 - 3 - 1) \sqrt{ 3 }\)

= \(\displaystyle 24 \sqrt{ 3 }\)

Solution:

 \(\displaystyle \sqrt{ 7 }\) − \(\displaystyle \frac {3}{5} \sqrt{ 7 }\) + \(\displaystyle 2\sqrt{ 7 }\)

= \(\displaystyle \left[ 1 - \frac {3}{5} + 2\right]\sqrt{ 7 }\)

= \(\displaystyle \left[ \frac {5 - 3 + 10}{5} \right]\sqrt{ 7 }\)

= \(\displaystyle \frac {12}{5} \sqrt{ 7 }\)

Solution:

 \(\displaystyle 3 \sqrt{ 12 } \times \sqrt{ 18 }\)

= \(\displaystyle 3 \sqrt{ 4 \times 3 } \times \sqrt{ 9 \times 2 }\)

= \(\displaystyle 3 \times 2 \sqrt{ 3 } \times 3 \sqrt{ 2 }\)

= \(\displaystyle 6 \sqrt{ 3 } \times 3 \sqrt{ 2 }\)

= \(\displaystyle 18 \sqrt{ 6 } \)

Solution:

 \(\displaystyle 3 \sqrt{ 12 } \times 7 \sqrt{ 15 }\)

= \(\displaystyle 3 \sqrt{ 4 \times 3 } \times 7 \sqrt{ 15 }\)

= \(\displaystyle 3 \times 2 \sqrt{ 3 } \times 7 \sqrt{ 15 }\)

= \(\displaystyle 6 \sqrt{ 3 } \times 7 \sqrt{ 15 }\)

= \(\displaystyle 42 \sqrt{ 45 } \)

= \(\displaystyle 42 \sqrt{ 9 \times 5 } \)

= \(\displaystyle 42 \times 3 \sqrt{ 5 } \)

= \(\displaystyle 42 \times 3 \sqrt{ 5 } \)

= \(\displaystyle 126 \sqrt{ 5 } \)

Solution:

 \(\displaystyle 3 \sqrt{ 8 } \times \sqrt { 5 }\)

= \(\displaystyle 3 \sqrt{ 4 \times 2 } \times \sqrt { 5 }\)

= \(\displaystyle 3 \times 2 \sqrt{ 2 } \times \sqrt { 5 }\)

= \(\displaystyle 6 \sqrt{ 10}\)

Solution:

 \(\displaystyle 5 \sqrt{ 8 } \times 2 \sqrt { 8 }\)

= \(\displaystyle 10 \sqrt{ 64 }\)

= \(\displaystyle 10 \times 8\)

= \(\displaystyle 80\)

Solution:

 \(\displaystyle \sqrt { 98 } \div \sqrt { 2 }\)

= \(\displaystyle \frac {\sqrt { 98 }}{ \sqrt { 2 } }\)

= \(\displaystyle \sqrt { \frac { 98 }{ 2 } }\)

= \(\displaystyle \sqrt{ 49 }\)

= \(\displaystyle 7 \)

Solution:

 \(\displaystyle \sqrt { 125 } \div \sqrt { 50 }\)

= \(\displaystyle \frac {\sqrt { 125 }}{ \sqrt { 50 } }\)

= \(\displaystyle \sqrt {\frac {125}{50}}\)

= \(\displaystyle \sqrt {\frac {5}{2}}\)

Solution:

 \(\displaystyle \sqrt { 54 } \div \sqrt { 27 }\)

= \(\displaystyle \frac {\sqrt { 54 }}{ \sqrt { 27 } }\)

= \(\displaystyle \sqrt {\frac {54}{27}}\)

= \(\displaystyle \sqrt { 2 }\)

Solution:

 \(\displaystyle \sqrt { 310 } \div \sqrt { 5 }\)

= \(\displaystyle \frac {\sqrt { 310 }}{ \sqrt { 5 } }\)

= \(\displaystyle \sqrt {\frac {310}{5}}\)

= \(\displaystyle \sqrt { 62 }\)

Solution:

 \(\displaystyle \frac { 3 }{ \sqrt { 5 } }\)

Multiplying the numerator and denominator by \(\displaystyle \sqrt { 5 }\),

 \(\displaystyle \frac { 3 }{ \sqrt { 5 } } \times \displaystyle \frac {\sqrt { 5 }} {\sqrt { 5 }}\)

= \(\displaystyle \frac { 3 \sqrt { 5 }}{\sqrt { 5 } \times {\sqrt { 5 }} }\)

= \(\displaystyle \frac { 3 \sqrt { 5 }}{\sqrt { 25 }}\)

= \(\displaystyle \frac { 3 \sqrt { 5 }}{5}\)

Solution:

 \(\displaystyle \frac { 1 }{ \sqrt { 14 } }\)

Multiplying the numerator and denominator by \(\displaystyle \sqrt { 14 }\),

 \(\displaystyle \frac { 1 }{ \sqrt { 14 } } \times \displaystyle \frac {\sqrt { 14 }} {\sqrt { 14 }}\)

= \(\displaystyle \frac { 1 \sqrt { 14 }}{\sqrt { 14 } \times {\sqrt { 14 }} }\)

= \(\displaystyle \frac { \sqrt { 14 }}{\sqrt { 196 }}\)

= \(\displaystyle \frac { \sqrt { 14 }}{14}\)

Solution:

 \(\displaystyle \frac { 5 }{ \sqrt { 7 } }\)

Multiplying the numerator and denominator by \(\displaystyle \sqrt { 7 }\),

 \(\displaystyle \frac { 5 }{ \sqrt { 7 } } \times \displaystyle \frac {\sqrt { 7 }} {\sqrt { 7 }}\)

= \(\displaystyle \frac { 5 \sqrt { 7 }}{\sqrt { 7 } \times {\sqrt { 7 }} }\)

= \(\displaystyle \frac { 5 \sqrt { 7 }}{\sqrt { 49 }}\)

= \(\displaystyle \frac { 5 \sqrt { 7 }}{7}\)

Solution:

 \(\displaystyle \frac { 6 }{ 9 \sqrt { 3 } }\)

Multiplying the numerator and denominator by \(\displaystyle \sqrt { 3 }\),

 \(\displaystyle \frac { 6 }{ 9 \sqrt { 3 } } \times \frac {\sqrt { 3 }} {\sqrt { 3 }}\)

= \(\displaystyle \frac { 6 \sqrt { 3 }}{9 \sqrt { 3 } \times {\sqrt { 3 }} }\)

= \(\displaystyle \frac { 6 \sqrt { 3 }}{ 9 \times 3 }\)

= \(\displaystyle \frac { 6 \sqrt { 3 }}{ 27 }\)

= \(\displaystyle \frac { 2 \sqrt { 3 }}{ 9 }\)

Solution:

 \(\displaystyle \frac { 11 }{ \sqrt { 3 } }\)

Multiplying the numerator and denominator by \(\displaystyle \sqrt { 3 }\),

 \(\displaystyle \frac { 11 }{ \sqrt { 3 } } \times \displaystyle \frac {\sqrt { 3 }} {\sqrt { 3 }}\)

= \(\displaystyle \frac { 11 \sqrt { 3 }}{\sqrt { 3 } \times {\sqrt { 3 }} }\)

= \(\displaystyle \frac { 11 \sqrt { 3 }}{\sqrt { 9 }}\)

= \(\displaystyle \frac { 11 \sqrt { 3 }}{3}\)