Samacheer Kalvi Class 9 Maths Solutions Chapter 2 Real Numbers Exercise 2.7

Question 1. Rationalise the denominator:
(i) \( \frac{1}{\sqrt{50}} \)
(ii) \( \frac{5}{3\sqrt{5}} \)
(iii) \( \frac{\sqrt{75}}{\sqrt{18}} \)
(iv) \( \frac{3\sqrt{5}}{\sqrt{6}} \)
Answer:
(i) To rationalise \( \frac{1}{\sqrt{50}} \), we first simplify the denominator:
\( \frac{1}{\sqrt{50}} = \frac{1}{\sqrt{25 \times 2}} = \frac{1}{5\sqrt{2}} \)
Now, we multiply the numerator and denominator by \( \sqrt{2} \):
\( \frac{1}{5\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{5 \times 2} = \frac{\sqrt{2}}{10} \)
(ii) To rationalise \( \frac{5}{3\sqrt{5}} \), we multiply the numerator and denominator by \( \sqrt{5} \):
\( \frac{5}{3\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{5\sqrt{5}}{3 \times 5} = \frac{\sqrt{5}}{3} \)
(iii) To rationalise \( \frac{\sqrt{75}}{\sqrt{18}} \), we first simplify both numerator and denominator:
\( \frac{\sqrt{75}}{\sqrt{18}} = \frac{\sqrt{25 \times 3}}{\sqrt{9 \times 2}} = \frac{5\sqrt{3}}{3\sqrt{2}} \)
Now, we multiply the numerator and denominator by \( \sqrt{2} \):
\( \frac{5\sqrt{3}}{3\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{5\sqrt{6}}{3 \times 2} = \frac{5\sqrt{6}}{6} \)
(iv) To rationalise \( \frac{3\sqrt{5}}{\sqrt{6}} \), we multiply the numerator and denominator by \( \sqrt{6} \):
\( \frac{3\sqrt{5}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{3\sqrt{30}}{6} = \frac{\sqrt{30}}{2} \)
In simple words: Rationalising the denominator means removing any square root signs from the bottom part of a fraction. You do this by multiplying both the top and bottom by the square root from the denominator, or its conjugate. This makes calculations easier to handle.

🎯 Exam Tip: Always simplify the radical expressions first before rationalising, as it often makes the numbers smaller and calculations easier.

Question 2. Rationalise the denominator and simplify:
(i) \( \frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}} \)
(iii) \( \frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}} \)
(iv) \( \frac{\sqrt{5}}{\sqrt{6}+2} – \frac{\sqrt{5}}{\sqrt{6}-2} \)
Answer:
(i) First, simplify the terms in the expression:
\( \frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}} = \frac{\sqrt{16 \times 3}+\sqrt{16 \times 2}}{\sqrt{9 \times 3}-\sqrt{9 \times 2}} \)
\( = \frac{4\sqrt{3}+4\sqrt{2}}{3\sqrt{3}-3\sqrt{2}} = \frac{4(\sqrt{3}+\sqrt{2})}{3(\sqrt{3}-\sqrt{2})} \)
Next, multiply the numerator and denominator by the conjugate of the denominator, which is \( (\sqrt{3}+\sqrt{2}) \):
\( = \frac{4(\sqrt{3}+\sqrt{2})}{3(\sqrt{3}-\sqrt{2})} \times \frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})} \)
\( = \frac{4(\sqrt{3}+\sqrt{2})^2}{3((\sqrt{3})^2 - (\sqrt{2})^2)} \)
Using the identity \( (a+b)^2 = a^2+2ab+b^2 \) and \( (a-b)(a+b) = a^2-b^2 \):
\( = \frac{4((\sqrt{3})^2 + 2\sqrt{3}\sqrt{2} + (\sqrt{2})^2)}{3(3-2)} \)
\( = \frac{4(3 + 2\sqrt{6} + 2)}{3(1)} = \frac{4(5+2\sqrt{6})}{3} \)

(iii) To rationalise \( \frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}} \), multiply the numerator and denominator by the conjugate of the denominator, which is \( (3\sqrt{5}+2\sqrt{6}) \):
\( \frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}} \times \frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}} \)
Using the identity \( (a-b)(a+b) = a^2-b^2 \) for the denominator:
\( = \frac{(2\sqrt{6})(3\sqrt{5}) + (2\sqrt{6})(2\sqrt{6}) - (\sqrt{5})(3\sqrt{5}) - (\sqrt{5})(2\sqrt{6})}{(3\sqrt{5})^2 - (2\sqrt{6})^2} \)
\( = \frac{6\sqrt{30} + 4 \times 6 - 3 \times 5 - 2\sqrt{30}}{9 \times 5 - 4 \times 6} \)
\( = \frac{6\sqrt{30} + 24 - 15 - 2\sqrt{30}}{45 - 24} \)
\( = \frac{(6-2)\sqrt{30} + 9}{21} = \frac{4\sqrt{30} + 9}{21} \)

(iv) First, combine the two fractions. The common denominator is \( (\sqrt{6}+2)(\sqrt{6}-2) \):
\( \frac{\sqrt{5}}{\sqrt{6}+2} – \frac{\sqrt{5}}{\sqrt{6}-2} = \sqrt{5} \left( \frac{1}{\sqrt{6}+2} - \frac{1}{\sqrt{6}-2} \right) \)
\( = \sqrt{5} \left( \frac{(\sqrt{6}-2) - (\sqrt{6}+2)}{(\sqrt{6}+2)(\sqrt{6}-2)} \right) \)
Using the identity \( (a-b)(a+b) = a^2-b^2 \) for the denominator:
\( = \sqrt{5} \left( \frac{\sqrt{6}-2-\sqrt{6}-2}{(\sqrt{6})^2 - 2^2} \right) \)
\( = \sqrt{5} \left( \frac{-4}{6-4} \right) = \sqrt{5} \left( \frac{-4}{2} \right) \)
\( = \sqrt{5}(-2) = -2\sqrt{5} \)
In simple words: When you need to get rid of square roots in the denominator, especially with two terms, multiply by its conjugate. A conjugate just means changing the plus to a minus, or vice versa. This uses a special math rule that helps clear the roots.

🎯 Exam Tip: Remember that \( (a+b)(a-b) = a^2-b^2 \) is key for rationalising binomial denominators. Also, simplify all square roots at the beginning to avoid large numbers.

Question 3. Find the value of a and b if \( \frac{\sqrt{7}-2}{\sqrt{7}+2} = a\sqrt{7} + b \).
Answer: We start by rationalising the left-hand side (LHS) of the equation:
\( \frac{\sqrt{7}-2}{\sqrt{7}+2} = \frac{\sqrt{7}-2}{\sqrt{7}+2} \times \frac{\sqrt{7}-2}{\sqrt{7}-2} \)
\( = \frac{(\sqrt{7}-2)^2}{(\sqrt{7})^2 - (2)^2} \)
Using \( (a-b)^2 = a^2-2ab+b^2 \) for the numerator and \( (a+b)(a-b) = a^2-b^2 \) for the denominator:
\( = \frac{(\sqrt{7})^2 - 2(\sqrt{7})(2) + 2^2}{7-4} \)
\( = \frac{7 - 4\sqrt{7} + 4}{3} \)
\( = \frac{11 - 4\sqrt{7}}{3} \)
Now, we rewrite this expression to match the form \( a\sqrt{7} + b \):
\( = \frac{11}{3} - \frac{4\sqrt{7}}{3} \)
\( = -\frac{4}{3}\sqrt{7} + \frac{11}{3} \)
By comparing this with \( a\sqrt{7} + b \), we can find the values of \( a \) and \( b \).
Comparing the coefficient of \( \sqrt{7} \): \( a = -\frac{4}{3} \)
Comparing the constant term: \( b = \frac{11}{3} \)
Thus, the value of \( a = -\frac{4}{3} \) and \( b = \frac{11}{3} \).
In simple words: First, remove the square root from the bottom of the left fraction. Then, rearrange the result so it looks exactly like \( a\sqrt{7} + b \). Once it matches, you can easily see what numbers 'a' and 'b' stand for by comparing both sides.

🎯 Exam Tip: Remember to express the simplified fraction in the form \( (\text{constant}) + (\text{coefficient})\sqrt{\text{root}} \) to correctly identify 'a' and 'b' by comparing terms.

Question 4. If \( x = \sqrt{7} + 2 \), then find the value of \( x^2 + \frac{1}{x^2} \).
Answer: Given \( x = \sqrt{5} + 2 \). (Note: The solution is based on \( \sqrt{5} \) as shown in the original steps)
First, find \( x^2 \):
\( x^2 = (\sqrt{5}+2)^2 \)
Using the identity \( (a+b)^2 = a^2+2ab+b^2 \):
\( x^2 = (\sqrt{5})^2 + 2(\sqrt{5})(2) + 2^2 \)
\( = 5 + 4\sqrt{5} + 4 = 9 + 4\sqrt{5} \)
Next, find \( \frac{1}{x} \):
\( \frac{1}{x} = \frac{1}{\sqrt{5}+2} \)
Rationalise the denominator by multiplying by the conjugate \( (\sqrt{5}-2) \):
\( \frac{1}{\sqrt{5}+2} \times \frac{\sqrt{5}-2}{\sqrt{5}-2} = \frac{\sqrt{5}-2}{(\sqrt{5})^2 - 2^2} \)
\( = \frac{\sqrt{5}-2}{5-4} = \frac{\sqrt{5}-2}{1} = \sqrt{5}-2 \)
Now, find \( \frac{1}{x^2} \):
\( \frac{1}{x^2} = (\frac{1}{x})^2 = (\sqrt{5}-2)^2 \)
Using the identity \( (a-b)^2 = a^2-2ab+b^2 \):
\( = (\sqrt{5})^2 - 2(\sqrt{5})(2) + 2^2 \)
\( = 5 - 4\sqrt{5} + 4 = 9 - 4\sqrt{5} \)
Finally, add \( x^2 \) and \( \frac{1}{x^2} \):
\( x^2 + \frac{1}{x^2} = (9+4\sqrt{5}) + (9-4\sqrt{5}) \)
\( = 9 + 4\sqrt{5} + 9 - 4\sqrt{5} = 18 \)
The value of \( x^2 + \frac{1}{x^2} = 18 \).
In simple words: First, work out \( x \) squared. Then, find what \( 1/x \) is by getting rid of the square root at the bottom. Square that result to get \( 1/x^2 \). Finally, add the two results together. Notice how the \( \sqrt{5} \) terms cancel out, making the answer a whole number.

🎯 Exam Tip: When faced with \( x + \frac{1}{x} \) or \( x^2 + \frac{1}{x^2} \) problems, remember to rationalise the denominator of \( \frac{1}{x} \) or \( \frac{1}{x^2} \) first, as it often simplifies the expression significantly.

Question 5. Given \( \sqrt{2} = 1.414 \), find the value of \( \frac{8 – 5\sqrt{2}}{3 – 2\sqrt{2}} \) (to 3 decimal places).
Answer: We need to find the value of \( \frac{8 – 5\sqrt{2}}{3 – 2\sqrt{2}} \).
First, rationalise the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is \( (3+2\sqrt{2}) \):
\( \frac{8 – 5\sqrt{2}}{3 – 2\sqrt{2}} = \frac{8 – 5\sqrt{2}}{3 – 2\sqrt{2}} \times \frac{3+2\sqrt{2}}{3+2\sqrt{2}} \)
Using the identity \( (a-b)(a+b) = a^2-b^2 \) for the denominator:
\( = \frac{(8)(3) + (8)(2\sqrt{2}) - (5\sqrt{2})(3) - (5\sqrt{2})(2\sqrt{2})}{(3)^2 - (2\sqrt{2})^2} \)
\( = \frac{24 + 16\sqrt{2} - 15\sqrt{2} - 10 \times 2}{9 - 4 \times 2} \)
\( = \frac{24 + \sqrt{2} - 20}{9 - 8} \)
\( = \frac{4 + \sqrt{2}}{1} = 4 + \sqrt{2} \)
Now, substitute the given value \( \sqrt{2} = 1.414 \):
\( = 4 + 1.414 = 5.414 \)
The value of the expression is \( 5.414 \).
In simple words: To solve this, first get rid of the square root from the bottom of the fraction by multiplying by its special opposite. Once the fraction is simpler, put in the number for \( \sqrt{2} \) that was given. Then, just add the numbers to get your final answer.

🎯 Exam Tip: Always rationalise the denominator before substituting numerical values for square roots; it simplifies the calculation and reduces chances of error.