Ganit Prakash - Class-X - Quadratic Surd
Let us work out 9.2 Algebra
📘 Let us work out 9.2 Solutions
Quadratic Surd(a) Let us find the product of \( 3^{\frac{1}{2}} \) and \( \sqrt{3} \)
(b) Let us write what should be multiplied with \( 2\sqrt{2} \) to get the product 4.
(c) Let us calculate the product of \( 3\sqrt{5} \) and \( 5\sqrt{3} \)
(d) If \( \sqrt{6} \times \sqrt{15} = x\sqrt{10} \), then let us write by calculating the value of \( x \).
(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 \( x \).
(a)
\[\begin{array}{l} 3^{\frac{1}{2}} \times \sqrt{3} \\ = \sqrt{3} \times \sqrt{3} \\ = 3 \end{array}\](b)
Let the required number be \( x \).
\[\begin{array}{l} x \times 2\sqrt{2} = 4 \\ \Rightarrow x = \frac{4}{2\sqrt{2}} \\ \Rightarrow x = \frac{2}{\sqrt{2}} \\ \Rightarrow x = \sqrt{2} \end{array}\]Therefore, \(\sqrt{2}\) should be multiplied.
(c)
\[\begin{array}{l} 3\sqrt{5} \times 5\sqrt{3} \\ = (3 \times 5) \times \sqrt{5 \times 3} \\ = 15\sqrt{15} \end{array}\](d)
\[\begin{array}{l} \sqrt{6} \times \sqrt{15} = x\sqrt{10} \\ \Rightarrow \sqrt{90} = x\sqrt{10} \\ \Rightarrow \sqrt{9 \times 10} = x\sqrt{10} \\ \Rightarrow 3\sqrt{10} = x\sqrt{10} \\ \Rightarrow x = 3 \end{array}\](e)
\[\begin{array}{l} (\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3}) = 25 - x^2 \\ \Rightarrow (\sqrt{5})^2 - (\sqrt{3})^2 = 25 - x^2 \quad [\text{Since } (a+b)(a-b) = a^2 - b^2] \\ \Rightarrow 5 - 3 = 25 - x^2 \\ \Rightarrow 2 = 25 - x^2 \\ \Rightarrow x^2 = 25 - 2 \\ \Rightarrow x^2 = 23 \\ \Rightarrow x = \pm\sqrt{23} \end{array}\](a) \( \sqrt{7} \times \sqrt{14} \)
(b) \( \sqrt{12} \times 2\sqrt{3} \)
(c) \( \sqrt{5} \times \sqrt{15} \times \sqrt{3} \)
(d) \( \sqrt{2} (3 + \sqrt{5}) \)
(e) \( (\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3}) \)
(f) \( (2\sqrt{3} + 3\sqrt{2})(4\sqrt{2} + \sqrt{5}) \)
(g) \( (\sqrt{3} + 1)(\sqrt{3} - 1)(2 - \sqrt{3})(4 + 2\sqrt{3}) \)
(a)
\[\begin{array}{l} \sqrt{7} \times \sqrt{14} \\ = \sqrt{7 \times 14} \\ = \sqrt{7 \times 7 \times 2} \\ = 7\sqrt{2} \end{array}\](b)
\[\begin{array}{l} \sqrt{12} \times 2\sqrt{3} \\ = \sqrt{4 \times 3} \times 2\sqrt{3} \\ = 2\sqrt{3} \times 2\sqrt{3} \\ = 4 \times (\sqrt{3})^2 \\ = 4 \times 3 \\ = 12 \end{array}\](c)
\[\begin{array}{l} \sqrt{5} \times \sqrt{15} \times \sqrt{3} \\ = \sqrt{5 \times 15 \times 3} \\ = \sqrt{5 \times (5 \times 3) \times 3} \\ = \sqrt{5 \times 5 \times 3 \times 3} \\ = 5 \times 3 \\ = 15 \end{array}\](d)
\[\begin{array}{l} \sqrt{2} (3 + \sqrt{5}) \\ = 3\sqrt{2} + \sqrt{2 \times 5} \\ = 3\sqrt{2} + \sqrt{10} \end{array}\](e)
\[\begin{array}{l} (\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3}) \\ = (\sqrt{2})^2 - (\sqrt{3})^2 \\ = 2 - 3 \\ = -1 \end{array}\](f)
\[\begin{array}{l} (2\sqrt{3} + 3\sqrt{2})(4\sqrt{2} + \sqrt{5}) \\ = 2\sqrt{3}(4\sqrt{2} + \sqrt{5}) + 3\sqrt{2}(4\sqrt{2} + \sqrt{5}) \\ = 8\sqrt{6} + 2\sqrt{15} + 12(\sqrt{2})^2 + 3\sqrt{10} \\ = 8\sqrt{6} + 2\sqrt{15} + 12(2) + 3\sqrt{10} \\ = 8\sqrt{6} + 2\sqrt{15} + 24 + 3\sqrt{10} \end{array}\](g)
\[\begin{array}{l} (\sqrt{3} + 1)(\sqrt{3} - 1)(2 - \sqrt{3})(4 + 2\sqrt{3}) \\ = \{(\sqrt{3})^2 - 1^2\} \times (2 - \sqrt{3}) \times 2(2 + \sqrt{3}) \\ = (3 - 1) \times 2 \times (2 - \sqrt{3})(2 + \sqrt{3}) \\ = 2 \times 2 \times \{2^2 - (\sqrt{3})^2\} \\ = 4 \times (4 - 3) \\ = 4 \times 1 \\ = 4 \end{array}\](a) If \( \sqrt{x} \) is the rationalising factor of \( \sqrt{5} \), let us write by calculating what is the smallest value of x (where x is an integer)
(b) Let us calculate the value of \( 3\sqrt{2} \div 3 \)
(c) Let us write which factor should we multiply the denominator to rationalise the denominator of \( 7 \div \sqrt{48} \).
(d) Let us calculate the rationalising factor of \( (\sqrt{5} + 2) \) which is also its conjugate surd.
(e) If \( (\sqrt{5} + \sqrt{2}) \div \sqrt{7} = \frac{1}{7} (\sqrt{35} + a) \), Let us calculate the value of a.
(f) Let us write a rationalising factor of the denominator of \( \frac{5}{\sqrt{3}-2} \), which is not its conjugate surd.
(a) For \( \sqrt{x} \) to be a rationalising factor of \( \sqrt{5} \), their product must be rational.
\[\begin{array}{l} \sqrt{5} \times \sqrt{x} = \sqrt{5x} \end{array}\]For \( \sqrt{5x} \) to be a rational number, \( 5x \) must be a perfect square. The smallest positive integer value for \( x \) that makes \( 5x \) a perfect square is 5 (since \( 5 \times 5 = 25 \)).
Answer: The smallest integer value of \( x \) is 5.
(b)
\[\begin{array}{l} 3\sqrt{2} \div 3 = \frac{3\sqrt{2}}{3} = \sqrt{2} \end{array}\]Answer: \( \sqrt{2} \)
(c)
\[\begin{array}{l} 7 \div \sqrt{48} = \frac{7}{\sqrt{16 \times 3}} = \frac{7}{4\sqrt{3}} \end{array}\]To rationalise the denominator \( 4\sqrt{3} \), we need to multiply it by the smallest irrational factor, which is \( \sqrt{3} \).
Answer: We should multiply by \( \sqrt{3} \).
(d) A conjugate surd is obtained by changing the sign of the irrational part of the mixed surd. Both the sum and the product of a surd and its conjugate are rational.
The irrational part of \( (\sqrt{5} + 2) \) or \( (2 + \sqrt{5}) \) is \( \sqrt{5} \).
Changing its sign, the conjugate surd is \( -\sqrt{5} + 2 \) or \( 2 - \sqrt{5} \).
Answer: \( 2 - \sqrt{5} \)
(e)
\[\begin{array}{l} \frac{\sqrt{5} + \sqrt{2}}{\sqrt{7}} = \frac{(\sqrt{5} + \sqrt{2}) \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} \\ = \frac{\sqrt{35} + \sqrt{14}}{7} \\ = \frac{1}{7}(\sqrt{35} + \sqrt{14}) \end{array}\]Comparing this with \( \frac{1}{7}(\sqrt{35} + a) \), we get \( a = \sqrt{14} \).
Answer: \( a \) = \( \sqrt{14} \)
(f) The denominator is \( \sqrt{3} - 2 \). Its conjugate surd is obtained by changing the sign of the surd part, which is \( -\sqrt{3} - 2 \).
A rationalising factor that is not its conjugate is formed by changing the sign of the rational part.
Rationalising factor (not conjugate) = \( \sqrt{3} + 2 \).
(Verification: Product = \( (\sqrt{3} - 2)(\sqrt{3} + 2) = 3 - 4 = -1 \), which is rational, but sum = \( 2\sqrt{3} \), which is irrational).
Answer: \( \sqrt{3} + 2 \)
To find the conjugate surd, we change the sign of the irrational (surd) part.
- The conjugate surd of \( 9 - 4\sqrt{5} \) is \( 9 + 4\sqrt{5} \).
- The conjugate surd of \( -2 - \sqrt{7} \) is \( -2 + \sqrt{7} \).
(i) \( \sqrt{5} + \sqrt{2} \) (ii) \( 13 + \sqrt{6} \) (iii) \( \sqrt{8} - 3 \) (iv) \( \sqrt{17} - \sqrt{15} \)
Note: The textbook asks for two rationalising factors for each mixed surd (often loosely referred to as conjugates in the English translation). We can find them by changing the sign of either of the terms.
(i) \( \sqrt{5} + \sqrt{2} \)
- Factor 1: \( \sqrt{5} - \sqrt{2} \)
- Factor 2: \( -\sqrt{5} + \sqrt{2} \)
(ii) \( 13 + \sqrt{6} \)
- Factor 1: \( 13 - \sqrt{6} \)
- Factor 2: \( -13 + \sqrt{6} \)
(iii) \( \sqrt{8} - 3 \)
- Factor 1: \( -\sqrt{8} - 3 \)
- Factor 2: \( \sqrt{8} + 3 \)
(iv) \( \sqrt{17} - \sqrt{15} \)
- Factor 1: \( \sqrt{17} + \sqrt{15} \)
- Factor 2: \( -\sqrt{17} - \sqrt{15} \)
(i) \(\frac{2\sqrt{3} + 3\sqrt{2}}{\sqrt{6}}\) (ii) \(\frac{\sqrt{2} - 1 + \sqrt{6}}{\sqrt{5}}\) (iii) \(\frac{\sqrt{3} + 1}{\sqrt{3} - 1}\)
(iv) \(\frac{3 + \sqrt{5}}{\sqrt{7} - \sqrt{3}}\) (v) \(\frac{3\sqrt{2} + 1}{2\sqrt{5} - 1}\) (vi) \(\frac{3\sqrt{2} + 2\sqrt{3}}{3\sqrt{2} - 2\sqrt{3}}\)
(i) Multiply numerator and denominator by \(\sqrt{6}\):
\[\begin{array}{l} \frac{2\sqrt{3} + 3\sqrt{2}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} \\ = \frac{\sqrt{6}(2\sqrt{3} + 3\sqrt{2})}{6} \\ = \frac{2\sqrt{18} + 3\sqrt{12}}{6} \\ = \frac{2\sqrt{9 \times 2} + 3\sqrt{4 \times 3}}{6} \\ = \frac{2(3\sqrt{2}) + 3(2\sqrt{3})}{6} \\ = \frac{6\sqrt{2} + 6\sqrt{3}}{6} \\ = \frac{6(\sqrt{2} + \sqrt{3})}{6} = \sqrt{2} + \sqrt{3} \end{array}\](ii) Multiply numerator and denominator by \(\sqrt{5}\):
\[\begin{array}{l} \frac{\sqrt{2} - 1 + \sqrt{6}}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} \\ = \frac{\sqrt{5}(\sqrt{2} - 1 + \sqrt{6})}{5} \\ = \frac{\sqrt{10} - \sqrt{5} + \sqrt{30}}{5} \end{array}\](iii) Multiply numerator and denominator by the conjugate of the denominator, i.e., \((\sqrt{3} + 1)\):
\[\begin{array}{l} \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \\ = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} \\ = \frac{(\sqrt{3})^2 + 2(\sqrt{3})(1) + 1^2}{3 - 1} \\ = \frac{3 + 2\sqrt{3} + 1}{2} \\ = \frac{4 + 2\sqrt{3}}{2} \\ = \frac{2(2 + \sqrt{3})}{2} = 2 + \sqrt{3} \end{array}\](iv) Multiply numerator and denominator by \((\sqrt{7} + \sqrt{3})\):
\[\begin{array}{l} \frac{3 + \sqrt{5}}{\sqrt{7} - \sqrt{3}} \times \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} + \sqrt{3}} \\ = \frac{(3 + \sqrt{5})(\sqrt{7} + \sqrt{3})}{(\sqrt{7})^2 - (\sqrt{3})^2} \\ = \frac{3\sqrt{7} + 3\sqrt{3} + \sqrt{35} + \sqrt{15}}{7 - 3} \\ = \frac{3\sqrt{7} + 3\sqrt{3} + \sqrt{35} + \sqrt{15}}{4} \end{array}\](v) Multiply numerator and denominator by \( (2\sqrt{5} + 1) \):
\[\begin{array}{l} \frac{3\sqrt{2} + 1}{2\sqrt{5} - 1} \times \frac{2\sqrt{5} + 1}{2\sqrt{5} + 1} \\ = \frac{(3\sqrt{2} + 1)(2\sqrt{5} + 1)}{(2\sqrt{5})^2 - 1^2} \\ = \frac{6\sqrt{10} + 3\sqrt{2} + 2\sqrt{5} + 1}{4(5) - 1} \\ = \frac{6\sqrt{10} + 3\sqrt{2} + 2\sqrt{5} + 1}{20 - 1} \\ = \frac{6\sqrt{10} + 3\sqrt{2} + 2\sqrt{5} + 1}{19} \end{array}\](vi) Multiply numerator and denominator by \( (3\sqrt{2} + 2\sqrt{3}) \):
\[\begin{array}{l} \frac{3\sqrt{2} + 2\sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} \times \frac{3\sqrt{2} + 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} \\ = \frac{(3\sqrt{2} + 2\sqrt{3})^2}{(3\sqrt{2})^2 - (2\sqrt{3})^2} \\ = \frac{(3\sqrt{2})^2 + 2(3\sqrt{2})(2\sqrt{3}) + (2\sqrt{3})^2}{9(2) - 4(3)} \\ = \frac{18 + 12\sqrt{6} + 12}{18 - 12} \\ = \frac{30 + 12\sqrt{6}}{6} \\ = \frac{6(5 + 2\sqrt{6})}{6} = 5 + 2\sqrt{6} \end{array}\](i) \( 3\sqrt{2} + \sqrt{5} \ , \ \sqrt{2} + 1 \)
(ii) \( 2\sqrt{3} - \sqrt{2} \ , \ \sqrt{2} - \sqrt{3} \)
(iii) \( 3 + \sqrt{6} \ , \ \sqrt{3} + \sqrt{2} \)
(i) Division expression: \(\frac{3\sqrt{2} + \sqrt{5}}{\sqrt{2} + 1}\)
Rationalising the divisor by multiplying with \( (\sqrt{2} - 1) \):
\[\begin{array}{l} = \frac{3\sqrt{2} + \sqrt{5}}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} \\ = \frac{(3\sqrt{2} + \sqrt{5})(\sqrt{2} - 1)}{(\sqrt{2})^2 - 1^2} \\ = \frac{3(\sqrt{2})^2 - 3\sqrt{2} + \sqrt{10} - \sqrt{5}}{2 - 1} \\ = \frac{3(2) - 3\sqrt{2} + \sqrt{10} - \sqrt{5}}{1} \\ = 6 - 3\sqrt{2} + \sqrt{10} - \sqrt{5} \end{array}\](ii) Division expression: \(\frac{2\sqrt{3} - \sqrt{2}}{\sqrt{2} - \sqrt{3}}\)
Rationalising the divisor by multiplying with \( (\sqrt{2} + \sqrt{3}) \):
\[\begin{array}{l} = \frac{2\sqrt{3} - \sqrt{2}}{\sqrt{2} - \sqrt{3}} \times \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{3}} \\ = \frac{(2\sqrt{3} - \sqrt{2})(\sqrt{2} + \sqrt{3})}{(\sqrt{2})^2 - (\sqrt{3})^2} \\ = \frac{2\sqrt{6} + 2(\sqrt{3})^2 - (\sqrt{2})^2 - \sqrt{6}}{2 - 3} \\ = \frac{\sqrt{6} + 2(3) - 2}{-1} \\ = \frac{\sqrt{6} + 6 - 2}{-1} \\ = \frac{\sqrt{6} + 4}{-1} \\ = -4 - \sqrt{6} \end{array}\](iii) Division expression: \(\frac{3 + \sqrt{6}}{\sqrt{3} + \sqrt{2}}\)
Rationalising the divisor by multiplying with \( (\sqrt{3} - \sqrt{2}) \):
\[\begin{array}{l} = \frac{3 + \sqrt{6}}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\ = \frac{(3 + \sqrt{6})(\sqrt{3} - \sqrt{2})}{(\sqrt{3})^2 - (\sqrt{2})^2} \\ = \frac{3\sqrt{3} - 3\sqrt{2} + \sqrt{18} - \sqrt{12}}{3 - 2} \\ = \frac{3\sqrt{3} - 3\sqrt{2} + \sqrt{9 \times 2} - \sqrt{4 \times 3}}{1} \\ = 3\sqrt{3} - 3\sqrt{2} + 3\sqrt{2} - 2\sqrt{3} \\ = \sqrt{3} \end{array}\](i) \( \frac{2\sqrt{5} + 1}{\sqrt{5} + 1} - \frac{4\sqrt{5} - 1}{\sqrt{5} - 1} \)
(ii) \( \frac{8 + 3\sqrt{2}}{3 + \sqrt{5}} - \frac{8 - 3\sqrt{2}}{3 - \sqrt{5}} \)
(i) Expression:
\[\begin{array}{l} = \frac{2\sqrt{5} + 1}{\sqrt{5} + 1} - \frac{4\sqrt{5} - 1}{\sqrt{5} - 1} \end{array}\]Taking the LCM of the denominators, which is \( (\sqrt{5} + 1)(\sqrt{5} - 1) = (\sqrt{5})^2 - 1^2 = 5 - 1 = 4 \):
\[\begin{array}{l} = \frac{(2\sqrt{5} + 1)(\sqrt{5} - 1) - (4\sqrt{5} - 1)(\sqrt{5} + 1)}{4} \\ = \frac{(2(5) - 2\sqrt{5} + \sqrt{5} - 1) - (4(5) + 4\sqrt{5} - \sqrt{5} - 1)}{4} \\ = \frac{(10 - \sqrt{5} - 1) - (20 + 3\sqrt{5} - 1)}{4} \\ = \frac{(9 - \sqrt{5}) - (19 + 3\sqrt{5})}{4} \\ = \frac{9 - \sqrt{5} - 19 - 3\sqrt{5}}{4} \\ = \frac{-10 - 4\sqrt{5}}{4} \\ = \frac{2(-5 - 2\sqrt{5})}{4} \\ = \frac{-5 - 2\sqrt{5}}{2} \end{array}\](ii) Expression:
\[\begin{array}{l} = \frac{8 + 3\sqrt{2}}{3 + \sqrt{5}} - \frac{8 - 3\sqrt{2}}{3 - \sqrt{5}} \end{array}\]Taking the LCM of the denominators, which is \( (3 + \sqrt{5})(3 - \sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4 \):
\[\begin{array}{l} = \frac{(8 + 3\sqrt{2})(3 - \sqrt{5}) - (8 - 3\sqrt{2})(3 + \sqrt{5})}{4} \\ = \frac{(24 - 8\sqrt{5} + 9\sqrt{2} - 3\sqrt{10}) - (24 + 8\sqrt{5} - 9\sqrt{2} - 3\sqrt{10})}{4} \\ = \frac{24 - 8\sqrt{5} + 9\sqrt{2} - 3\sqrt{10} - 24 - 8\sqrt{5} + 9\sqrt{2} + 3\sqrt{10}}{4} \\ = \frac{-16\sqrt{5} + 18\sqrt{2}}{4} \\ = \frac{2(-8\sqrt{5} + 9\sqrt{2})}{4} \\ = \frac{9\sqrt{2} - 8\sqrt{5}}{2} \end{array}\]
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