# Chapter – 9 : Quadratic Surd

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

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}+\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{2}}{(\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

1. 2
2. 2 \sqrt{3}
3. 4
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

1. 2
2. 18
3. 9
4. 8

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

1. 8
2. 4
3. 2
4. 1

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

1. \sqrt{80}
2. \sqrt{120}
3. \sqrt{100}
4. 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

1. 22
2. 44
3. 2
4. 11

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.

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