Chapter – 9 : Quadratic Surd

Ganit Prakash Solution 10

Book Name	: Ganit Prakash
Subject	: Mathematics
Class	: 10 (Madhyamik)
Publisher	: Prof Nabanita Chatterjee
Chapter Name	: Quadratic Surd (9th Chapter)

Table of Contents

Question 6

I write by understanding 4 pure quadratic surds and 4 mixed quadratic surds.

Solution:

4 pure quadratic surds are $\sqrt{3}, - \sqrt{5}, \sqrt{\frac{2}{3}}, \sqrt{2}.$

4 mixed quadratic surds are $2 - \sqrt{3}, 2 + \sqrt{6}, \frac{3}{5} - \sqrt{10}, 3 + \sqrt{5}.$

Question 7

Are $\sqrt{4}, \sqrt{25}$ quadratic surds ?

Solution:

Apparently $\sqrt{4}, \sqrt{25}$ are in the form of surds but they are not surds. Rational number, $\sqrt{4}$ = 2 and $\sqrt{25}$ = 5.

I apply Sreedhar Acharyya’s formula for solving the equation x² – 2 ax + (a² – b² = 0). We see that the roots are $a + \sqrt{b}$ and $a - \sqrt{b}$ , both of which are mixed surds, where b is a positive rational number which is not a square number of any rational number.

Question 8

What type of number we get by the addition, subtraction, multiplication, division and square of the two numbers 8 and 12 ? [Let me do it myself]

Solution:

8 + 12 = 20 (Integer)

8 – 12 = – 4 (Integer)

8 × 12 = 96 (Integer)

$\frac{8}{12} = \frac{2}{3}$ (Rational)

Application 3.

We write similar surds in a specific place from the following quadratic surds.

$\sqrt{45}, \sqrt{80}, \sqrt{147}, \sqrt{180} and \sqrt{500}$

Solution:

$\sqrt{45} = \sqrt{9 × 5} = 3 \sqrt{5} \quad \sqrt{147} = \sqrt{7 × 7 × 3} = 7 \sqrt{3}$

$\sqrt{80} = \sqrt{16 × 5} = 4 \sqrt{5} \quad \sqrt{180} = \sqrt{6 × 6 × 5} = 6 \sqrt{5} \\$

$\sqrt{500} = \sqrt{2 × 2 × 5 × 5 × 5} = 2 × 5 \sqrt{5} = 10 \sqrt{5}$

Application 4.

Let us write the similar surds among the quadratic surds $\sqrt{48} \sqrt{27}, \sqrt{20} \text{ and } \sqrt{75}$ [Let me do it myself].

Solution:

$\sqrt{48} = \sqrt{2 × 2 × 2 × 2 × 3} = 4 \sqrt{3}$

$\sqrt{27} = \sqrt{3 × 3 × 3} = 3 \sqrt{3} \\$

$\sqrt{20} = \sqrt{2 × 2 × 5} = 2 \sqrt{5} \\$

$\sqrt{75} = \sqrt{5 × 5 × 3} = 5 \sqrt{3} \\$

$\therefore \sqrt{48}, \sqrt{27}, \sqrt{75}$ are similar surds.

Application 6.

Let us write by calculating the value of $(\sqrt{12} + \sqrt{45}) \text{ and } (\sqrt{2} - \sqrt{8})$ and see whether they can be expressed in pure quardratic surds. [Let me do it myself]

Solution:

$\sqrt{2} + \sqrt{8} = \sqrt{2} + \sqrt{2 × 2 × 2} = \sqrt{2} + 2 \sqrt{2} = 3 \sqrt{2}$

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

$\therefore(\sqrt{2} - \sqrt{8})$ is a pure quadratic surd.

Application 9.

Let us write by calculating the sum of $\sqrt{12}, - 4 \sqrt{3} and \sqrt{3}$ [Let me do it myself].

Solution:

$(\sqrt{12}) + ( - 4 \sqrt{3}) + 6 \sqrt{3} = 2 \sqrt{3} - 4 \sqrt{3} + 6 \sqrt{3} = 4 \sqrt{3}.$

Application 11.

$(9 - 2 \sqrt{5}) + (12 + 7 \sqrt{5}) = \square$ [Let me do it myself]

Solution:

$(9 - 2 \sqrt{5}) + (12 + 7 \sqrt{5}) = 9 + 12 - 2 \sqrt{5} = 21 + 5 \sqrt{5}$

Application 13.

I write any other two quadratic surds whose sum is a rational number: [Let me do it myself]

Solution:

$(6 + \sqrt{7}) + (6 - \sqrt{7}) = 6 + 6 + \sqrt{7} - \sqrt{7}$ = 12

LET US WORK OUT – 9.1

Question 1

Let us write the following numbers in the form of product of rational and irrational numbers.

(i) $\sqrt{175}$

Solution:

$\sqrt{175} = \sqrt{5 × 5 × 7}$

$= 5 \sqrt{7}$

(ii) $2 \sqrt{112}$

Solution:

$2 \sqrt{112} = 2 \cdot \sqrt{4 × 4 × 7}$

$= 2 × 4 \sqrt{7} = 8 \sqrt{7}$

(iii) $\sqrt{108}$

Solution:

$\sqrt{108} = \sqrt{2 × 2 × 3 × 3 × 3}$

$= 2 × 3 \sqrt{3} = 6 \sqrt{3}$

(iv) $\sqrt{125}$

Solution:

$\sqrt{125} = \sqrt{5 × 5 × 5}$

$= 5 \sqrt{5}$

(v) $5 \sqrt{119}$

Solution:

$5 \sqrt{119} = 5 \sqrt{7 × 17}$

$= 5 \sqrt{119}$

Question 2

Let us show that $\sqrt{108} - \sqrt{75} = \sqrt{3}$

Solution:

$\sqrt{109} - \sqrt{75} = \sqrt{3}$

L.H.S $= \sqrt{108} - \sqrt{75} = \sqrt{6 × 6 × 3} - \sqrt{5 × 5 × 3} \\$

$= 6 \sqrt{3} - 5 \sqrt{3} = \sqrt{3}$ = R.H.S.

Question 3

Let us show that $\sqrt{98} + \sqrt{8} - 2 \sqrt{32} = \sqrt{2}$

Solution:

$\sqrt{98} + \sqrt{8} - 2 \sqrt{32} = \sqrt{2}$

L.H.S $= \sqrt{98} + \sqrt{8} - 2 \sqrt{32} = \sqrt{7 × 7 × 2} + \sqrt{2 × 2 × 2} - 2 \sqrt{4 × 4 × 2} \\$

$= 7 \sqrt{2} + 2 \sqrt{2} - 2 × 4 \sqrt{2} = 9 \sqrt{2} - 8 \sqrt{2} = \sqrt{2}$ = R.H.S

Question 4

Let us show that $3 \sqrt{48} - 4 \sqrt{75} + \sqrt{192}$ = 0

Solution:

$3 \sqrt{48} - 4 \sqrt{75} + \sqrt{192}$ = 0

L.H.S. $= 3 \sqrt{48} - 4 \sqrt{75} + \sqrt{192} \\$

$= 3 \sqrt{4 × 4 × 3} - 4 \sqrt{5 × 5 × 3} + \sqrt{8 × 8 × 3} \\$

$= 3 × 4 \sqrt{3} - 4 × 5 \sqrt{3} + 8 \sqrt{3} = 12 \sqrt{3} - 20 \sqrt{3} + 8 \sqrt{3} \\$

$= 20 \sqrt{3} - 20 \sqrt{3}$ = 0 = R.H.S

Question 5

Let us simplify: $\sqrt{12} + \sqrt{18} + \sqrt{27} - \sqrt{32}$

Solution:

$\sqrt{12} + \sqrt{18} + \sqrt{27} - \sqrt{32} \\$

$= \sqrt{2 × 2 × 3} + \sqrt{3 × 3 × 2} + \sqrt{3 × 3 × 3} - \sqrt{4 × 4 × 2} \\$

$= 2 \sqrt{3} + 3 \sqrt{2} + 3 \sqrt{3} - 4 \sqrt{2} \\$

$= 2 \sqrt{3} + 3 \sqrt{3} + 3 \sqrt{3} - 4 \sqrt{2} \\$

$= 5 \sqrt{3} - \sqrt{2}$

Question 6

(a) Let us write what should be added with $\sqrt{5} + \sqrt{3} \text{ to get the sum } 2 \sqrt{5}.$

Solution:

Required number $= 2 \sqrt{5} - (\sqrt{5} + \sqrt{3}) = 2 \sqrt{5} - \sqrt{5} - \sqrt{3} = \sqrt{5} - \sqrt{3}$

(b) Let us write what should be subtracted from $7 - \sqrt{3} \text{ to get the sum } 3 + \sqrt{3}.$

Solution:

Required number $= (7 - \sqrt{3}) - (3 + \sqrt{3})$

$= 7 - \sqrt{3} - 3 - \sqrt{3} = 7 - 3 - \sqrt{3} = 4 - 2 \sqrt{3}.$

(c) Let us write the sum of $2 + \sqrt{3}, \sqrt{3} + \sqrt{5} \text{ and } 2 + \sqrt{7}.$

Solution:

Required sum $= 2 + \sqrt{3} + \sqrt{3} + \sqrt{5} + 2 + \sqrt{7} = 2 + 2 + \sqrt{3} + \sqrt{3} + \sqrt{5} + \sqrt{7}$

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

$= 4 + 2 \sqrt{3} + \sqrt{5} + \sqrt{7}$

(d) Let us subtract $( - 5 + 3 \sqrt{11}) \text{ from } (10 + \sqrt{11})$ and let us write the value of the subtraction.

Solution:

Required subtraction $= (10 - \sqrt{11}) - ( - 5 + 3 \sqrt{11})$

$= 10 - \sqrt{11} + 5 - 3 \sqrt{11} = 15 - 4 \sqrt{11}$

(e) Let us subract $(5 + \sqrt{2} + \sqrt{7})$ from the sum of $( - 5 + \sqrt{7}) \text{ and } (\sqrt{7} + \sqrt{2})$ and find the value of the subtraction.

Solution:

Required value of subtraction $= ( - 5 + \sqrt{7}) + (\sqrt{7} + \sqrt{2}) - (5 + \sqrt{2} + \sqrt{7}) = - 5 + \sqrt{7} + \sqrt{7} + \sqrt{2} - 5 - \sqrt{2} - \sqrt{7} = - 10 + \sqrt{7}$

(f) I write two quadratic surds whose sum is a rational number.

Solution:

Two quadratic surds whose sum is a rational number, are:

$5 + \sqrt{3} ; 5 - \sqrt{3}.$

Application 16.

Let us write by calculating the product of $(3 + \sqrt{7} - \sqrt{5}) \text{ and } (2 \sqrt{2} - 1)$ [Let me do it myself].

Solution:

$(3 + \sqrt{7} - \sqrt{5}) ×(2 \sqrt{2} - 1)$

$= 6 \sqrt{2} - 3 + 2 \sqrt{14} - \sqrt{7} - 2 \sqrt{10} - \sqrt{5}$

Application 18.

Let us write two rationalising factors of $\sqrt{7}$ . [Let me do it myself]

Solution:

$\sqrt{7} \ \& \ 2 \sqrt{7}$

Application 19.

Let us see what will be the rationalising factor of $(5 + \sqrt{7}).$

Solution:

$(5 + \sqrt{7})$

$= (5 + \sqrt{7}) ×(5 - \sqrt{7}) = (5)^2 - (\sqrt{7})^2 = 25 - 7 = 18\left[\because(a + b)(a - b) = a^2 - b^2\right]$

Again, $(5 + \sqrt{7}) ×( - 5 + \sqrt{7}) = (\sqrt{7} + 5)(\sqrt{7} - 5) = (\sqrt{7})^2 - (5)^2$ = – 18

Application 20.

Let us write two rationalising factors of $7 - \sqrt{3}$ [Let me do it myself].

Solution:

$7 - \sqrt{3}$

$(7 + \sqrt{3}) ;( - 7 - \sqrt{3})$

Application 21.

Let us see the rationalising factors of $(\sqrt{11} - \sqrt{6}).$

Solution:

$(\sqrt{11} - \sqrt{6})(\sqrt{11} + \sqrt{6}) = (\sqrt{11})^2 - (\sqrt{6})^2$ = 11 – 6 = 5

Again, $(\sqrt{11} - \sqrt{6})( - \sqrt{11} - \sqrt{6}) = - \lfloor(\sqrt{11} - \sqrt{6})(\sqrt{11} + \sqrt{6})\rfloor = [11 - 6] = - 5$

Application 22.

Let us write two rationalising factors of $(\sqrt{15} + \sqrt{3})$ [Let me do it myself].

Solution:

$(\sqrt{15} + \sqrt{3})$

$(\sqrt{15} - \sqrt{3}) ;( - \sqrt{15} + \sqrt{3})$

Application 25.

Let us write the conjugate surds of the following mixed and pure surds :

(i) $2+\sqrt{3}$

Solution:

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

(ii) $5-\sqrt{2}$

Solution:

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

(iii) $\sqrt{5}-7$

Solution:

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

(iv) $\sqrt{11}+6$

Solution:

$\sqrt{11}+6 =(6-\sqrt{11})$

(v) $\sqrt{5}$

Solution:

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

Application 28.

Let us rationalise the denominator of

(i) $\frac{4 \sqrt{5}}{5 \sqrt{3}}$

Solution:

$\frac{4 \sqrt{5}}{5 \sqrt{3}} =\frac{4 \sqrt{5} \cdot \sqrt{3}}{5 \sqrt{3} \cdot \sqrt{3}}=\frac{4 \sqrt{3}}{5 \times 3}=\frac{4 \sqrt{3}}{15}$

(ii) $\frac{3 \sqrt{7}}{\sqrt{6}}$

Solution:

$\frac{3 \sqrt{7}}{\sqrt{6}}$

$=\frac{3 \sqrt{7}}{\sqrt{6}}=\frac{3 \sqrt{7} \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}}=\frac{3 \sqrt{42}}{6}=\frac{\sqrt{42}}{2}$

Application 30.

Let us rationalise the denominator of

(i) $(4 + 2 \sqrt{3}) \div(2 - \sqrt{3})$

Solution:

$\frac{4 + 2 \sqrt{3}}{2 - \sqrt{3}}$

$= \frac{(4 + 2 \sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{8 + 4 \sqrt{3} + 4 \sqrt{3} + 6}{(2)^2 - (\sqrt{3})^2} = \frac{14 + 8 \sqrt{3}}{4 - 3}$

$= 14 + 8 \sqrt{3}$

(ii) $(\sqrt{5} + \sqrt{3}) \div(\sqrt{5} - \sqrt{3})$

Solution:

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

$= \frac{(\sqrt{5} + \sqrt{3})(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})} = \frac{5 + 2 \sqrt{5} \cdot \sqrt{3} + 3}{(\sqrt{5})^2 - (\sqrt{3})^2} = \frac{8 + 2 \sqrt{15}}{5 - 3} \\$

$= \frac{2(4 + \sqrt{15})}{2} \\$

$= 4 + \sqrt{15}$

LET US WORK OUT – 9.2

Question 1

(a) Let us find the product of $3^{1 / 2} \ and \ \sqrt{3}.$

Solution:

$3^{1 / 2} \times \sqrt{3}$

$= 3^{1 / 2} \times 3^{1 / 2} = 3^{1 / 2+1 / 2}$ = 3¹ = 3.

(b) Let us write what should be multiplied with $2 \sqrt{2}$ to get the product 4.

Solution:

Required Number $= \frac{4}{2 \sqrt{2}} = \frac{2 \times 2}{2 \sqrt{2}} = \frac{2}{\sqrt{2}} = \frac{2 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}$

$= \frac{2 \sqrt{2}}{2} = \sqrt{2}$

(c) Let us calculate the product of $3 \sqrt{5} \ and \ 5 \sqrt{3}.$

Solution:

$3 \sqrt{5} \times 5 \sqrt{3}$

$= 3 \times 5 \times \sqrt{5} \times \sqrt{3} = 15 \sqrt{15}$

(d) If $\sqrt{6} \times \sqrt{15} = x \sqrt{10}$ , then let us write by calculating the value of x.

Solution : If $\sqrt{6} \times \sqrt{15} = x \sqrt{10}$

or, $\sqrt{90} = x \sqrt{10}$

or, $\sqrt{9} \times \sqrt{10} = x \cdot \sqrt{10}$

$\therefore \ x = \sqrt{9}$ = 3

(e) If $(\sqrt{5}+\sqrt{3})(\sqrt{5} - \sqrt{3}) = 25 - x^2$ be an equation, then let us write by calculating the value of x.

Solution:

If $(\sqrt{5}+\sqrt{3})(\sqrt{5} - \sqrt{3}) = 25 - x^2$

or, $(\sqrt{5})^2 - (\sqrt{3})^2$ = 25 – x^2

or, 5 – 3 = 25 – x²

or, 2 = 25 – x²

or, x² = 25 – 2 = 23

$\therefore x = \pm \sqrt{23}$

Question 2

Let us calculate the product :

(a) $\sqrt{7} \times \sqrt{14}$

Solution:

$\sqrt{7} \times \sqrt{14} = \sqrt{7} \times \sqrt{7 \times 2} = \sqrt{7} \times \sqrt{7} \times \sqrt{2} = 7 \sqrt{2}$ Ans.

(b) $\sqrt{12} \times 2 \sqrt{3}$

Solution:

$\sqrt{12} \times 2 \sqrt{3} = \sqrt{2 \times 2 \times 3} \times 2 \sqrt{3} = 2 \sqrt{3} \times 2 \sqrt{3} = 2 \times 2 \times \sqrt{3} \cdot \sqrt{3}$ = 4 × 3 = 12 Ans.

(c) $\sqrt{5} \times \sqrt{15} \times \sqrt{3}$

Solution:

$\sqrt{5} \times \sqrt{15} \times \sqrt{3} = \sqrt{15} \times \sqrt{15} = \sqrt{15 \times 15}$ = 15 Ans.

(d) $\sqrt{2} \times(3+\sqrt{5})$

Solution:

$\sqrt{2} \times(3+\sqrt{5}) = 3 \sqrt{2}+\sqrt{2} \times \sqrt{5} = 3 \sqrt{2}+\sqrt{10}$ Ans.

(e) $(\sqrt{2}+\sqrt{3})(\sqrt{2} - \sqrt{3})$

Solution:

$(\sqrt{2}+\sqrt{3})(\sqrt{2} - \sqrt{3}) = (\sqrt{2})^2 - (\sqrt{3})^2$ = 2 – 3 = – 1 Ans.

(f) $(2 \sqrt{3}+3 \sqrt{2})(4 \sqrt{2}+\sqrt{5})$

Solution:

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

$= 8 \sqrt{6}+2 \sqrt{15}+12 \sqrt{4}+3 \sqrt{10} \\$

$= 8 \sqrt{6}+2 \sqrt{15}+12.2+3 \sqrt{10} \\$

$= 8 \sqrt{6}+2 \sqrt{15}+24+3 \sqrt{10}$

(g) $(\sqrt{3}+1)(\sqrt{3} - 1)(2 - \sqrt{3})(4+2 \sqrt{3})$

Solution:

$(\sqrt{3}+1)(\sqrt{3} - 1)(2 - \sqrt{3})(4+2 \sqrt{3})$

$= \left\{(\sqrt{3})^2 - (1)^2\right\}(2 - \sqrt{3})(4+2 \sqrt{3}) \\$

$= (3 - 2)(2 - \sqrt{3}) \times 2(2+\sqrt{3}) \\$

$= 2 \times 2 \times(2 - \sqrt{3})(2+\sqrt{3}) \\$

$= 4 \times\left\{(2)^2 - (\sqrt{3})^2\right\} \\$

$= 4 \times(4 - 3) \\$

= 4 × 1 = 4 Ans.

Question 3.

(a) If $\sqrt{x}$ is the rationalising factor of $\sqrt{5}$ , let us write by calculating the smallest value of x (where x is an integer).

Solution:

x = $\sqrt{5}$

(b) Let us calculate the value of $3 \sqrt{2} \div 3.$

Solution:

$3 \sqrt{2} \div 3 = \frac{3 \sqrt{2}}{3} = \sqrt{2}$

(c) Let us write which smallest factor we should multiply with the denominator to rationalise the denominator of $7 \div \sqrt{48}.$

Solution:

$7 \div \sqrt{48} = \frac{7}{\sqrt{48}} = \frac{7}{\sqrt{4 \times 4 \times 3}} = \frac{7}{4 \sqrt{3}}$

Ans. Required smallest factor $= \sqrt{3}$

(d) Let us calculate the rationalising factor of $(\sqrt{5}+2)$ which is also its conjugate surd.

Solution:

$(\sqrt{5}+2)$

Conjugate surd of $(\sqrt{5}+2) is 2 - \sqrt{5}.$

(e) If $(\sqrt{5}+\sqrt{2}) \div \sqrt{7} = \frac{1}{7}(\sqrt{35}+a)$ , let us calculate the value of a.

Solution:

$(\sqrt{5}+\sqrt{2}) \div \sqrt{7} = \frac{1}{7}(\sqrt{35}+a)$

If $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{7}} = \frac{\sqrt{35}+a}{7} \\$

$\text { or, } \frac{(\sqrt{5}+\sqrt{2}) \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{\sqrt{35}+a}{7} \\$

$\text { or, } \frac{\sqrt{35}+\sqrt{14}}{7} = \frac{\sqrt{35}+a}{7} \\$

$\text { or, } \sqrt{35}+\sqrt{14} = \sqrt{35}+a \\$

$\therefore a = \sqrt{14}$

(f) Let us write a rationalising factor of $\frac{5}{\sqrt{3} - 2}$ , which its not its conjugate surd.

Solution:

$\frac{5}{\sqrt{3} - 2}$

The rationalising factor of the denominator of $\frac{5}{\sqrt{3} - 2} \ is \ (\sqrt{3}+2).$

Question 4

Let us write the conjugate surds of mixed quadratic surds $(9 - 4 \sqrt{3}) \ and \ ( - 2 - \sqrt{7})$

Solution:

Conjugate surd of $(9 - 4 \sqrt{5}) \ is \ (9+4 \sqrt{5})$ & conjugate surd of $( - 2 - \sqrt{7}) \ is \ ( - 2+\sqrt{7}).$

Question 5

Let us write two conjugate surds of each of the mixed quadratic surds given below.

(i) $\sqrt{5}+\sqrt{2}$

Solution:

$\sqrt{5}+\sqrt{2}$

Two conjugate surds of $\sqrt{5}+\sqrt{2} \ are \ (\sqrt{5} - \sqrt{2}) ( - \sqrt{5}+\sqrt{2}).$

(ii) $13+\sqrt{6}$

Solution:

$13+\sqrt{6}$

Two conjugate surds of $13+\sqrt{6} \ are \ (13 - \sqrt{6}) ( - 13+\sqrt{6}).$

(iii) $\sqrt{8} - 3$

Solution:

$\sqrt{8} - 3$

Two conjugate surds of $\sqrt{8} - 3 \ are \ ( - \sqrt{8} - 3) (\sqrt{8}+3).$

(iv) $\sqrt{17} - \sqrt{15}$

Solution:

$\sqrt{17} - \sqrt{15}$

Two conjugate surds of $\sqrt{17} - \sqrt{15} \ are \ (\sqrt{17}+\sqrt{15}) ( - \sqrt{17} - \sqrt{15}).$

Question 6

Let us rationalise the denominators of the following surds :

(i) $\frac{2 \sqrt{3}+3 \sqrt{2}}{\sqrt{6}}$

Solution:

$\frac{2 \sqrt{3}+3 \sqrt{2}}{\sqrt{6}} = \frac{(2 \sqrt{3}+3 \sqrt{2}) \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}} = \frac{2 \sqrt{18}+3 \sqrt{12}}{6} = \frac{2 \times 3 \sqrt{2}+3 \times 2 \sqrt{3}}{6}$

$= \frac{6 \sqrt{2}+6 \sqrt{3}}{6} = \frac{6(\sqrt{2}+\sqrt{3})}{6} = \sqrt{2}+\sqrt{3}.$ Ans.

(ii) $\frac{\sqrt{2} - 1+\sqrt{6}}{\sqrt{5}}$

Solution:

$\frac{\sqrt{2} - 1+\sqrt{6}}{\sqrt{5}} = \frac{(\sqrt{2} - 1+\sqrt{6}) \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{\sqrt{10} - \sqrt{5}+\sqrt{30}}{5}$ Ans.

(iii) $\frac{\sqrt{3}+1}{\sqrt{3} - 1}$

Solution:

$\frac{\sqrt{3}+1}{\sqrt{3} - 1} = \frac{(\sqrt{3}+1) \times(\sqrt{3}+1)}{(\sqrt{3} - 1) \times(\sqrt{3}+1)} = \frac{3+1+2 \sqrt{3}}{3 - 1} = \frac{4+2 \sqrt{3}}{2} = \frac{2(2+\sqrt{3})}{2} = (2+\sqrt{3})$ Ans.

(iv) $\frac{3+\sqrt{5}}{\sqrt{7} - \sqrt{3}}$

Solution:

$\frac{3+\sqrt{5}}{\sqrt{7} - \sqrt{3}} = \frac{(3+\sqrt{5})(\sqrt{7}+\sqrt{3})}{(\sqrt{7} - \sqrt{3})(\sqrt{7}+\sqrt{3})} = \frac{3 \sqrt{7}+\sqrt{35}+3 \sqrt{3}+\sqrt{15}}{(\sqrt{7})^2(\sqrt{3})^2}$

$= \frac{3 \sqrt{7}+\sqrt{35}+3 \sqrt{3}+\sqrt{15}}{7-3} = \frac{3 \sqrt{7}+\sqrt{35}+3 \sqrt{3}+\sqrt{15}}{4}$ Ans.

(v) $\frac{3 \sqrt{2}+1}{2 \sqrt{5} - 1}$

Solution:

$\frac{3 \sqrt{2}+1}{2 \sqrt{5} - 1} = \frac{(3 \sqrt{2}+1) \times(2 \sqrt{5}+1)}{(2 \sqrt{5} - 1)(2 \sqrt{5}+1)} = \frac{6 \sqrt{10}+3 \sqrt{2}+2 \sqrt{5}+1}{(2 \sqrt{5})^2 - (1)^2}$

$= \frac{6 \sqrt{10}+3 \sqrt{2}+2 \sqrt{5}+1}{4 \times 5 - 1} = \frac{6 \sqrt{10}+3 \sqrt{2}+2 \sqrt{5}+1}{19}$ Ans.

(vi) $\frac{3 \sqrt{2}+2 \sqrt{3}}{3 \sqrt{2} - 2 \sqrt{3}}$

Solution:

$\frac{3 \sqrt{2}+2 \sqrt{3}}{3 \sqrt{2} - 2 \sqrt{3}} = \frac{(3 \sqrt{2}+1) \times(3 \sqrt{2}+2 \sqrt{3})}{(3 \sqrt{2} - 2 \sqrt{3}) \times(3 \sqrt{2}+2 \sqrt{3})} = \frac{9 \times 2+6 \sqrt{6}+6 \sqrt{6}+4 \times 3}{(3 \sqrt{2})^2 - (2 \sqrt{3})^2} = \frac{18+12 \sqrt{6}+10}{9 \times 2 - 4 \times 3} = \frac{30+12 \sqrt{6}}{18 - 12} = \frac{6(5+2 \sqrt{3})}{6} = 5+2 \sqrt{3}.$ Ans.

Question 7

Let us divide first by second and rationalise the divisor.

(i) $3 \sqrt{2}+\sqrt{5}, \sqrt{2}+1$

Solution:

$\frac{3 \sqrt{2}+\sqrt{5}}{\sqrt{2}+1} = \frac{(3 \sqrt{2}+\sqrt{5}) \times(\sqrt{2} - 1)}{(\sqrt{2}+1) \times(\sqrt{2} - 1)} = \frac{3 \times 2 - 3 \sqrt{2}+\sqrt{10} - \sqrt{5}}{(\sqrt{2})^2 - (1)^2}$

$= \frac{6 - 3 \sqrt{2}+\sqrt{10} - \sqrt{5}}{2 - 1} = 6 - 3 \sqrt{2}+\sqrt{10} - \sqrt{5}$ Ans.

(ii) $2 \sqrt{3}-\sqrt{2}, \sqrt{2}-\sqrt{3}$

Solution:

$\frac{2 \sqrt{3}-\sqrt{2}}{\sqrt{2}-\sqrt{3}}=\frac{(2 \sqrt{3}-\sqrt{2}) \times(\sqrt{2}+\sqrt{3})}{(\sqrt{2}-\sqrt{3}) \times(\sqrt{2}+\sqrt{3})}=\frac{2 \sqrt{6}+2 \times 3-2-\sqrt{6}}{(\sqrt{2})^2-(\sqrt{3})^2}$

$=\frac{\sqrt{6}+6-2}{-1}=\frac{\sqrt{6}+4}{-1}=-(\sqrt{6}-4)$

(iii) $3+\sqrt{6}, \sqrt{3}+\sqrt{2}$

Solution:

$\frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\frac{(3+\sqrt{6})(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\frac{3 \sqrt{3}-3 \sqrt{2}+\sqrt{18}-\sqrt{12}}{(\sqrt{3})^2-(\sqrt{2})^2}$

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

$=\sqrt{3}$ Ans.

Question 8

Let us find the value of

(i) $\frac{2 \sqrt{5}+1}{\sqrt{5}+1}-\frac{4 \sqrt{5}-1}{\sqrt{5}-1}$

Solution:

$\frac{2 \sqrt{5}+1}{\sqrt{5}+1}-\frac{4 \sqrt{5}-1}{\sqrt{5}-1}$

$=\frac{(2 \sqrt{5}+1)(\sqrt{5}-1)-(4 \sqrt{5}-1)(\sqrt{5}+1)}{(\sqrt{5}+1)(\sqrt{5}-1)}$

$=\frac{(2 \times 5-2 \sqrt{5}+\sqrt{5}-1)-(4 \times 5+4 \sqrt{5}-\sqrt{5}-1)}{(\sqrt{5})^2-(1)^2}$

$=\frac{10-\sqrt{5}-1-20-3 \sqrt{5}+1}{5-1}=\frac{-10-4 \sqrt{5}}{4}=\frac{2(-5-2 \sqrt{5})}{4}$

$=\frac{-5-2 \sqrt{5}}{2}$ Ans.

(ii) $\frac{8+3 \sqrt{2}}{3+\sqrt{5}}-\frac{8-3 \sqrt{2}}{3-\sqrt{5}}$

Solution:

$\frac{8+3 \sqrt{2}}{3+\sqrt{5}}-\frac{8-3 \sqrt{2}}{3-\sqrt{5}}$

$=\frac{(8+3 \sqrt{2})(3-\sqrt{5})-(8-3 \sqrt{2})(3-+\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})} \\$

$=\frac{(24-8 \sqrt{5}+9 \sqrt{2}-3 \sqrt{10})-(24+8 \sqrt{5}-9 \sqrt{2}-3 \sqrt{10})}{(3)^2(\sqrt{5})^2} \\$

$=\frac{24-8 \sqrt{5}+9 \sqrt{2}-3 \sqrt{10}-24-8 \sqrt{5}+9 \sqrt{2}+3 \sqrt{10}}{9-5}$

$=\frac{18 \sqrt{2}-16 \sqrt{5}}{4}=\frac{2(9 \sqrt{2}-8 \sqrt{5})}{2} \\$

$=\frac{9 \sqrt{2}-8 \sqrt{5}}{2}$

Application 32.

Let us simplify : $\frac{3 \sqrt{20}+2 \sqrt{28}+\sqrt{12}}{5 \sqrt{45}+2 \sqrt{175}+\sqrt{75}}$

Solution:

$\frac{3 \sqrt{20}+2 \sqrt{28}+\sqrt{12}}{5 \sqrt{45}+2 \sqrt{175}+\sqrt{75}}$

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

$=\frac{6 \sqrt{5}+4 \sqrt{7}+4 \sqrt{3}}{15 \sqrt{5}+10 \sqrt{7}+5 \sqrt{3}}=\frac{2(3 \sqrt{5}+2 \sqrt{7}+\sqrt{3})}{5(3 \sqrt{5}+2 \sqrt{7}+\sqrt{3})} \\$

$=\frac{2}{5} \text { (Ans.) }$

Application 34.

Let us simplify : $\frac{5}{\sqrt{2}+\sqrt{3}}-\frac{1}{\sqrt{2}-\sqrt{3}}$ [Do it yourself]

Solution:

$\frac{5}{\sqrt{2}+\sqrt{3}}-\frac{1}{\sqrt{2}-\sqrt{3}}$

$=\frac{5(\sqrt{2}-\sqrt{3})-1(\sqrt{2}+\sqrt{3})}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}=\frac{5 \sqrt{2}-5 \sqrt{3}-\sqrt{2}-\sqrt{3}}{(\sqrt{2})^{2}-(\sqrt{3})^{2}}=\frac{4 \sqrt{2}-6 \sqrt{3}}{2-3} \\$

$=\frac{4 \sqrt{2}-6 \sqrt{3}}{-1}$

$=6 \sqrt{3}-4 \sqrt{2}$ Ans.

Application 36.

If $x=\sqrt{3}+\sqrt{2}$ let us calculate the simplified value of $\left(x-\frac{1}{x}\right) , \left(x^{3}-\frac{1}{x^{3}}\right) \text{ and } \left(x^{2}-\frac{1}{x^{2}}\right)$ .[ Do it yourself ]

Solution:

If $x=\sqrt{3}+\sqrt{2}$ , find the values of $\left(x-\frac{1}{x}\right),\left(x^{3}+\frac{1}{x^{3}}\right) \&\left(x^{2}+\frac{1}{x^{2}}\right)$

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

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

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

Now, $x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)^{3}-3 \cdot x \cdot \frac{1}{x}\left(x+\frac{1}{x}\right)$

$=(2 \sqrt{3}) 3-3 \times 2 \sqrt{3}=8 \times 3 \sqrt{3}-6 \sqrt{3}=24 \sqrt{3}-6 \sqrt{3}=18 \sqrt{3} \\$

$\& x^{2}-\frac{1}{x^{2}}=\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right) \\$

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

$=4 \sqrt{6} \text { Ans. }$

LET US WORK OUT – 9.3

(a) If $m+\frac{1}{m}=\sqrt{3}$ , let us calculate the simplified value of (i) $m^{2}+\frac{1}{m^{2}}$ and (ii) $m^{3} +\frac{1}{m^{3}}$

Solution:

(i) $m^{2}+\frac{1}{m^{2}}=\left(m+\frac{1}{m}\right)^{2}-2 \cdot m \cdot \frac{1}{m}$

$=(\sqrt{3})^{2}-2=3-2=1.$ Ans.

Solution (ii) : $m^{3}+\frac{1}{m^{3}}=\left(m+\frac{1}{m}\right)^{3}-3 \cdot m \cdot \frac{1}{m}\left(m+\frac{1}{m}\right)$

$=(\sqrt{3})^{3}-3 \sqrt{3}$

$=3 \sqrt{3}-3 \sqrt{3}$ =0 Ans.

(b) Let us show that $\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2 \sqrt{15}.$

Solution:

$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2 \sqrt{15}$

L.H.S. $=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

$=\frac{(\sqrt{5}+\sqrt{3})^{2}-(\sqrt{5}-\sqrt{3})^{2}}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{(5+3+2 \sqrt{15})-(5+3-2 \sqrt{15})}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}$

$=\frac{8+2 \sqrt{15}-8+2 \sqrt{15}}{5-3}=\frac{4 \sqrt{15}}{2}=2 \sqrt{15}$ = R.H.S.

Question 2

Let us simplify :

(a) $\frac{\sqrt{2}(2+\sqrt{3})}{\sqrt{3}(\sqrt{3}+1)}-\frac{\sqrt{2}(2-\sqrt{3})}{\sqrt{3}(\sqrt{3}-1)}$

Solution:

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

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

$=\frac{\sqrt{2}}{\sqrt{3}}\left[\frac{(2 \sqrt{3}-2+3-\sqrt{3})-(2 \sqrt{3}+2-3-\sqrt{3})}{(\sqrt{3})^{2}-(1)^{2}}\right]$

$=\frac{\sqrt{2}}{\sqrt{3}}\left[\frac{2 \sqrt{3}-2+3-\sqrt{3}-2 \sqrt{3}-2+3+\sqrt{3}}{3-1}\right]$

$=\frac{\sqrt{2}}{\sqrt{3}} \times \frac{6-4}{2}=\frac{\sqrt{2}}{\sqrt{3}} \times \frac{2}{2}=\frac{\sqrt{2} \times \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}=\frac{\sqrt{6}}{3}$ Ans.

(b) $\frac{3 \sqrt{7}}{\sqrt{5}+\sqrt{2}}-\frac{5 \sqrt{5}}{\sqrt{2}+\sqrt{7}}+\frac{2 \sqrt{2}}{\sqrt{7}+\sqrt{5}}$

Solution:

$\frac{3 \sqrt{7}}{\sqrt{5}+\sqrt{2}}-\frac{5 \sqrt{5}}{\sqrt{2}+\sqrt{7}}+\frac{2 \sqrt{2}}{\sqrt{7}+\sqrt{5}}$

$=\frac{3 \sqrt{7}(\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}-\frac{5 \sqrt{5}(\sqrt{7}-\sqrt{2})}{(\sqrt{7}+\sqrt{2})(\sqrt{7}-\sqrt{2})}+\frac{2 \sqrt{5}}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}$

$=\frac{3 \sqrt{7}(\sqrt{5}-\sqrt{2})}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}-\frac{5 \sqrt{5}(\sqrt{7}-\sqrt{2})}{(\sqrt{7})^{2}-(\sqrt{2})^{2}}+\frac{2 \sqrt{2}}{(\sqrt{7})^{2}-(\sqrt{5})^{2}}$

$=\frac{3 \sqrt{7}(\sqrt{5}-\sqrt{2})}{5-2}-\frac{5 \sqrt{5}(\sqrt{7}-\sqrt{2})}{7-2}+\frac{2 \sqrt{2}(\sqrt{7}-\sqrt{5})}{7-5}$

$=\frac{3 \sqrt{2}(\sqrt{5}-\sqrt{2})}{3}-\frac{5 \sqrt{5}(\sqrt{7}-\sqrt{2})}{5}+\frac{2 \sqrt{2}(\sqrt{7}-\sqrt{5})}{2}$

$=\sqrt{35}-\sqrt{14}-\sqrt{35}+\sqrt{10}+\sqrt{14}-\sqrt{10}$

=0 Ans.

(c) $\frac{4 \sqrt{3}}{2-\sqrt{2}}-\frac{30}{4 \sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}$

Solution:

$\frac{4 \sqrt{3}}{2-\sqrt{2}}-\frac{30}{4 \sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}$

$=\frac{4 \sqrt{3}(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})}-\frac{30(4 \sqrt{3}+\sqrt{18})}{(4 \sqrt{3}-\sqrt{18})(4 \sqrt{3}+\sqrt{18})}-\frac{\sqrt{18}(3+\sqrt{12})}{(3-\sqrt{12})(3+\sqrt{12})}$

$=\frac{4 \sqrt{3}(2+\sqrt{2})}{(2)^{2}(\sqrt{2})^{2}}-\frac{30(4 \sqrt{3}+\sqrt{2})}{(4 \sqrt{3})^{2}-(\sqrt{18})^{2}}-\frac{3 \sqrt{2}(3+2 \sqrt{3})}{(3)^{2}-(\sqrt{12})^{2}}$

$=\frac{4 \sqrt{3}(2+\sqrt{2})}{4-2}-\frac{30(4 \sqrt{3}+\sqrt{2})}{16 \times 3-18}-\frac{3 \sqrt{2}(3+2 \sqrt{3})}{9-12}$

$=\frac{4 \sqrt{3}(2+\sqrt{2})}{2}-\frac{30(4 \sqrt{3}+3 \sqrt{2})}{30}-\frac{3(3+2 \sqrt{3})}{-3}$

$=4 \sqrt{3}+2 \sqrt{6}-4 \sqrt{3}-3 \sqrt{2}+2 \sqrt{6}$

$=4 \sqrt{6}$ Ans.

(d) $\frac{3 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

Solution:

$\frac{3 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

$=\frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{(\sqrt{6}+\sqrt{3})(\sqrt{6}-\sqrt{3})}-\frac{4 \sqrt{2}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}+\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$

$=\frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{6-3}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{6-2}+\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{3-2}$

$=\frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{3}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{4}+\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{1}$

$=\sqrt{12}-\sqrt{6}-\sqrt{18}+\sqrt{6}+\sqrt{18}-\sqrt{12}$

= 0 Ans.

Question 3

If x = 2, y = 3 and z = 6 let us write by calculating the value of $\frac{3 \sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4 \sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}.$

Solution:

$\frac{3 \sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4 \sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}} =\frac{3 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

$=\frac{3 \sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}$

$=\frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{(\sqrt{6}+\sqrt{3})(\sqrt{6}-\sqrt{3})}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}+\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$

$=\frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{6-3}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{6-2}+\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{3-2}$

$=\frac{3 \sqrt{2}(\sqrt{6}-\sqrt{3})}{3}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{4}+\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{1}$

$=\sqrt{12}-\sqrt{6}-\sqrt{18}+\sqrt{6}+\sqrt{18}-\sqrt{12}$

= 0 Ans.

Question 4

If x $=\sqrt{7}+\sqrt{6}$ let us calculate the simplified values of :

(i) $x-\frac{1}{x}$

Solution:

$x-\frac{1}{x}=(\sqrt{7}+\sqrt{6})-(\sqrt{7}-\sqrt{6})=\sqrt{7}+\sqrt{6}-\sqrt{7}+\sqrt{6} =2 \sqrt{6}$ Ans.

(ii) $x+\frac{1}{x}$

Solution:

$x+\frac{1}{x}=(\sqrt{7}+\sqrt{6})+(\sqrt{7}-\sqrt{6})=\sqrt{7}+\sqrt{6}+\sqrt{7}-\sqrt{6} =2 \sqrt{7}$ Ans.

(iii) $\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}$

Solution:

$x^{2}+\frac{1}{x^{2}}=\left(x+\frac{1}{x}\right)^{2}-2 \cdot x \cdot \frac{1}{x}=(2 \sqrt{7})^{2}-2 \cdot 1=28-2$ =26 Ans.

(iv) $x^{3}+\frac{1}{x^{3}}$

Solution:

$x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)^{3}-3 \cdot x \cdot \frac{1}{x}\left(x+\frac{1}{x}\right)=(2 \sqrt{7})^{3}-3 \cdot 1 \cdot 2 \sqrt{7} =8 \times 7 \sqrt{7}-6 \sqrt{7}=56 \sqrt{7}-6 \sqrt{7}=50 \sqrt{7}$ Ans.

Question 5

Let us simplify: $\frac{x+\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}+\frac{x-\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}}$ . If the simplified value is 14, let us write by calculating what is the value of x.

Solution:

$\frac{x+\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}+\frac{x-\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}}$

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

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

$=\frac{4 x^{2}-2}{x^{2}-\left(x^{2}-1\right)}=\frac{4 x^{2}-2}{x^{2}-x^{2}+1}=4 x^{2}-2$

According to the problem,

$4 x^{2}$ – 2 = 14

or, $4 x^{2}$ =14 + 2 = 16

$x^{2}=\frac{16}{4}$ = 4

$x= \pm \sqrt{4}= \pm 2$

Question 6

If $a=\frac{\sqrt{5}+1}{\sqrt{5}-1} \text{ and } b=\frac{\sqrt{5}-1}{\sqrt{5}+1}$ , let us calculate the following expressions:

a + b $=\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{(\sqrt{5}+1)^{2}+(\sqrt{5}-1)^{2}}{(\sqrt{5}-1)(\sqrt{5}+1)}$

$=\frac{5+1+2 \sqrt{5}+5+1-2 \sqrt{5}}{5-1}=\frac{12}{4}=3$

$\ \& \ a-b=\frac{\sqrt{5}+1}{\sqrt{5}-1}-\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{(\sqrt{5}+1)^{2}-(\sqrt{5}-1)^{2}}{(\sqrt{5}-1)(\sqrt{5}+1)}$

ab $=\frac{\sqrt{5}+1}{\sqrt{5}-1} \times \frac{\sqrt{5}-1}{\sqrt{5}+1}=1$

(i) $\frac{a^{2}+a b+b}{a^{2}-a b+b^{2}}$

Solution:

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

$=\frac{(a+b)^{2}-a b}{(a+b)^{2}-3 a b} =\frac{(3)^{2}-1}{(3)^{2}-3.1}=\frac{9-1}{9-3}=\frac{8}{6}=\frac{4}{3}$ Ans.

(ii) $\frac{(a-b)^{3}}{(a+b)^{3}}$

Solution:

$\frac{(a-b)^{3}}{(a+b)^{3}}=\frac{(\sqrt{5})^{3}}{(3)^{3}}=\frac{5 \sqrt{5}}{27}$ Ans.

(iii) $\frac{3 a^{2}+5 a b+3 b^{2}}{3 a^{2}-5 a b+3 b^{2}}$

Solution:

$\frac{3 a^{2}+5 a b+3 b^{2}}{3 a^{2}-5 a b+3 b^{2}}=\frac{3 a^{2}+6 a b+3 b^{2}-a b}{3 a^{2}-6 a b+3 b^{2}+a b}=\frac{3\left(a^{2}+2 a b+b^{2}\right)-1}{3\left(a^{2}-2 a b+b^{2}\right)+1}$

$=\frac{3(a+b)^{2}-a b}{3(a-b)^{2}+a b}=\frac{3(3)^{2}-1}{3(\sqrt{5})^{2}+1}=\frac{27-1}{3.5+1}=\frac{26}{16}=\frac{13}{8}$

$=1 \frac{5}{8}$ Ans.

(iv) $\frac{a^{3}+b^{3}}{a^{3}-b^{3}}$

Solution:

$\frac{a^{3}+b^{3}}{a^{3}-b^{3}}=\frac{(a+b)^{3}-3 a b(a+b)}{(a-b)^{3}+3 a b(a-b)}=\frac{(3)^{3}-3 \cdot 1 \cdot 3}{(\sqrt{5})^{3}+3 \cdot 1 \cdot \sqrt{5}}=\frac{27-9}{8 \sqrt{5}}=\frac{18 \sqrt{5}}{8 \sqrt{5} \cdot \sqrt{5}}$

$=\frac{18 \sqrt{5}}{8 \times 5}=\frac{9 \sqrt{5}}{20}$ Ans.

Question 7

If x = 2 + $\sqrt{3}$ , y = 2 – $\sqrt{3},$ let us calculate the simplified value of :

x = 2 + $\sqrt{3}$

$\therefore \frac{1}{x}=\frac{1}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=\frac{2-\sqrt{3}}{4-3}=\frac{2-\sqrt{3}}{1}=2-\sqrt{3}.$

Again, y = 2 – $\sqrt{3}$ :

$\therefore \frac{1}{y}=\frac{1}{2-\sqrt{3}}=\frac{(2+\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}=\frac{2+\sqrt{3}}{4-3}=\frac{2+\sqrt{3}}{1}=2+\sqrt{3}.$

(a) (i) $x-\frac{1}{x}$

Solution:

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

(ii) $y^{2}+\frac{1}{y^{2}}$

Solution:

$y^{2}+\frac{1}{y^{2}}=\left(y+\frac{1}{y}\right)^{2}-2 \cdot y \cdot \frac{1}{y}$

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

$=\left\{(4)^{2}-2\right\}=16-2=14 \text { Ans. }$

(iii) $x^{3}-\frac{1}{x^{3}}$

Solution:

$x^{3}-\frac{1}{x^{3}}=\left(x-\frac{1}{x}\right)^{3}+3 \cdot x \cdot \frac{1}{x}\left(x-\frac{1}{x}\right)$

$=(2 \sqrt{3})^{3}+3 \cdot 1 \cdot 2 \sqrt{3}=8 \cdot 3 \cdot \sqrt{3}+\dot{6} \cdot \sqrt{3}$

$=24 \sqrt{3}+\sqrt{3}=30 \sqrt{3} \text { Ans. }$

(iv) $x y+\frac{1}{x y}$

Solution:

$x y+\frac{1}{x y}=x \cdot y=(2+\sqrt{3})(2-\sqrt{3})=(2)^{2}-(\sqrt{3})^{2}$ = 4 – 3 = 1

$\therefore x y+\frac{1}{x y}=1+\frac{1}{1}$ = 1 + 1 = 2 Ans.

(b) $3 x^{2}-5 x y+3 y^{2}$

Solution:

$3 x^{2}-5 x y+3 y^{2}=3 x^{2}-6 x y+3 y^{2}+x y=3\left(x^{2}-2 x y+y^{2}\right)+x y$

$=3(x-y)^{2}+x y$

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

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

$=3(2 \sqrt{3})^{2}+1=3 \times 4 \times 3+1=36+1$

= 37. Ans.

Question 8

If x $=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}$ and xy = 1, let us show that $\frac{x^{2}+x y+y^{2}}{x^{2}-x y+y^{2}}=\frac{12}{11}.$

Solution:

$\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} \ \& \ x y=1, \text{ show that } \frac{x^{2}+x y+y^{2}}{x^{2}-x y+y^{2}}.$

$x=\frac{(\sqrt{7}+\sqrt{3})(\sqrt{7}+\sqrt{3})}{(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3})}=\frac{(\sqrt{7}+\sqrt{3})^{2}}{(\sqrt{7})^{2}(\sqrt{3})^{2}}=\frac{7+3+2 \cdot \sqrt{7} \cdot \sqrt{3}}{7-3}=\frac{10+2 \sqrt{21}}{4} \\$

$=\frac{2(5+\sqrt{21})}{4}=\frac{5+\sqrt{21}}{2}$

As $xy = 1 \therefore y=\frac{1}{x}=\frac{2}{5+\sqrt{21}}=\frac{2(5-\sqrt{21})}{(5+\sqrt{21})(5-\sqrt{21})}$

$=\frac{2(5-\sqrt{21})}{(5)^{2}-(\sqrt{21})^{2}}=\frac{2(5-\sqrt{21})}{25-21}=\frac{2(5-\sqrt{21})}{4}=\frac{5-\sqrt{21}}{2}$

$x+y=\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=\frac{5+\sqrt{21}+5-\sqrt{21}}{2}=\frac{10}{2}=5$

Now, $\frac{x^{2}+x y+y^{2}}{x^{2}-x y+y^{2}}=\frac{x^{2}+y^{2}+x y}{x^{2}+y^{2}-x y}=\frac{x^{2}+y^{2}+x y}{x^{2}+y^{2}-x y}$

$=\frac{(x+y)^{2}-x y}{(x+y)^{2}-3 x y}=\frac{(5)^{2}-1}{(5)^{2}-3.1}=\frac{25-1}{25-3}=\frac{24}{22}=\frac{12}{11} \text { Ans. }$

Question 9

Let us write which one is greater between $(\sqrt{7}+1) \text{ and } (\sqrt{5}+\sqrt{3}).$

Solution:

$(\sqrt{7}+1)^{2}=7+2 \sqrt{7}=8+2 \sqrt{7}$

$(\sqrt{5}+\sqrt{3})^{2}=5+3+2 \sqrt{5} \cdot \sqrt{3}=8+2 \sqrt{15} \\$

$\text { as }(\sqrt{5}+\sqrt{3})^{2}>(\sqrt{7}+1)^{2} \\$

$\therefore(\sqrt{5}+\sqrt{3})>(\sqrt{7}+1)$

Ans. $(\sqrt{5}+\sqrt{3})$ is greater.

Very short answer type questions :

(A) M.C.Q. :

Question 1

If $x = 2+\sqrt{3}$ , the value of $x+\frac{1}{x}$ is

2
$2 \sqrt{3}$
4
$2-\sqrt{3}$

Solution:

$\therefore \frac{1}{x} = \frac{1}{2+\sqrt{3}} = \frac{1 \times(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} = \frac{2-\sqrt{3}}{4-3} = \frac{2-\sqrt{3}}{1} = 2-\sqrt{3}$

$\therefore x+\frac{1}{x} = 2+\sqrt{3}+2-\sqrt{3}$ = 4

Ans. (c) 4

Question 2

If p + q $= \sqrt{13}$ and p – q $= \sqrt{5}$ then the value of pq is

Solution:

We know, pq $= \frac{(\mathrm{p}+q)^{2}-(\mathrm{p}-\mathrm{q})^{2}}{4} = \frac{(\sqrt{13})^{2}-(\sqrt{5})^{2}}{4} = \frac{13-5}{4} = \frac{8}{4}$ = 2

Ans. (a) 2

Question 3

If $a+b \cong \sqrt{5}$ and $a-b = \sqrt{3},$ the value of $\left(a^{2}+b^{2}\right)$ is

Solution:

$a^{2}+b^{2} = \frac{(a+b)^{2}+(a-b)^{2}}{2} = \frac{(\sqrt{5})^{2}+(\sqrt{3})^{2}}{2} = \frac{5+3}{2} = \frac{8}{2}$ = 2

Ans. (a) 8

Question 4

If we subtract $\sqrt{5} \text{ from } \sqrt{125}$ , the value is

$\sqrt{80}$
$\sqrt{120}$
$\sqrt{100}$
none of these

Solution:

$\sqrt{125}-\sqrt{5} = \sqrt{5 \times 5 \times 5}-\sqrt{5}-\sqrt{5} = 4 \sqrt{5} = \sqrt{16 \times 5} = \sqrt{80}$

Ans. (a) $\sqrt{80}$

Question 5

The product of the bracketed terms $(5-\sqrt{3}),(\sqrt{3}-1),(5+\sqrt{3}),(\sqrt{3}+1)$ is

Solution:

$= (5-\sqrt{3})(5+\sqrt{3})(\sqrt{3}-1)(\sqrt{3}+1)$

$= \left\{(5)^{2}-(\sqrt{3})^{2}\right\}\left\{(\sqrt{3})^{2}-(1)^{2}\right\} = (25-3) \times(3-1) = 22 \times 2 = 44$

Ans. (b) 44

Let us write true or false of the following statements :

Question 1

$\sqrt{75} \text{ and } \sqrt{147}$ are similar surds.

Solution:

$\sqrt{75} \ \& \ \sqrt{5 \times 5 \times 3} = 5 \sqrt{3} \ \& \ \sqrt{147} = \sqrt{7 \times 7 \times 3} = 7 \sqrt{3}$

Ans. True

Question 2

$\sqrt{\pi}$ is a quadratic surd.

Ans. False

Let us fill in the blanks :

Question 1

$5 \sqrt{11}$ is a __________________ number. (rational / irrational)

Solution:

Irrational.

Question 2

Conjugate surd of $(\sqrt{3}-5)$ is ______________.

Solution:

$-\sqrt{3}-5$

Question 3

If the product and sum of two quadratic surds is a rational number, then the surds are surd.

Solution:

Irrational

11. Short answer type questions :

Question 1

If $x = 3+2 \sqrt{2}, \text { let us write the value of } x+\frac{1}{x}.$

Solution:

$\frac{1}{x} = \frac{1}{3+2 \sqrt{2}} = \frac{1 \times(3-2 \sqrt{2})}{(3+2 \sqrt{2})(3-2 \sqrt{2})} = \frac{3-2 \sqrt{2}}{9-8} = 3-2 \sqrt{2}$

$\therefore x+\frac{1}{x} = 3+2 \sqrt{2}+3-2 \sqrt{2}$ = 6

Question 2

Let us write which one is greater between $(\sqrt{15}+\sqrt{3}) \text{ and } (\sqrt{10}+\sqrt{8}).$

Solution:

Now, $(\sqrt{15}+\sqrt{3})^{2} = (\sqrt{15})^{2}+(\sqrt{3})^{2}+2 \cdot \sqrt{15} \cdot \sqrt{3} = 15+3+2 \sqrt{45} = 18+ 2 \sqrt{45}$

& $(\sqrt{10}+\sqrt{8})^{2} = (\sqrt{10})^{2}+(\sqrt{8})^{2}+2 \cdot \sqrt{10} \cdot \sqrt{8} = 10+8+2 \sqrt{80} = 18+2 \sqrt{80}$

As 2 $\sqrt{80} \text{ is greater than } 2 \sqrt{45},$

$\therefore(\sqrt{10}+\sqrt{8})^{2}>(\sqrt{15}+\sqrt{3})^{2}$

$\therefore(\sqrt{10}+\sqrt{8}) \text{ is greater than } (\sqrt{15}+\sqrt{3}).$

Question 3

Let us write two quardratic surds whose product is a rational number.

Solution:

$(5+2 \sqrt{6}) \ \& \ (5-2 \sqrt{6})$

Question 4

Let us write what should be subtracted from $\sqrt{72} \text{ to get } \sqrt{32}.$

Solution:

Required number $= \sqrt{72}-\sqrt{32} = \sqrt{6 \times 6 \times 2}-\sqrt{4 \times 4 \times 2} = 6 \sqrt{2}-4 \sqrt{2} = 2 \sqrt{2}$ Ans.

Question 5

Let us write the simplified value of $\left(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}\right).$

Solution:

$= \frac{1(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}+\frac{1(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}+\frac{1(\sqrt{4}-\sqrt{3})}{(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})}$

$= \frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}$

$= \sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}$

$= \sqrt{4}-1 = 2-1.$

= 1.