## Surds

Subject: Compulsory Maths

#### Overview

Surds are numbers left in 'square root form' (or 'cube root form' etc). They are therefore irrational numbers. The reason we leave them as surds is because in a decimal form they would go on forever and so this is a very clumsy way of writing them.

##### Surds

A surd is a square root which cannot be reduced to the whole number. If we can't simplify a number to remove a square root (or cube root) then it is a surd.

The numbers left in the square root form or cube root form etc. is called surds. The reason we leave them as surds is because in the decimal form. They would go on forever and this is a clumsy way of writing them.

Example: √3 ( square root of 3 ) can't be simplified further.

### Addition and Subtraction of Surds

Adding and subtracting surds are simple- however we need the numbers being square rooted ( or cube rooted) to be the same.

5√8 + 2√8 = 10√8

6√7 - 3√7 = 3√7

However, if the number in the square root sign isn't prime, we might be able to split it up in order to simplify an expression. For example :

√12 + √27 = $\sqrt{4×3}$ + $\sqrt{9×3}$

= $\sqrt{4×3}$ + $\sqrt{9×3}$

= 2√3 + 3√3 = 5√3

= 5√3

### Rationalising the Denominator

It is untidy to have to have the fraction which has the surd denominator. This can be tidied up by multiplying the top and bottom of the fraction by the particular expression. This is known as rationalising the denominator. For example: $\frac{1}{√2}$ has an irrational denominator. We multiply the top and bottom by √2.

$\frac{1}{√2}$ = $\frac{1}{√2}$× $\frac{√2}{√2}$ = $\frac{√2}{2}$

Now the denominator has the rational number.

##### Things to remember
• Surds are square roots which can't be reduced to rational numbers.
• Surds are number left in root form to express its exact value.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.

Solution:

$\frac{2}{3√5}$ × $\frac{√5}{√5}$

= $\frac{2√5}{3×5}$

= $\frac{2√5}{15}$

Solution:

$\frac{√6+√10}{√2}$

=$\frac{√6+√10}{√2}$×$\frac{√2}{√2}$

= $\frac{√2(√6+√10)}{2}$

= $\frac{(√2+√20)}{√2}$

= $\frac{√4×3+√4×5}{2}$

= $\frac{2√3+2√5}{2}$

= $\frac{2(√3+√5)}{2}$

= √3+√5

Solution:

$\frac{3}{√2}$×$\frac{√2}{√2}$

=$\frac{3√2}{√2^2}$

=$\frac{3√2}{2}$

Solution:

$\frac{5+√3}{√5}$×$\frac{√5}{√5}$

=$\frac{5√5+√15}{5}$

Solution:

3√5+6√5

=(3+6)√5

=9√5

Solution:

3√10 - 3√10

= (3-3)√10

= 0×√10

= 0

Solution:

3√20+2√45

= 3$\sqrt{2×2×5}$+2$\sqrt{3×3×5}$

= 3$\sqrt{2^2×5}$+2$\sqrt{3^2×5}$

= 3×2√5+2×3√5

= 6√5+6√5

= (6+6)√5

= 12√5

Soutioln:

√288 - √72 + √8

= $\sqrt{2×2×2×2×2×3×3}$ - $\sqrt{2×2×2×3×3}$ + $\sqrt{2×2×2}$

= $\sqrt{2^2 × 2^2 × 2 × 3^2}$ - $\sqrt{2^2× 2 × 3^2}$ + $\sqrt{2^2 ×2}$

= 2 × 2 × 3√2 - 2 × 3√2 + 2√2

= (12-6+2)√2

= (14-6)√2

= 8√2

Solution:

√128-√50

= $\sqrt{2×2×2×2×2×2×2}$-$\sqrt{2×5^2}$

=$\sqrt{2^2×2^2×2^2×2}$-$\sqrt{2×5^2}$

= 2 × 2 × 2√2 - 5√2

= 8√2 - 5√2

= (8-5)√2

= 3√2

Solution:

$\frac{3}{√2}$+ 5

= $\frac{3×√2}{√2×√2}$+ 5

= $\frac{3√2}{(√2)^2}$+ 5

= $\frac{3√2}{2}$+ 5

= $\frac{3×√2+10}{2}$

Solution:

(5√7 × 3√5) × 4√3

= 15$\sqrt{7×5}$ × 4√3

= 15√35 × 4√3

= 60$\sqrt{35×3}$

= 60√105

Solution:

(2√3 × 3√5) + 5√15

= (6$\sqrt{3×5}$ + 5√15

= (6$\sqrt{15}$ + 5√15

= (6+5)√15

=11√15

Solution:

$\frac{2}{√5}$+$\frac{3}{√2}$

=$\frac{2}{√5}$×$\frac{√5}{√5}$+$\frac{3}{√2}$+$\frac{√2}{√2}$ (Rationalise the denominator of each term)

=$\frac{2√5}{5}$+$\frac{3√2}{2}$

=$\frac{4√5+15√2}{10}$ (LCM of 5 and 2 = 10)

Solution:

√125-√45

= $\sqrt{25×5}$ - $\sqrt{9×5}$

= 5√5 - 3√5

= 2√5

Solution:

3√2 - 4√2 + 5√2

= 3$\sqrt{2}$ + 5$\sqrt{2}$ - 4√2

= 8√2 - 4√2

= 4√2

Solution:

$\sqrt{63}$ - 2$\sqrt{28}$ + 5 $\sqrt{7}$

= $\sqrt{3\times3\times7}$ - 2$\sqrt{2\times2\times7}$ + 5$\sqrt{7}$

=$\sqrt{3^2\times7}$ - 2$\sqrt{2^2\times7}$ + 5$\sqrt{7}$

= 3$\sqrt{7}$ - 2$\times$2$\sqrt{7}$ + 5$\sqrt{7}$

= 3 $\sqrt{7}$ - 4$\sqrt{7}$ + 5$\sqrt{7}$

= (3-4+5)$\sqrt{7}$

= (8-4)$\sqrt{7}$

= 4$\sqrt{7}$

Solution:

(3 $\sqrt{7}$ + 2$\sqrt{28}$ $\times$ 4$\sqrt{7}$

=( 3 $\sqrt{7}$ + 2 $\sqrt{2\times2\times7}$) $\times$ 4$\sqrt{7}$

= (3 $\sqrt{7}$ + 2$\times$2$\sqrt{7}$ $\times$ 4$\sqrt{7}$

= (3$\sqrt{7}$ + 4$\sqrt{7}$ ) $\times$ 4$\sqrt{7}$

= (3+4) $\sqrt{7}$ $\times$ 4$\sqrt{7}$

= 7 $\sqrt{7}$ $\times$ 4$\sqrt{7}$

= 28$\sqrt{7\times7}$

= 28$\sqrt{7^2}$

= 28$\times$7

= 196

Solution:

$\sqrt{128}$ - $\sqrt{50}$

= $\sqrt{2\times2\times2\times2\times2\times2\times2}$ - $\sqrt{2\times5\times5}$

= $\sqrt{2^2\times2^2\times2^2\times2}$ - $\sqrt{2\times5^2}$

= 2$\times$2$\times$2$\sqrt{2}$ - 5$\sqrt{2}$

= 8$\sqrt{2}$ - 5$\sqrt{2}$

= (8 - 5) $\sqrt{2}$

= 3$\sqrt{2}$

Solution:

21$\sqrt{7}$ - 3$\sqrt{28}$ + $\sqrt{63}$

= 21$\sqrt{7}$ - 3$\sqrt{2\times2\times7}$ + $\sqrt{3\times3\times7}$

= 21$\sqrt{7}$ - 3$\sqrt{2^2\times7}$ + $\sqrt{3^2\times7}$

= 21$\sqrt{7}$ - 3$\times$2$\sqrt{7}$ + 3$\sqrt{7}$

= 21$\sqrt{7}$ - 6$\sqrt{7}$ + 3$\sqrt{7}$

= (21 - 6 + 3)$\sqrt{7}$

= 18$\sqrt{7}$

Solution:

3$\sqrt{20}$ + 2 $\sqrt{45}$

= 3$\sqrt{2\times2\times5}$ + 2$\sqrt{3\times3\times5}$

= 3$\times$2 $\sqrt{5}$ + 2$\times$3$\sqrt{5}$

= 3$\sqrt{2^2\times5}$ + 2$\sqrt{3^2\times5}$

= 3$\times$2$\sqrt{5}$ + 2$\times$3$\sqrt{5}$

= 6$\sqrt{5}$ + 6$\sqrt{5}$

= (6+6)$\sqrt{5}$

= 12$\sqrt{5}$