### Roots

\begin{align*} \text {We know that} \: 3 \times 3 &= 9 \\ or, \: 3^2 &= 9 \\ or, \: 3^{2 \times \frac {1}{2}} &= 9^{\frac{1}{2}} \\ or, \: 3 &=9^{\frac{1}{2}} \\ or, \: 3 &= \sqrt{9} \\ \end{align*}

The square root of 9 is 3.

\begin{align*} similarly, \: 5 \times 5\times 5 &= 125 \\ or, \: 5^3 &= 125 \\ or, 5^{3 \times \frac {1}{3}} &= (125)^{\frac{1}{3}} \\ or, \: 5 &= (125)^{\frac{1}{3}} \\ or, \: 5 &= \sqrt[3]{125} \end{align*}

The cube root of 125 is 5.
In case of equation, root indicates the value of variable.
e.g. x + 2 = 0, x = -2,
So, -2 is the root of x
e.g. x2 - 4 = 0,
(x - 2)(x +2) = 0

Either, (x - 2) = 0 or, x = 2
or, (x + 2) = 0, x = -2

So, -2 and 2 are roots of x2 -4 =0

### Surds

Surds are numerical expressions containing an irrational number. Surds may be quadratic, bi-quadratic, cubic etc.

For example: $$\sqrt{2}, \sqrt[3]{3}, \sqrt [4]{4}, \sqrt [5] {5}$$

The surds cannot be written in the form of $$\frac {p}{q}$$ q≠ 0, so they are irrational numbers.

#### Types of surds

1. Pure surds:
If the natural number is completely inside the root or radical, the surd is called pure surd. $$\sqrt {2}, \sqrt [3]{5}$$ are pure surds.
2. Mixed surd:
If the integers are inside and outside the radical, the surds is called mixed surd. 2$$\sqrt [2] {5}$$ is a mixed surd.
3. Simple surd:
The single surd which may be pure or mixed is called simple surd. $$\sqrt [4]{4} \: and \:3 \sqrt [4]{6}$$ are simple surds.
4. Compound surd
The sum or difference of two pure or mixed surds is called compound surd. $$\sqrt {5} + 2 , \: \sqrt[3]{2} + 6$$ are compoud surds.
5. Like surds
If the power (degree) of surds and number inside the root is same the surds are called like surds.$$\sqrt {5} , 2\sqrt {5}$$ are like surds.
6. Unlike surds
If a power of root is different or numbers inside the root are different the surds are called, unlike surds.$$\sqrt {5} , \sqrt [4]{4}\:and \sqrt {2}$$ are unlike surds.

#### Four fundamental operations on surds

1. Additional and subtraction of surds:
The addition and subtraction of like surds can exist, unlike surds are neither added nor subtracted. For example: $$7 \sqrt{2} + 8 \sqrt{2} = (7 + 8) \sqrt{2} = 15 \sqrt{2}$$
2. Multiplication/division of surds:
If the order of surds is same, we can put them within common root and perform the multiplication and division just like in arithmetic.
For example: a. $$\sqrt {2} \times \sqrt{3} \\ = \sqrt {2 \times 3} \\ = \sqrt {6}$$ b. $$\sqrt{10}÷ \sqrt{2} \\ = \sqrt {\frac{10}{2}} \\ = \sqrt{}5$$
If the order of surds are not same then we shall reduce these surds in the same order.
For example: $$\sqrt{3} \times \sqrt[3]{4} \\ = \sqrt [2 \times 3]{3^3} \times \sqrt [3 \times 2]{4^2} \\ = \sqrt[6]{27} \times \sqrt [6] {16}\\ = \sqrt [6]{27 \times 16} \\ = \sqrt[6] {432}$$

$$\boxed{\text {Note: Same process can be applied for division also}}$$

Surds are numerical expressions containing an irrational number. Surds may be quadratic, bi-quadratic, cubic etc.

For example: $$\sqrt2$$, $$\sqrt[3]{2}$$, $$\sqrt [4]{4}$$, $$\sqrt [5] {5}$$

The surds cannot be written in the form of $$\frac {p}{q}$$ q≠ 0, so they are irrational numbers.

Solution:

\begin{align*} &= 3\sqrt{2} + \sqrt [4] {2500} + \sqrt [4] {64} +6 \sqrt{8}\\&= 3\sqrt{2}+ \sqrt[4]{5 \times 5 \times 5 \times 5 \times 2 \times 2}+ \sqrt [4]{2 \times 2 \times 2 \times 2\times 2 \times 2}+ 6\sqrt{2 \times 2 \times 2}\\&= 3\sqrt{2}+5\sqrt[4]{4}+2\sqrt[4]{4}+12\sqrt{2}\\&=15\sqrt{2}+7\sqrt[4]{4} :_\text {Ans.} \end{align*}

\begin{align*} \sqrt [3] {16} + \sqrt [3] {54} - \sqrt [3] {250} &= \sqrt [3] {2^3 \times 2} + \sqrt [3] {3^3 \times 2} - \sqrt [3] {5^3 \times 2}\\ &= 2\sqrt [3]{2} + 3\sqrt [3]2 - 5\sqrt [3]2\\ &= 5\sqrt [3]2 - 5\sqrt [3]3\\ &= 0_{Ans}\end{align*}

\begin{align*} \frac {\sqrt {a + b} - \sqrt {a - b}}{\sqrt {a + b} + \sqrt {a - b}} &= \frac {\sqrt {a + b} - \sqrt {a - b}}{\sqrt {a + b} + \sqrt {a - b}} \times \frac {\sqrt {a + b} - \sqrt {a - b}}{\sqrt {a + b} - \sqrt {a - b}}\\ &= \frac {(\sqrt {a + b} - \sqrt {a - b})^2}{(\sqrt {a + b})^2 - (\sqrt {a - b})^2}\\ &= \frac {a + b + a - b - 2 \sqrt {a + b}\sqrt {a - b}}{a + b - a + b}\\ &= \frac {2a - 2\sqrt {a^2 - b^2}}{2b}\\ &= \frac {2(a - \sqrt {a^2 - b^2})}{2b}\\ &= \frac {a - \sqrt {a^2 - b^2}}b_{Ans}\\ \end{align*}

$$\frac {\sqrt y + \sqrt 5}{\sqrt y - \sqrt 5}$$ = 3

or, $$\sqrt y$$ + $$\sqrt 5$$ = 3$$\sqrt y$$ - 3$$\sqrt 5$$

or, 3$$\sqrt y$$ - $$\sqrt y$$ = $$\sqrt 5$$ + 3$$\sqrt 5$$

or, 2$$\sqrt y$$ = 4$$\sqrt 5$$

or, $$\sqrt y$$ = $$\frac {4\sqrt 5}{2}$$

or, $$\sqrt y$$ = 2$$\sqrt 5$$

Squaring on both sides,

($$\sqrt y$$)2= (2$$\sqrt 5$$)2

or, y = 4× 5

∴ y = 20Ans

$$\frac {2y + 3}{\sqrt y - 1}$$ = $$\frac 13$$ ($$\sqrt y$$ + 1)

or, 6y + 9 = ($$\sqrt y$$ + 1) ($$\sqrt y$$ - 1)

or, 6y + 9 = ($$\sqrt y$$)2 - (1)2

or, 6y + 9 = y - 1

or, 6y - y = -1 -9

or, 5y = - 10

or, y = $$\frac {-10}5$$

∴ y = -2Ans

$$\sqrt x$$ + 1 = 5 - $$\frac {\sqrt x -1}2$$

or, $$\sqrt x$$ + 1 - 5 = $$\frac {- (\sqrt x - 1)}{2}$$

or, $$\sqrt x$$ - 4 = $$\frac {-\sqrt x + 1}{2}$$

or, 2$$\sqrt x$$ - 8 + $$\sqrt x$$ = 1

or, 3$$\sqrt x$$ = 1 + 8

or, 3$$\sqrt x$$ = 9

or, $$\sqrt x$$ = $$\frac 93$$

or, $$\sqrt x$$ = 3

Squaring on both sides,

($$\sqrt x$$)2 = 32

∴ x = 9 Ans

$$\sqrt x$$ + $$\sqrt {x - 20}$$ = 10

or, $$\sqrt {x - 20}$$ = 10 - $$\sqrt x$$

Squaring on both sides,

($$\sqrt {x - 20}$$)2 = (10 - $$\sqrt x$$)2

or, x - 20 = 100 - 20$$\sqrt x$$ + x

or, 20$$\sqrt x$$ = 100 + 20

or, $$\sqrt x$$ = $$\frac {120}{20}$$

or, $$\sqrt x$$ = 6

Squaring on both sides,

($$\sqrt x$$)2 = 62

∴ x = 36Ans

$$\sqrt {4 (x + 1)}$$ = $$\sqrt {4x}$$ + 1

Squaring on both sides,

($$\sqrt {4 (x + 1)}$$)2 = ($$\sqrt {4x}$$ + 1)2

or, 4(x + 1) = 4x + 2$$\sqrt {4x}$$ + 1

or, 4x + 4 - 4x - 1 = 2$$\sqrt {4x}$$

or, $$\frac 32$$ = $$\sqrt {4x}$$

Again,

Squaring on both sides,

($$\sqrt {4x}$$)2 = ($$\frac 32$$)2

or, 4x = $$\frac 94$$

or, x = $$\frac 9{4 \times 4}$$

∴ x = $$\frac 9{16}_{Ans}$$

\begin{align*} \frac {3\sqrt 2}{\sqrt 6 - \sqrt 3} - \frac {4\sqrt 3}{\sqrt 6 - \sqrt 2} + \frac {2\sqrt 3}{\sqrt 6 + \sqrt 2} &= \frac {3\sqrt 2}{\sqrt 6 - \sqrt 3} \times \frac {\sqrt 6 + \sqrt 3}{\sqrt 6 + \sqrt 3} - \frac {4\sqrt 3}{\sqrt 6 - \sqrt 2} \times \frac {\sqrt 6 + \sqrt 2}{\sqrt 6 + \sqrt 2} + \frac {2\sqrt 3}{\sqrt 6 + \sqrt 2} \times \frac {\sqrt 6 - \sqrt 2}{\sqrt 6 - \sqrt 2}\\ &= \frac {3\sqrt {12} + 3\sqrt 6}{(\sqrt 6)^2 - (\sqrt 3)^2} - \frac {4\sqrt {18} + 4\sqrt {6}}{(\sqrt 6)^2 - (\sqrt 2)^2} + \frac {2\sqrt {18} - 2\sqrt 6}{(\sqrt 6)^2 - (\sqrt 2)^2}\\ &= \frac {3 \times 2\sqrt 3 + 3\sqrt 6}{6 - 3} - \frac {4 \times 3\sqrt 2 + 4\sqrt 6}{6 - 2} + \frac {2 \times 3\sqrt 2 - 2\sqrt 6}{6 - 2}\\ &= \frac {3(2\sqrt 3 + \sqrt 6)}{3} - \frac {4(3\sqrt 2 + \sqrt 6)}{4} + \frac {2(3\sqrt 2 - \sqrt 6)}{4}\\ &= 2\sqrt 3 + \sqrt 6 - 3\sqrt 2 - \sqrt 6 + \frac {3\sqrt 2 - \sqrt 6}2\\ &= \frac {2\sqrt 3 - 3\sqrt 2}{1} + \frac {3\sqrt 2 - \sqrt 6}{2}\\ &= \frac {4\sqrt 3 - 6\sqrt 2 + 3\sqrt 2 - \sqrt 6}{2}\\ &= \frac {4\sqrt 3 - 3\sqrt 2 - \sqrt 6}2_{Ans} \end{align*}

\begin{align*} \frac {1 + \sqrt 3}{1 - \sqrt 3} + \frac {2 + \sqrt 3}{2 - \sqrt 3} - \frac {1 + 2\sqrt 3}{2 + \sqrt 3} &= \frac {1 + \sqrt 3}{1 - \sqrt 3} \times \frac {1 + \sqrt 3}{1 + \sqrt 3}+ \frac {2 + \sqrt 3}{2 - \sqrt 3} \times \frac {2 + \sqrt 3}{2 + \sqrt 3} - \frac {1 + 2\sqrt 3}{2 + \sqrt 3} \times \frac {2 - \sqrt 3}{2 - \sqrt 3}\\ &= \frac {(1 + \sqrt 3)^2}{(1)^2 - (\sqrt 3)^2} + \frac {(2 + \sqrt 3)^2}{(2)^2 - (\sqrt 3)^2} - \frac {(2 - \sqrt 3 + 4\sqrt 3 - 2(\sqrt 3)^2)}{(2)^2 - (\sqrt 3)^2}\\ &= \frac {1^2 + 2\sqrt 3 +(\sqrt 3)^2}{1 - 3} + \frac {2^2 + 4\sqrt 3 + (\sqrt 3)^2}{4 - 3} - \frac {(2 + 3\sqrt 3 - 6)}{4 - 3}\\ &= \frac {1 + 3 + 2\sqrt 3}{-2} + \frac {4 + 4\sqrt 3 + 3}{1} - \frac {(3\sqrt 3 - 4)}{1}\\ &= \frac {4 + 2\sqrt 3}{-2} + 7 + 4\sqrt 3 - 3\sqrt 3 + 4\\ &= \frac {2(2 + \sqrt 3)}{-2} + 11 + \sqrt 3\\ &= -2 -\sqrt 3 + 11 + \sqrt 3\\ &= 9_{Ans}\\ \end{align*}

$$\frac {3\sqrt x - 4}{\sqrt x + 2}$$ = $$\frac {15 + 3\sqrt x}{\sqrt x + 40}$$

or, (3$$\sqrt x$$ - 4) ($$\sqrt x$$ + 40) = (15 + 3$$\sqrt x$$) ($$\sqrt x$$ + 2)

or, 3x + 120$$\sqrt x$$ - 4$$\sqrt x$$ - 160 = 15$$\sqrt x$$ + 30 + 3x + 6$$\sqrt x$$

or, 3x - 3x + 116$$\sqrt x$$ - 160 = 21$$\sqrt x$$ + 30

or, 116$$\sqrt x$$ - 21$$\sqrt x$$ = 160 + 30

or, 95$$\sqrt x$$ = 190

or, $$\sqrt x$$ = $$\frac {190}{95}$$

or, $$\sqrt x$$ = 2

Squaring on both sides,

($$\sqrt x$$)2 = (2)2

∴ x = 4Ans

$$\frac {x - 1}{\sqrt x + 1}$$ = 4 + $$\frac {\sqrt x - 1}2$$

or, $$\frac {x - 1}{\sqrt x + 1}$$ = $$\frac {8 + \sqrt x - 1}2$$

or,$$\frac {x - 1}{\sqrt x + 1}$$ = $$\frac {7 + \sqrt x}2$$

or, 2x - 2 = 7$$\sqrt x$$ + 7 + x + $$\sqrt x$$

or, 2x - x - 2 = 8$$\sqrt x$$ + 7

or, x - 2 - 7 = 8$$\sqrt x$$

or, x - 9 = 8$$\sqrt x$$

Squaring on both sides,

(x - 9)2 = (8$$\sqrt x$$)2

or, x2 - 18x + 81 = 64x

or, x2 - 18x + 81 - 64x = 0

or, x2 - 82x + 81 = 0

or, x2 - 81x - x + 81 = 0

or, x(x - 81) - 1 (x - 81) = 0

or, (x - 81) (x - 1) = 0

Either,

x - 81 = 0

∴ x = 81

Or,

x - 1 = 0

∴ x = 1

Putting the value of x = 1 is false.

∴ x = 81Ans

$$\frac {\sqrt {x + 4} + \sqrt {x - 4}}{\sqrt {x + 4} - \sqrt {x - 4}}$$ = 3

or, $$\sqrt {x + 4}$$ + $$\sqrt {x - 4}$$ = 3$$\sqrt {x + 4}$$ - 3$$\sqrt {x - 4}$$

or, $$\sqrt {x - 4}$$ + 3$$\sqrt {x - 4}$$ = 3$$\sqrt {x + 4}$$ - $$\sqrt {x + 4}$$

or, 4$$\sqrt {x - 4}$$ = 2$$\sqrt {x + 4}$$

or, $$\frac {\sqrt {x + 4}}{\sqrt {x - 4}}$$ = $$\frac 42$$

or, $$\frac {\sqrt {x + 4}}{\sqrt {x - 4}}$$ = 2

Squaring on both sides,

( $$\frac {\sqrt {x + 4}}{\sqrt {x - 4}}$$)2 = 22

or, $$\frac {x + 4}{x - 4}$$ = 4

or, x + 4 = 4x - 16

or, 4x - x = 16 + 4

or, 3x = 20

∴ x = $$\frac {20}3_{Ans}$$

$$\sqrt x$$ + $$\sqrt {x + 13}$$ = $$\frac {91}{\sqrt {x + 13}}$$

or, $$\sqrt {x + 13}$$ ($$\sqrt x$$ + $$\sqrt {x + 13}$$) = 91

or, $$\sqrt x$$ $$\sqrt {x + 13}$$ + ($$\sqrt {x + 13}$$)2 = 91

or, $$\sqrt x$$ $$\sqrt {x + 13}$$ + x + 13 = 91

or, $$\sqrt x$$ $$\sqrt {x + 13}$$ = 91 - 13 - x

or, $$\sqrt x$$ $$\sqrt {x + 13}$$ = 78 - x

Squaring on both sides,

($$\sqrt x$$ $$\sqrt {x + 13}$$)2 = (78 - x)2

or, x (x + 13) = (78)2 - 2 . 78 . x + x2

or, x2 + 13x = 6084 - 156x + x2

or, x2 + 13x + 156x - x2 = 6084

or, 169x = 6084

or, x = $$\frac {6084}{169}$$

∴ x = 36Ans

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

Squaring on both sides,

($$\sqrt {4x + 5}$$ - $$\sqrt x$$)2 = ($$\sqrt {x + 3}$$)

or, ($$\sqrt {4x + 5}$$)2 - 2$$\sqrt {4x + 5}$$ . $$\sqrt x$$ + ($$\sqrt x$$)2 = x + 3

or, 4x + 5 - 2$$\sqrt {4x^2 - 5x}$$ + x - x - 3 = 0

or, 4x + 2 =2$$\sqrt {4x^2 - 5x}$$

Again,

Squaring on both sides,

(4x + 2)2 = (2$$\sqrt {4x^2 - 5x}$$)2

or, 16x2 + 16x + 4 = 4(4x2 + 5x)

or, 16x2 + 16x + 4 = 16x2 + 20x

or, 16x2 + 16x + 4 - 16x2 - 20x = 0

or, -4x + 4 = 0

or, -4x = -4

or, x = $$\frac 44$$

∴ x = 1Ans

$$\sqrt {x - \sqrt {1 - x}}$$ = 1 - $$\sqrt x$$

Squaring on both sides,

($$\sqrt {x - \sqrt {1 - x}}$$)2 = (1 - $$\sqrt x$$)2

or, x - $$\sqrt {1 - x}$$ = 1 - 2$$\sqrt x$$ + ($$\sqrt x$$)2

or, x - $$\sqrt {1 - x}$$ = 1 - 2$$\sqrt x$$ + x

or, x - x - $$\sqrt {1 - x}$$ = 1 - 2$$\sqrt x$$

or, -$$\sqrt {1 - x}$$ = 1 - 2$$\sqrt x$$

Again,

Squaring on both sides,

(-$$\sqrt {1 - x}$$)2 = (1 - 2$$\sqrt x$$)2

or, 1 - x = 1 - 4$$\sqrt x$$ + 4x

or, 1 - x - 1 - 4x = -4$$\sqrt x$$

or, -5x = -4$$\sqrt x$$

Again,

Squaring on both sides,

(-5x)2 = (-4$$\sqrt x$$)2

or, 25x2 = 16x

or, 25x2 - 16x = 0

or, x(25x - 16) = 0

Either,

x = 0 (impossible)

Or,

25x - 16 = 0

or, 25x = 16

∴ x = $$\frac {16}{25}$$

∴ x = $$\frac {16}{25}_{Ans}$$

$$\sqrt {x^2 - 3x + 3}$$ + $$\sqrt {x^2 - x + 1}$$ = 2

or,$$\sqrt {x^2 - 3x + 3}$$ = 2 -$$\sqrt {x^2 - x + 1}$$

Squaring on both sides,

($$\sqrt {x^2 - 3x + 3}$$)2 = (2 -$$\sqrt {x^2 - x + 1}$$)2

or, x2 - 3x + 3 = 4 - 4$$\sqrt {x^2 - x + 1}$$ + ($$\sqrt {x^2 - x + 1}$$)2

or, x2 - 3x + 3 - 4 = - 4$$\sqrt {x^2 - x + 1}$$ + x2 - x + 1

or, x2 - 3x - 1 - x2 + x - 1 =- 4$$\sqrt {x^2 - x + 1}$$

or, - 2x - 2 =- 4$$\sqrt {x^2 - x + 1}$$

or, -2 (x + 1) =- 2$$\sqrt {x^2 - x + 1}$$

or, x + 1 = $$\sqrt {x^2 - x + 1}$$

Again,

Squaring on both sides,

(x + 1)2 = ($$\sqrt {x^2 - x + 1}$$)2

or, x2 + 2x + 1 = 4(x2 - x + 1)

or, x2 + 2x + 1 = 4x2 - 4x + 4

or, 4x2 - x2 - 4x - 2x + 4 - 1 = 0

or, 3x2 - 6x + 3 = 0

or, 3(x2- 2x + 1) = 0

or, x2 - 2x + 1 = 0

or, (x)2 - 2 . x . 1 + (1)2 = 0

or, (x - 1)2 = 0

Removing square from both sides,

x - 1 = 0

∴ x = 1Ans

$$\frac {a - 2}{a^2 - 2a + 4}$$ + $$\frac {a + 2}{a^2 + 2a + 4}$$ - $$\frac {16}{a^4 + 16 + 4a^2}$$

= $$\frac {(a - 2) (a^2 + 2a + 4) + (a + 2) (a^2 - 2a + 4)}{(a^2 - 2a + 4) (a^2 + 2a + 4)}$$ - $$\frac {16}{(a^2)^2 + 2 . a^2 . 4 + (4)^2 - 4a^2}$$

= $$\frac {a^3 - 2^3 + a^3 + 2^3}{(a^2 - 2a + 4) (a^2 + 2a + 4)}$$ - $$\frac {16}{(a^2 + 4)^2 - (2a)^2}$$

= $$\frac {2a^3}{(a^2 - 2a + 4) (a^2 + 2a + 4)}$$ - $$\frac {16}{(a^2 + 2a + 4) (a^2 - 2a + 4)}$$

= $$\frac {2a^3 - 16}{(a^2 - 2a + 4) (a^2 + 2a + 4)}$$

= $$\frac {2(a^3 - 8)}{(a^2 - 2a + 4) (a^2 + 2a + 4)}$$

= $$\frac {2(a^3 - 2^3)}{(a^2 - 2a + 4) (a^2 + 2a + 4)}$$

= $$\frac {2(a - 2) (a^2 + 2a + 4)}{(a^2 - 2a + 4) (a^2 + 2a + 4)}$$

= $$\frac {2(a - 2)}{(a^2 - 2a + 4)}_{Ans}$$

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