Here is a list of solutions from Algebra chapter form Vedanta Excel in Mathematics Book 10:

Exercise 11.2

Important Formulae we will use:
  1. (a + b)² + (a - b)² = 2(a² + b²)
  2. (a + b)² - (a - b)² = 4ab

General Section


1. Rationalize the denominators and simplify.

a) \dfrac{1 }{√2 +1}
Solution:

= \dfrac{1}{√2+1}
= \dfrac{1}{√2+1} × \dfrac{√2-1}{√2-1}

= \dfrac{√2-1}{(√2)^2 -1^2}

= \dfrac{√2-1}{2-1}
= √2-1


c) \dfrac{3}{√5-√2}
Solution:

= \dfrac{3}{√5-√2}
= \dfrac{3}{√5-√2} × \dfrac{√5+√2}{√5+√2}

= \dfrac{3(√5+√2)}{(√5-√2)(√5+√2)}

= \dfrac{3(√5+√2)}{(√5)^2 -(√2)^2}

= \dfrac{3(√5+√2)}{5-2}
= \dfrac=3(√5+√2)}{3}
= √5 + √2

e) \dfrac{5}{3√7 + 2√3}
Solution:

= \dfrac{5}{3√7 + 2√3}
= \dfrac{5}{3√7+2√3} × \dfrac{3√7 - 2√3}{3√7-2√3}

= \dfrac{5(3√7 -2√3)}{(3√7)^2 - (2√3)^2}

= \dfrac{5(3√7 - 2√3)}{9×7 - 4×3}

= \dfrac{5(3√7 - 2√3)}{63 - 12}

= \dfrac{5(3√7 -2√3)}{51}

= \dfrac{5}{51} (3√7 -2√3)


g) \dfrac{5√3}{2√3-√2}
Solution:

= \dfrac{5√3}{2√3 - √2}
= \dfrac{5√3}{2√3 -√2} × \dfrac{2√2 +√2}{2√3+√2}

= \dfrac{5√3(2√3 +√2)}{(2√3-√2)(2√3+√2)}

= \dfrac{5√3(2√3 +√2)}{(2√3)^2 -(√2)^2}

= \dfrac{5√3(2√3 +√2)}{4×3 - 2}

= \dfrac{5√3 (2√3 +√2)}{10}

= \dfrac{1}{2} √3(2√3+√2)

= \dfrac{1}{2} (2 \sqrt{3×3} + \sqrt{2×3})

= \dfrac{1}{2} (2×3 + \sqrt{6})

= \dfrac{1}{2} (6 + \sqrt{6})



2. Rationalize the denominators and simplify.

a) \dfrac{√2+1}{√2-1}
Solution:

= \dfrac{√2+1}{√2-1}

= \dfrac{√2+1}{√2-1} × \dfrac{√2+1}{√2+1}

= \dfrac{(√2+1)^2}{(√2)^2 -(1)^2}

= \dfrac{2 + 2√2 +1}{2 -1}

= \dfrac{3 +2√2}{1}
= 3 + 2√2


c) \dfrac{√3 -√2}{√3+√2}
Solution:

= \dfrac{√3-√2}{√3 +√2}
= \dfrac{√3 -√2}{√3+√2} × \dfrac{√3-√2}{√3 -√2}

= \dfrac{(√3-√2)^2}{(√3)^2 -(√2)^2}

= \dfrac{3 - 2√6 + 2}{3 -2}

= \dfrac{5 - 2√6}{1}
= 5 - 2√6


e) \dfrac{2√3 -3√2}{2√3+3√2}
Solution:

= \dfrac{2√3 - 3√2}{2√3 + 3√2}

= \dfrac{2√3 - 3√2}{2√3 + 3√2} × \dfrac{2√3 - 3√2}{2√3 - 3√2}

= \dfrac{(2√3 - 3√2)^2}{(2√3)^2 - (3√2)^2}

= \dfrac{(2√3)^2 - 2√3×3√2 + (3√2)^2}{4×3 - 9×2}

= \dfrac{4×3 - 6√6 + 9×2}{12 - 18}

= \dfrac{12 + 18 - 6√6}{-6}

=  - \dfrac{30 - 6√6}{6}

= - \dfrac{6(5 - √6)}{6}
= -(5 - √6)
= √6 - 5


g) \dfrac{\sqrt{a + b} - \sqrt{a - b}}{\sqrt{a + b} + \sqrt{a -b}}
Solution:

= \dfrac{ \sqrt{a + b} - \sqrt{ a -b}}{ \sqrt{a + b} + \sqrt{ a - b}}

$= \dfrac{ \sqrt{a + b} - \sqrt{a - b}}{\sqrt{a +b} + \sqrt{a -b}} × \dfrac{ \sqrt{a +b} - \sqrt{a - b}}{ \sqrt{a + b} - \sqrt{ a - b}} $

= \dfrac{ ( \sqrt{ a + b} - \sqrt{a - b})^2}{(\sqrt{a + b})^2 - (\sqrt{a - b})^2}

= \dfrac{ ( \sqrt{a + b})^2 - 2 × \sqrt{a + b} × \sqrt{ a -b} + (\sqrt{ a -b})^2}{a + b - (a -b)}

= \dfrac{ a + b - 2\sqrt{(a +b)(a -b)} + a - b}{ a + b - a +b}

= \dfrac{2a - 2\sqrt{a^2 - b^2}}{2b}

= \dfrac{2(a - \sqrt{a^2 - b^2})}{2b}

= \dfrac{a - \sqrt{a^2 - b^2}}{b}



i) \dfrac{\sqrt{a +2} - \sqrt{a -2}}{\sqrt{ a +2} + \sqrt{a -2}}
Solution:

= \dfrac{ \sqrt{a +2} - \sqrt{a -2}}{\sqrt{a +2} + \sqrt{a -2}}

= \dfrac{ \sqrt{a +2} - \sqrt{a -2}}{ \sqrt{a +2} + \sqrt{a -2}} × \dfrac{ \sqrt{a +2} - \sqrt{a -2}}{\sqrt{a -2} - \sqrt{a +2}}

= \dfrac{ ( \sqrt{a +2} - \sqrt{a -2})^2 }{ (\sqrt{a +2})^2 - (\sqrt{a -2})^2}

= \dfrac{ (\sqrt{a +2})^2 - 2×\sqrt{a+2}×\sqrt{a-2} + (\sqrt{a-2})^2}{ (a +2) - (a -2)}

= \dfrac{ a +2 - 2\sqrt{(a+2)(a-2} + a -2}{a +2 - a +2}

= \dfrac{2a-2\sqrt{a^2 - 4}}{4}

= \dfrac{ 2(a- \sqrt{a^2 - 4})}{4}

= \dfrac{a - \sqrt{a^2 - 4}}{2}


k) \dfrac{√2}{√2+√3-√5}
Solution:

= \dfrac{ √2}{√2 +√3 -√5}
= \dfrac{ √2}{(√2 +√3) -√5}

= \dfrac{√2}{(√2 +√3 )-√5} × \dfrac{(√2 +√3)+√5}{(√2 +√3)+√5}

= \dfrac{√2(√2 +√3 +√5)}{(√2+√3)^2 - (√5)^2}

= \dfrac{√2(√2 + √3+ √5)}{(2 + 2√6 + 3) - 5}

= \dfrac{ √2(√2 + √3 + √5)}{5 + 2√6 - 5}

= \dfrac{√2(√2 + √3 + √5)}{2 × √2 × √3}

= \dfrac{√2 + √3 + √5}{2√3}

= \dfrac{√2 + √3 + √5}{2√3} × \dfrac{√3}{√3}

= \dfrac{√3(√2+√3+√5)}{2√3 × √3}

= \dfrac{√6 +3 + √15}{2×3}

= \dfrac{3 + √6 + √15}{6}


3. Simplify
a) 3√5 - \dfrac{1}{√5}
Solution:

= 3√5 - \dfrac{1}{√5}
= 3√5 - \dfrac{1}{√5} × \dfrac{√5}{√5}

= 3√5 - \dfrac{√5}{5}
= \dfrac{3√5 × 5 - √5}{5}

= \dfrac{15√5 - √5}{5}
= \dfrac{14√5}{5}


c) \dfrac{3}{√5} + \dfrac{√5}{2}
Solution:

= \dfrac{3}{√5} + \dfrac{√5}{2}

= \dfrac{3}{√5} × \dfrac{√5}{√5} + \dfrac{√5}{2}

= \dfrac{3√5}{5} + \dfrac{√5}{2}

= \dfrac{3√5 × 2 + \dfrac{√5 × 5}{5×2}

= \dfrac{6√5 + 5√5}{10}
= \dfrac{11√5}{10}

e) \dfrac{√2}{5} + \dfrac{3}{√2}
Solution:

= \dfrac{√2}{5} + \dfrac{3}{√2}

= \dfrac{√2}{5} + \dfrac{3}{√2} × \dfrac{√2}{√2}

= \dfrac{√2}{5} + \dfrac{3√2}{2}

= \dfrac{√2 × 2 + 3√2 × 5}{5×2}

= \dfrac{2√2 + 15√2}{10}

= \dfrac{17√2}{10}

Creative Section A
4. Simplify.
a) \dfrac{\sqrt{3} + 1}{\sqrt{3} -1} + \dfrac{\sqrt{3}-1}{\sqrt{3} +1}
Solution:
= \dfrac{√3 +1}{√3 -1} + \dfrac{√3 -1}{√3+2}

= \dfrac{(√3+1)(√3-1) + (√3-1)(√3-1)}{(√3-1)(√3+1)}

= \dfrac{(√3+1)^2 + (√3 -1)^2}{(√3)^2 -1^2}

Using (a +b)² + (a -b)² = 2(a² +b²)

= \dfrac{2 [ (√3)^2 +1^2]}{3 -1}

= \dfrac{2(3+1)}{2}
= 4
Answer
------------

c) \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} + \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}
Solution:
= \dfrac{√3+√2}{√3-√2} + \dfrac{√3-√2}{√3+√2}

= \dfrac{(√3+√2)(√3+√2) + (√3-√2)(√3-√2)}{(√3-√2)(√3+√2)}

= \dfrac{2[(√3)^2 + (√2)^2]}{(√3)^2 - (√2)^2}

= \dfrac{2[3+2]}{3-2}
= \dfrac{2×5}{1}
= 10
Answer
------------


e) \dfrac{x + \sqrt{x^2 -1}}{x - \sqrt{x^2-1}} - \dfrac{ x - \sqrt{x^2 -1}}{x + \sqrt{x^2 -1}}
Solution:
= \dfrac{x + \sqrt{x^2 -1}}{x - \sqrt{x^2-1}} - \dfrac{ x - \sqrt{x^2 -1}}{x + \sqrt{x^2 -1}}

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

Using (a+b)² - (a -b)² = 4ab

= \dfrac{4×x× \sqrt{x^2-1}}{(x)^2 - (\sqrt{x^2 -1})^2}

= \dfrac{4x \sqrt{x^2-1}}{x^2 - x^2 +1}

= \dfrac{4x\sqrt{x^2-1}}{1}

= 4x\sqrt{x^2 -1}

Answer
------------

Creative Section B


5. Simplify:

a) 3\sqrt{20} + \dfrac{4}{\sqrt{5}} + \dfrac{\sqrt{5} + 3}{\sqrt{5} - 3}

Solution:
= \dfrac{3\sqrt{20} × \sqrt{5} + 4}{\sqrt{5}} + \dfrac{\sqrt{5}+3}{\sqrt{5}-3} × \dfrac{\sqrt{5}+3}{\sqrt{5}+3}

= \dfrac{3\sqrt{20×5} + 4}{\sqrt{5}} + \dfrac{(\sqrt{5} + 3)^2}{(\sqrt{5})^2 - 3^2}

= \dfrac{3\sqrt{10^2} + 4}{\sqrt{5}} + \dfrac{(\sqrt{5})^2 + 6\sqrt{5} + 3^2}{5 - 9}

= \dfrac{3×10 + 4}{\sqrt{5}} + \dfrac{5 + 9 + 6\sqrt{5}}{-4}

= \dfrac{34}{\sqrt{5}} - \dfrac{14 + 6\sqrt{5})}{4}

= \dfrac{34}{\sqrt{5}} × \dfrac{\sqrt{5}}{\sqrt{5}}  - \dfrac{2(7 + 3\sqrt{5})}{2×2}

= \dfrac{34 \sqrt{5}}{5} - \dfrac{7 + 3\sqrt{5}}{2}

= \dfrac{34\sqrt{5}×2- (7 + 3\sqrt{5})5}{5×2}

= \dfrac{68 \sqrt{5} - 35 - 15\sqrt{5}}{10}

= \dfrac{53\sqrt{5} - 35}{10}



c) \frac{2√10}{√3+1} - \frac{2√5}{√6+2} - \frac{√10}{√2 +1}

This answer is not yet confirmed. Please do not copy this.

= \frac{2√10}{√3+1} × \frac{√3-1}{√3-1} - \frac{2√5}{√6+2} × \frac{√6-2}{√6-2} - \frac{√10}{√2+1} × \frac{√2-1}{√2-1}

= \frac{2√10(√3-1)}{(√3)^2 -1^2} - \frac{2√5(√6-2)}{(√6)^2 - 2^2} - \frac{√10(√2-1)}{(√2)^2 -1^2}

= \frac{2√30 - 2√10}{3-1} - \frac{2√30 - 2√10}{6-2} - \frac{√20 - √10}{2 -1}

= \frac{2√30 - 2√10}{2} - \frac{2√30 - 2√10}{4} - \frac{√20 - √10 }{1}

= \frac{2(√30 - √10)}{2} - \frac{2(√30 - √10)}{2×2} - (√20 - √10)

= \frac{√30 - √10}{1} - \frac{√30 - √10}{2} - √20 + √10

= √30 - √10 - √20 + √10 - \frac{√30 - √10}{2}

= √30 - √20 - \frac{√30 - √10}{2}

= \frac{2(√30 - √20) - (√30 - √10)}{2}

= \frac{2√30 - 2√20 - √30 + √10}{2}

= \frac{√30 - 2√20 + √10}{2}


About vedanta EXCEL in MATHEMATICS Book 10

Author: Hukum Pd. Dahal
Editor: Tara Bahadur Magar

Vanasthali, Kathmandu, Nepal
+977-10-4382404, 01-4362082
vedantapublication@gmail.com

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Exercise 11.2 | Rationalisation of Surds | vedanta Excel in Mathematics | Class 10 is a collection of the solutions related to exercises of rationalisation from Surds Chapter for Nepal's Secondary Education Examination (SEE) appearing students.

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