Chapter 3 of RD Sharma's Class 9 mathematics textbook focuses on the concept of rationalization, a crucial algebraic technique used to simplify expressions containing surds or irrational numbers. Exercise 3.2 specifically deals with the rationalization of denominators, which is an essential skill for simplifying complex fractions and solving equations involving irrational numbers. This set of problems aims to reinforce students' understanding of rationalization techniques and their application in various mathematical contexts.
Question 6. In each of the following determine rational numbers a and b:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Solution:
(i) We know that rationalisation factor for
\sqrt3+1\ is\ \sqrt3-1 . We will multiply numerator and denominator of the given expression\frac{\sqrt3-1}{\sqrt3+1}\ by\ \sqrt3-1 , to get
\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3-1}{\sqrt3-1}=\frac{\sqrt3^2+1^2-2\times\sqrt3\times1}{\sqrt3^2-1^2}\\ =\frac{3+1-2\sqrt3}{3-1}\\ =\frac{4-2\sqrt3}{2}\\ =2-\sqrt3 On equating rational and irrational terms, we get
a-b\sqrt3=2-\sqrt3\\ =2-1\sqrt3 Hence, we get a = 2, b = 1
(ii) We know that rationalisation factor for
2+\sqrt2\ is\ 2-\sqrt2 . We will multiply numerator and denominator of the given expression\frac{4+\sqrt2}{2+\sqrt2}\ by\ 2-\sqrt2 , to get
\frac{4+\sqrt2}{2+\sqrt2}\times\frac{2-\sqrt2}{2-\sqrt2}=\frac{4\times2-4\times\sqrt2+2\times\sqrt2-\sqrt2^2}{2^2-\sqrt2^2}\\ =\frac{8-4\sqrt2+2\sqrt2-2}{4-2}\\ =\frac{6-2\sqrt2}{2}\\ =3-\sqrt2 On equating rational and irrational terms, we get
a-\sqrt{b}=3-\sqrt2 Hence, we get a = 3, b = 2
(iii) We know that rationalisation factor for
3-\sqrt2\ is\ 3+\sqrt2 . We will multiply numerator and denominator of the given expression, to get
\frac{3+\sqrt2}{3-\sqrt2}\times\frac{3+\sqrt2}{3+\sqrt2}=\frac{3^2+\sqrt2^2+2\times3\times\sqrt2}{3^2-\sqrt2^2}\\ =\frac{9+2+6\sqrt2}{9-2}\\ =\frac{11+6\sqrt2}{7}\\ =\frac{11}{7}+\frac{6}{7}\sqrt2 On equating rational and irrational terms, we get
a+b\sqrt2=\frac{11}{7}+\frac{6}{7}\sqrt2 Hence, we get a =
\frac{11}7{} , b =\frac{6}{7} (iv) We know that rationalisation factor for
7+4\sqrt3\ is\ 7-4\sqrt3 . We will multiply numerator and denominator of the given expression\frac{5+3\sqrt3}{7+4\sqrt3}\ by\ 7-4\sqrt3 , to get
\frac{5+3\sqrt3}{7+4\sqrt3}\times\frac{7-4\sqrt3}{7-4\sqrt3}=\frac{5\times7-5\times4\times\sqrt3+3\times7\times\sqrt3-3\times4\times\sqrt3^2}{7^2-(4\sqrt3)^2}\\ =\frac{35-20\sqrt3+21\sqrt3-36}{49-48}\\ =\frac{\sqrt3-1}{1}\\ =\sqrt3-1 On equating rational and irrational terms, we get
a+b\sqrt3=\sqrt3-1\\ =-1+1\sqrt3 Hence, we get a = -1, b = 1
(v) We know that rationalisation factor for
\sqrt{11}+\sqrt7\ is\ \sqrt{11}-\sqrt7 . We will multiply numerator and denominator of the given expression\frac{\sqrt{11}-\sqrt7}{\sqrt{11}+\sqrt7}\ by\ \sqrt{11}-\sqrt7 , to get
\frac{\sqrt{11}-\sqrt7}{\sqrt{11}+\sqrt7}\times\frac{\sqrt{11}-\sqrt7}{\sqrt{11}-\sqrt7}=\frac{\sqrt{11}^2+\sqrt7^2-2\times\sqrt{11}\times\sqrt7}{\sqrt{11}^2-\sqrt7^2}\\ =\frac{11+7-2\sqrt{77}}{11-7}\\ =\frac{18-2\sqrt{77}}{4}\\ =\frac{9}{2}-\frac{1}{2}\sqrt{77} On equating rational and irrational terms, we get
a-b\sqrt{77}=\frac{9}{2}-\frac{1}{2}\sqrt{77} Hence, we get a =
\frac{9}{2} , b =\frac{1}{2} (vi) We know that rationalisation factor for
4-3\sqrt5\ in\ 4+3\sqrt5 . We will multiply numerator and denominator of the given expression\frac{4+3\sqrt5}{4-3\sqrt5}\ by\ 4+3\sqrt5 , to get
\frac{4+3\sqrt5}{4-3\sqrt5}\times\frac{4+3\sqrt5}{4+3\sqrt5}=\frac{4^2+(3\sqrt5)^2+2\times4\times3\sqrt5}{4^2-(3\sqrt5)^2}\\ =\frac{16+45+24\sqrt5}{16-45}\\ =\frac{61+24\sqrt5}{-29}\\ =-\frac{61}{29}-\frac{24}{29}\sqrt5 On equating rational and irrational terms, we get
a+b\sqrt5=-\frac{61}{29}-\frac{24}{29}\sqrt5 Hence, we get a =
-\frac{61}{29} , b =-\frac{24}{29}
Question 7. If x=2+\sqrt3 , find the value ofx^3+\frac{1}{x^3}
Solution:
We know that
x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)\left(x^2-1+\frac{1}{x^2}\right) . We have to find the value ofx^3+\frac{1}{x^3} As
x=2+\sqrt3 Therefore,
\frac{1}{x}=\frac{1}{2+\sqrt3} We know that rationalization factor for
2+\sqrt3\ is\ 2-\sqrt3 . We will multiply numerator and denominator of the given expression\frac{1}{2+\sqrt3}\ by\ 2-\sqrt3 to get,
\frac{1}{x}=\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}\\ =\frac{2-\sqrt3}{2^2-\sqrt3^2}\\ =\frac{2-\sqrt3}{4-3}\\ =2-\sqrt3 Putting the value of
x\ and\ \frac{1}{x} , we get
x^3+\frac{1}{x^3}=(2+\sqrt3+2-\sqrt3)((2+\sqrt3)^2-1+(2-\sqrt3)^2)\\ =4(2^2+(\sqrt3)^2+2\times2\times\sqrt3-1+2^2+(\sqrt3)^2-2\times2\times\sqrt3)\\ =4(4+3+4\sqrt3-1+4+3-4\sqrt3)\\ =52 Hence the value of the given expression is 52.
Question 8. If x=3+\sqrt8 . Find the value ofx^3+\frac{1}{x^3}
Solution:
We know that
x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2 . We have to find the value ofx^2+\frac{1}{x^2}\ as\ x=3+\sqrt8 Therefore,
\frac{1}{x}=\frac{1}{3+\sqrt8} We know that rationalization factor for
3+\sqrt8\ is\ 3-\sqrt8 .We will multiply numerator and denominator of the given expression
\frac{1}{3+\sqrt8}\ by\ 3-\sqrt8 to get,
\frac{1}{x}=\frac{1}{3+\sqrt8}\times\frac{3-\sqrt8}{3-\sqrt8}\\ =\frac{3-\sqrt8}{3^2-\sqrt8^2}\\ =\frac{3-\sqrt8}{9-8}\\ =3-\sqrt8 Putting the value of
x\ and\ \frac{1}{x} We get,
x^2+\frac{1}{x^2}=(3+\sqrt8+3-\sqrt8)^2-2\\ =(6)^2-2\\ =36-2\\ =34 Hence the given expression is simplified to 34.
Question 9. Find the value of \frac{6}{\sqrt5-\sqrt3} , it being given that\sqrt3=1.732\ and\ \sqrt5=2.236
Solution:
We know that for rationalization factor we will multiply denominator and numerator of the given expression
\frac{6}{\sqrt5-\sqrt3}\ by\ \sqrt5+\sqrt3 to get,
\frac{6}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}=\frac{6\sqrt5+6\sqrt3}{\sqrt5^2-\sqrt3^2}\\ =\frac{6\sqrt5+6\sqrt3}{5-3}\\ =\frac{6\sqrt5+6\sqrt3}{2}\\ =3\sqrt5+3\sqrt3 Putting the values of
\sqrt5\ and\ \sqrt3 we get,
3\sqrt5+3\sqrt3=3(2.236)+3(1.732)\\ =6.708+5.196\\ =11.904 Hence, value of the given expression is 11.904.
Question 10. Find the value of each of the following correct to three place of decimals, it being given that\sqrt2=1.4142,\ \sqrt3=1.732,\ \sqrt5=2.2360,\ \sqrt6=2.4495\ and\ \sqrt{10}=3.162
(i)
(ii)
Solution:
(i) We know that rationalization factor for
3+2\sqrt5\ is\ 3-2\sqrt5 We will multiply numerator and denominator of the given expression
\frac{3-\sqrt5}{3+2\sqrt5}\ by\ 3-2\sqrt5 to get
\frac{3-\sqrt5}{3+2\sqrt5}\times\frac{3-2\sqrt5}{3-2\sqrt5}=\frac{3^2-3\times2\times\sqrt5-3\times\sqrt5+2\times\sqrt5^2}{3^2-(2\sqrt5)^2}\\ =\frac{9-9\sqrt5+10}{9-20}\\ =\frac{19-9\sqrt5}{-11}\\ =\frac{9\sqrt5-19}{11} Putting the values of
\sqrt5 We get
\frac{9\sqrt5-19}{11}=\frac{9(2.236)-19}{11}\\ =\frac{20.124-19}{11}\\ =\frac{1.124}{11}\\ =0.102 Hence, the given expression is simplified to 0.102.
(ii) We know that rationalization factor for
3-2\sqrt2\ is\ 3+2\sqrt2 We will multiply numerator and denominator of the given expression
\frac{1+\sqrt2}{3-2\sqrt2}\ by\ 3+2\sqrt2 to get
\frac{1+\sqrt2}{3-2\sqrt2}\times\frac{3+2\sqrt2}{3+2\sqrt2}=\frac{3+2\times\sqrt2+3\times\sqrt2+2\times\sqrt2^2}{3^2-(2\sqrt2)^2}\\ =\frac{3+2\sqrt2+3\sqrt2+4}{9-8}\\ =\frac{7+5\sqrt2}{1}\\ =7+5\sqrt2 Putting the values of
\sqrt2 We get
7+5\sqrt2=7+5(1.4142)\\ =7+7.071\\ =14.071 Hence, the given expression is simplified to 14.071.
Question 11. Ifx=\frac{\sqrt3+1}{2} , find the value of4x^3+2x^2-8x+7
Solution:
We have,
x=\frac{\sqrt3+1}{2} It can be simplified as
2x-1=\sqrt3 On squaring both sides, we get
(2x-1)^2=\sqrt3^2\\ (2x)^2+1-2\times2x=3\\ 4x^2+1-4x=3\\ 4x^2-4x-2=0 The given equation can be rewritten as
4x^2+2x^2-8x+7=x(4x^2-4x-2)+\frac{6}{4}(4x^2-4x-2)+3+7 Therefore, we have
4x^3+2x^2-8x+7=x(0)+\frac{6}{4}(0)+3+7\\ =3+7\\ =10 Hence, the value of given expression is 10.
Summary
The practice questions in this set cover a range of rationalisation scenarios, including simple surds, compound surds, and expressions with multiple terms in the denominator. Students are challenged to rationalize denominators containing square roots, cube roots, and higher-order roots. Some problems involve algebraic manipulation alongside rationalisation, requiring students to apply multiple mathematical skills simultaneously. The questions progress in difficulty, starting with straightforward examples and advancing to more complex expressions that demand a deeper understanding of algebraic principles and rationalisation techniques.
