# Operation on Rational & Irrational Numbers | Q5

###### Number System | Extra Questions For Class 9 Maths Chapter 1 | Polynomial expansion

Question 5: Simplify the following:
(a) {8 + $sqrt{5})}) $${8 - $sqrt{5}$})$b) {10 + $sqrt{3})}) $${6 + $sqrt{2}$})$c) ${$$sqrt {3} + $sqrt {11}$}^2) + ${$$sqrt {3} - \sqrt {11}$}^2) ## Target Centum in CBSE 10th Maths #### Free Online CBSE Coaching online.maxtute.com ### Video Explanation ## NCERT Solution to Class 10 Maths #### With Videos ### Explanatory Answer | Chapter Extra Question 5$a) {8 + $sqrt{5})}) $${8 - $sqrt{5}$}) The expression is of the form$x + y) (x – y) = x2 – y2
So, {8 + $sqrt{5})}) $${8 - $sqrt{5}$} =${8^2} - $sqrt{5}^2)) = 64 – 5 = 59$b) {10 + $sqrt{3})}) $${6 + $sqrt{2}$}) Simplifying in such terms is the same as expanding the terms of the expression = ${$10 $times 6$} + 10 $times \sqrt{2} + \sqrt{3} \times 6 + \sqrt{2} \times \sqrt{3}$ = $60 + 10\sqrt{2} + 6\sqrt{3} + \sqrt{6}$$c) ${$$sqrt {3} + $sqrt {11}$}^2 + {$$sqrt {3} - $sqrt {11}$}^2) ${$$sqrt {3} + \sqrt {11}$}^2) = ${$$sqrt {3}$}^2) + ${$$sqrt {11}$}^2 + 2 $times \sqrt{3} \times \sqrt{11}$ $= 3 + 11 + 2\sqrt{33} = 14 + 2\sqrt{33}$ ${$$sqrt {3} - \sqrt {11}$}^2) = ${$$sqrt {3}$}^2) + ${$$sqrt {11}$}^2 - 2 $times \sqrt{3} \times \sqrt{11}$ $= 3 + 11 - 2\sqrt{33} = 14 - 2\sqrt{33}$ Thus, ${$$sqrt {3} + \sqrt {11}$}^2) + ${$\sqrt {3} - \sqrt {11}$}^2 = 14 + 2$sqrt{33} + 14 - 2\sqrt{33}$
= 28

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