Ganit Prakash - Class-X - Quadratic Surd
Let us work out 9.3 Algebra
📘 Let us work out 9.3 Solutions
Quadratic Surd(a) If \( \text{m} + \frac{1}{\text{m}} = \sqrt{3} \), let us calculate simplified values of (i) \( \text{m}^2 + \frac{1}{\text{m}^2} \) and (ii) \( \text{m}^3 + \frac{1}{\text{m}^3} \)
(b) Let us show that, \( \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} - \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} = 2\sqrt{15} \)
(a) Given: \( m + \frac{1}{m} = \sqrt{3} \)
(i) \( m^2 + \frac{1}{m^2} \)
\[\begin{array}{l} = \left(m + \frac{1}{m}\right)^2 - 2 \cdot m \cdot \frac{1}{m} \\ = (\sqrt{3})^2 - 2 \\ = 3 - 2 \\ = 1 \end{array}\](ii) \( m^3 + \frac{1}{m^3} \)
\[\begin{array}{l} = \left(m + \frac{1}{m}\right)^3 - 3 \cdot m \cdot \frac{1}{m}\left(m + \frac{1}{m}\right) \\ = (\sqrt{3})^3 - 3(\sqrt{3}) \\ = 3\sqrt{3} - 3\sqrt{3} \\ = 0 \end{array}\](b)
L.H.S:
\[\begin{array}{l} = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} - \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} \\ = \frac{(\sqrt{5} + \sqrt{3})^2 - (\sqrt{5} - \sqrt{3})^2}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})} \\ = \frac{4 \cdot \sqrt{5} \cdot \sqrt{3}}{(\sqrt{5})^2 - (\sqrt{3})^2} \quad [\text{Using } (a+b)^2 - (a-b)^2 = 4ab] \\ = \frac{4\sqrt{15}}{5 - 3} \\ = \frac{4\sqrt{15}}{2} \\ = 2\sqrt{15} \end{array}\]R.H.S: \( 2\sqrt{15} \)
\( \therefore \) L.H.S. = R.H.S. [Proved]
(a) \( \frac{\sqrt{2}(2+\sqrt{3})}{\sqrt{3}(\sqrt{3}+1)} - \frac{\sqrt{2}(2-\sqrt{3})}{\sqrt{3}(\sqrt{3}-1)} \)
(b) \( \frac{3\sqrt{7}}{\sqrt{5}+\sqrt{2}} - \frac{5\sqrt{5}}{\sqrt{2}+\sqrt{7}} + \frac{2\sqrt{2}}{\sqrt{7}+\sqrt{5}} \)
(c) \( \frac{4\sqrt{3}}{2-\sqrt{2}} - \frac{30}{4\sqrt{3}-\sqrt{18}} - \frac{\sqrt{18}}{3-\sqrt{12}} \)
(d) \( \frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}} - \frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}} + \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} \)
(a)
\[\begin{array}{l} \frac{\sqrt{2}(2+\sqrt{3})}{\sqrt{3}(\sqrt{3}+1)} - \frac{\sqrt{2}(2-\sqrt{3})}{\sqrt{3}(\sqrt{3}-1)} \\ = \frac{\sqrt{2}}{\sqrt{3}} \left[ \frac{2+\sqrt{3}}{\sqrt{3}+1} - \frac{2-\sqrt{3}}{\sqrt{3}-1} \right] \\ = \frac{\sqrt{2}}{\sqrt{3}} \left[ \frac{(2+\sqrt{3})(\sqrt{3}-1) - (2-\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3}+1)(\sqrt{3}-1)} \right] \\ = \frac{\sqrt{2}}{\sqrt{3}} \left[ \frac{(2\sqrt{3} - 2 + 3 - \sqrt{3}) - (2\sqrt{3} + 2 - 3 - \sqrt{3})}{(\sqrt{3})^2 - 1^2} \right] \\ = \frac{\sqrt{2}}{\sqrt{3}} \left[ \frac{(\sqrt{3} + 1) - (\sqrt{3} - 1)}{3 - 1} \right] \\ = \frac{\sqrt{2}}{\sqrt{3}} \left[ \frac{\sqrt{3} + 1 - \sqrt{3} + 1}{2} \right] \\ = \frac{\sqrt{2}}{\sqrt{3}} \left[ \frac{2}{2} \right] \\ = \frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{6}}{3} \end{array}\](b)
Rationalising each term separately:
\[\begin{array}{l} \text{1st term} = \frac{3\sqrt{7}(\sqrt{5}-\sqrt{2})}{(\sqrt{5})^2-(\sqrt{2})^2} = \frac{3\sqrt{7}(\sqrt{5}-\sqrt{2})}{5-2} = \frac{3(\sqrt{35}-\sqrt{14})}{3} = \sqrt{35} - \sqrt{14} \\ \text{2nd term} = \frac{5\sqrt{5}}{\sqrt{7}+\sqrt{2}} = \frac{5\sqrt{5}(\sqrt{7}-\sqrt{2})}{(\sqrt{7})^2-(\sqrt{2})^2} = \frac{5\sqrt{5}(\sqrt{7}-\sqrt{2})}{7-2} = \frac{5(\sqrt{35}-\sqrt{10})}{5} = \sqrt{35} - \sqrt{10} \\ \text{3rd term} = \frac{2\sqrt{2}}{\sqrt{7}+\sqrt{5}} = \frac{2\sqrt{2}(\sqrt{7}-\sqrt{5})}{(\sqrt{7})^2-(\sqrt{5})^2} = \frac{2\sqrt{2}(\sqrt{7}-\sqrt{5})}{7-5} = \frac{2(\sqrt{14}-\sqrt{10})}{2} = \sqrt{14} - \sqrt{10} \end{array}\]Now, substitute back:
\[\begin{array}{l} = (\sqrt{35} - \sqrt{14}) - (\sqrt{35} - \sqrt{10}) + (\sqrt{14} - \sqrt{10}) \\ = \sqrt{35} - \sqrt{14} - \sqrt{35} + \sqrt{10} + \sqrt{14} - \sqrt{10} \\ = 0 \end{array}\](c)
Rationalising each term separately:
\[\begin{array}{l} \text{1st term} = \frac{4\sqrt{3}(2+\sqrt{2})}{2^2-(\sqrt{2})^2} = \frac{8\sqrt{3}+4\sqrt{6}}{4-2} = \frac{2(4\sqrt{3}+2\sqrt{6})}{2} = 4\sqrt{3} + 2\sqrt{6} \\ \text{2nd term} = \frac{30(4\sqrt{3}+\sqrt{18})}{(4\sqrt{3})^2-(\sqrt{18})^2} = \frac{30(4\sqrt{3}+3\sqrt{2})}{48-18} = \frac{30(4\sqrt{3}+3\sqrt{2})}{30} = 4\sqrt{3} + 3\sqrt{2} \\ \text{3rd term} = \frac{\sqrt{18}(3+\sqrt{12})}{3^2-(\sqrt{12})^2} = \frac{3\sqrt{2}(3+2\sqrt{3})}{9-12} = \frac{9\sqrt{2}+6\sqrt{6}}{-3} = -3\sqrt{2} - 2\sqrt{6} \end{array}\]Now, substitute back:
\[\begin{array}{l} = (4\sqrt{3} + 2\sqrt{6}) - (4\sqrt{3} + 3\sqrt{2}) - (-3\sqrt{2} - 2\sqrt{6}) \\ = 4\sqrt{3} + 2\sqrt{6} - 4\sqrt{3} - 3\sqrt{2} + 3\sqrt{2} + 2\sqrt{6} \\ = 4\sqrt{6} \end{array}\](d)
Rationalising each term separately:
\[\begin{array}{l} \text{1st term} = \frac{3\sqrt{2}}{\sqrt{6}+\sqrt{3}} = \frac{3\sqrt{2}(\sqrt{6}-\sqrt{3})}{(\sqrt{6})^2-(\sqrt{3})^2} = \frac{3(\sqrt{12}-\sqrt{6})}{6-3} = \frac{3(2\sqrt{3}-\sqrt{6})}{3} = 2\sqrt{3} - \sqrt{6} \\ \text{2nd term} = \frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}} = \frac{4\sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6})^2-(\sqrt{2})^2} = \frac{4(\sqrt{18}-\sqrt{6})}{6-2} = \frac{4(3\sqrt{2}-\sqrt{6})}{4} = 3\sqrt{2} - \sqrt{6} \\ \text{3rd term} = \frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}} = \frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3})^2-(\sqrt{2})^2} = \frac{\sqrt{18}-\sqrt{12}}{3-2} = 3\sqrt{2} - 2\sqrt{3} \end{array}\]Now, substitute back:
\[\begin{array}{l} = (2\sqrt{3} - \sqrt{6}) - (3\sqrt{2} - \sqrt{6}) + (3\sqrt{2} - 2\sqrt{3}) \\ = 2\sqrt{3} - \sqrt{6} - 3\sqrt{2} + \sqrt{6} + 3\sqrt{2} - 2\sqrt{3} \\ = 0 \end{array}\]\( \frac{3\sqrt{x}}{\sqrt{y}+\sqrt{z}} - \frac{4\sqrt{y}}{\sqrt{z}+\sqrt{x}} + \frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}} \)
Substituting \( x = 2 \), \( y = 3 \), and \( z = 6 \) in the given expression:
\[\begin{array}{l} = \frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}} - \frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}} + \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} \end{array}\]Notice that this is the exact same expression as Question 2(d).
We can solve it by rationalising the denominator of each term:
\[\begin{array}{l} \text{1st term} = \frac{3\sqrt{2}(\sqrt{6}-\sqrt{3})}{(\sqrt{6})^2-(\sqrt{3})^2} = \frac{3(\sqrt{12}-\sqrt{6})}{3} = 2\sqrt{3} - \sqrt{6} \\ \text{2nd term} = \frac{4\sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6})^2-(\sqrt{2})^2} = \frac{4(\sqrt{18}-\sqrt{6})}{4} = 3\sqrt{2} - \sqrt{6} \\ \text{3rd term} = \frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3})^2-(\sqrt{2})^2} = \frac{\sqrt{18}-\sqrt{12}}{1} = 3\sqrt{2} - 2\sqrt{3} \end{array}\]Putting these values back in the expression:
\[\begin{array}{l} = (2\sqrt{3} - \sqrt{6}) - (3\sqrt{2} - \sqrt{6}) + (3\sqrt{2} - 2\sqrt{3}) \\ = 2\sqrt{3} - \sqrt{6} - 3\sqrt{2} + \sqrt{6} + 3\sqrt{2} - 2\sqrt{3} \\ = 0 \end{array}\]Answer: The value of the given expression is 0.
If the simplified value is 14, let us write by calculating the value of x.
Part 1: Simplify the expression
Let \( A = x + \sqrt{x^2-1} \) and \( B = x - \sqrt{x^2-1} \)
We need to find \( \frac{A}{B} + \frac{B}{A} = \frac{A^2 + B^2}{AB} \)
\[\begin{array}{l} AB = (x + \sqrt{x^2-1})(x - \sqrt{x^2-1}) = x^2 - (x^2-1) = x^2 - x^2 + 1 = 1 \\ A^2 + B^2 = (x + \sqrt{x^2-1})^2 + (x - \sqrt{x^2-1})^2 \\ = 2[x^2 + (\sqrt{x^2-1})^2] \quad [\text{Using } (a+b)^2 + (a-b)^2 = 2(a^2+b^2)] \\ = 2[x^2 + x^2 - 1] \\ = 2(2x^2 - 1) \\ = 4x^2 - 2 \end{array}\]So, the simplified expression is \( \frac{4x^2 - 2}{1} = \) \( 4x^2 - 2 \).
Part 2: Find the value of x
Given that the simplified value is 14:
\[\begin{array}{l} 4x^2 - 2 = 14 \\ \Rightarrow 4x^2 = 14 + 2 \\ \Rightarrow 4x^2 = 16 \\ \Rightarrow x^2 = \frac{16}{4} \\ \Rightarrow x^2 = 4 \\ \Rightarrow x = \pm 2 \end{array}\]Answer: The value of x is \( \pm 2 \).
(i) \( \frac{a^2+ab+b^2}{a^2-ab+b^2} \) (ii) \( \frac{(a-b)^3}{(a+b)^3} \) (iii) \( \frac{3a^2+5ab+3b^2}{3a^2-5ab+3b^2} \) (iv) \( \frac{a^3+b^3}{a^3-b^3} \)
First, let's find the values of \( a+b \), \( a-b \), and \( ab \).
\[\begin{array}{l} ab = \left(\frac{\sqrt{5}+1}{\sqrt{5}-1}\right) \times \left(\frac{\sqrt{5}-1}{\sqrt{5}+1}\right) = 1 \\ \\ a+b = \frac{\sqrt{5}+1}{\sqrt{5}-1} + \frac{\sqrt{5}-1}{\sqrt{5}+1} \\ = \frac{(\sqrt{5}+1)^2 + (\sqrt{5}-1)^2}{(\sqrt{5}-1)(\sqrt{5}+1)} \\ = \frac{2[(\sqrt{5})^2 + 1^2]}{(\sqrt{5})^2 - 1^2} = \frac{2(5+1)}{5-1} = \frac{2 \times 6}{4} = \frac{12}{4} = 3 \\ \\ a-b = \frac{\sqrt{5}+1}{\sqrt{5}-1} - \frac{\sqrt{5}-1}{\sqrt{5}+1} \\ = \frac{(\sqrt{5}+1)^2 - (\sqrt{5}-1)^2}{(\sqrt{5}-1)(\sqrt{5}+1)} \\ = \frac{4(\sqrt{5})(1)}{(\sqrt{5})^2 - 1^2} = \frac{4\sqrt{5}}{4} = \sqrt{5} \end{array}\](i) \( \frac{a^2+ab+b^2}{a^2-ab+b^2} \)
\[\begin{array}{l} = \frac{(a+b)^2 - 2ab + ab}{(a+b)^2 - 2ab - ab} \\ = \frac{(a+b)^2 - ab}{(a+b)^2 - 3ab} \\ = \frac{3^2 - 1}{3^2 - 3(1)} \\ = \frac{9 - 1}{9 - 3} = \frac{8}{6} = \frac{4}{3} \end{array}\](ii) \( \frac{(a-b)^3}{(a+b)^3} \)
\[\begin{array}{l} = \frac{(\sqrt{5})^3}{3^3} \\ = \frac{5\sqrt{5}}{27} \end{array}\](iii) \( \frac{3a^2+5ab+3b^2}{3a^2-5ab+3b^2} \)
\[\begin{array}{l} = \frac{3(a^2+b^2) + 5ab}{3(a^2+b^2) - 5ab} \\ = \frac{3[(a+b)^2 - 2ab] + 5ab}{3[(a+b)^2 - 2ab] - 5ab} \\ = \frac{3[3^2 - 2(1)] + 5(1)}{3[3^2 - 2(1)] - 5(1)} \\ = \frac{3(9 - 2) + 5}{3(9 - 2) - 5} \\ = \frac{3(7) + 5}{3(7) - 5} \\ = \frac{21 + 5}{21 - 5} = \frac{26}{16} = \frac{13}{8} \end{array}\](iv) \( \frac{a^3+b^3}{a^3-b^3} \)
\[\begin{array}{l} = \frac{(a+b)(a^2 - ab + b^2)}{(a-b)(a^2 + ab + b^2)} \\ = \frac{(a+b)[(a+b)^2 - 3ab]}{(a-b)[(a-b)^2 + 3ab]} \\ = \frac{3 \times [3^2 - 3(1)]}{\sqrt{5} \times [(\sqrt{5})^2 + 3(1)]} \\ = \frac{3 \times (9 - 3)}{\sqrt{5} \times (5 + 3)} \\ = \frac{3 \times 6}{\sqrt{5} \times 8} \\ = \frac{18}{8\sqrt{5}} = \frac{9}{4\sqrt{5}} = \frac{9\sqrt{5}}{20} \end{array}\](a) (i) \( x - \frac{1}{x} \) (ii) \( y^2 + \frac{1}{y^2} \) (iii) \( x^3 - \frac{1}{x^3} \) (iv) \( xy + \frac{1}{xy} \)
(b) \( 3x^2 - 5xy + 3y^2 \)
Given: \( x = 2+\sqrt{3} \), \( y = 2-\sqrt{3} \)
First, find \( \frac{1}{x} \), \( \frac{1}{y} \), \( xy \), \( x-y \) and \( x+y \).
\[\begin{array}{l} \frac{1}{x} = \frac{1}{2+\sqrt{3}} = \frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})} = \frac{2-\sqrt{3}}{4-3} = 2-\sqrt{3} \\ \frac{1}{y} = \frac{1}{2-\sqrt{3}} = \frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})} = \frac{2+\sqrt{3}}{4-3} = 2+\sqrt{3} \\ xy = (2+\sqrt{3})(2-\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \\ x+y = (2+\sqrt{3}) + (2-\sqrt{3}) = 4 \\ x-y = (2+\sqrt{3}) - (2-\sqrt{3}) = 2\sqrt{3} \end{array}\](a)
(i) \( x - \frac{1}{x} \)
\[\begin{array}{l} = (2+\sqrt{3}) - (2-\sqrt{3}) \\ = 2 + \sqrt{3} - 2 + \sqrt{3} \\ = 2\sqrt{3} \end{array}\](ii) \( y^2 + \frac{1}{y^2} \)
\[\begin{array}{l} = \left(y + \frac{1}{y}\right)^2 - 2 \cdot y \cdot \frac{1}{y} \\ = ((2-\sqrt{3}) + (2+\sqrt{3}))^2 - 2 \\ = (4)^2 - 2 \\ = 16 - 2 = 14 \end{array}\](iii) \( x^3 - \frac{1}{x^3} \)
\[\begin{array}{l} = \left(x - \frac{1}{x}\right)^3 + 3 \cdot x \cdot \frac{1}{x} \left(x - \frac{1}{x}\right) \\ = (2\sqrt{3})^3 + 3(2\sqrt{3}) \quad [\text{Using value from (i)}] \\ = 8(3\sqrt{3}) + 6\sqrt{3} \\ = 24\sqrt{3} + 6\sqrt{3} \\ = 30\sqrt{3} \end{array}\](iv) \( xy + \frac{1}{xy} \)
\[\begin{array}{l} = 1 + \frac{1}{1} \quad [\text{Since } xy = 1] \\ = 1 + 1 = 2 \end{array}\](b) \( 3x^2 - 5xy + 3y^2 \)
\[\begin{array}{l} = 3(x^2 + y^2) - 5xy \\ = 3[(x+y)^2 - 2xy] - 5xy \\ = 3[4^2 - 2(1)] - 5(1) \\ = 3[16 - 2] - 5 \\ = 3[14] - 5 \\ = 42 - 5 = 37 \end{array}\]Given: \( x = \frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} \) and \( xy=1 \).
Since \( xy = 1 \), \( y = \frac{1}{x} = \frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}} \).
Let's find \( x+y \):
\[\begin{array}{l} x+y = \frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} + \frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}} \\ = \frac{(\sqrt{7}+\sqrt{3})^2 + (\sqrt{7}-\sqrt{3})^2}{(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3})} \\ = \frac{2[(\sqrt{7})^2 + (\sqrt{3})^2]}{(\sqrt{7})^2 - (\sqrt{3})^2} \\ = \frac{2[7 + 3]}{7 - 3} \\ = \frac{2 \times 10}{4} = \frac{20}{4} = 5 \end{array}\]L.H.S:
\[\begin{array}{l} = \frac{x^2+xy+y^2}{x^2-xy+y^2} \\ = \frac{(x+y)^2 - 2xy + xy}{(x+y)^2 - 2xy - xy} \\ = \frac{(x+y)^2 - xy}{(x+y)^2 - 3xy} \\ = \frac{5^2 - 1}{5^2 - 3(1)} \\ = \frac{25 - 1}{25 - 3} \\ = \frac{24}{22} \\ = \frac{12}{11} \end{array}\]R.H.S: \( \frac{12}{11} \)
\( \therefore \) L.H.S. = R.H.S. [Proved]
Let's compare the squares of the two expressions.
\[\begin{array}{l} (\sqrt{7}+1)^2 = (\sqrt{7})^2 + 2(\sqrt{7})(1) + 1^2 \\ = 7 + 2\sqrt{7} + 1 \\ = 8 + 2\sqrt{7} \\ = 8 + \sqrt{4 \times 7} = 8 + \sqrt{28} \end{array}\]\[\begin{array}{l} (\sqrt{5}+\sqrt{3})^2 = (\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{3}) + (\sqrt{3})^2 \\ = 5 + 2\sqrt{15} + 3 \\ = 8 + 2\sqrt{15} \\ = 8 + \sqrt{4 \times 15} = 8 + \sqrt{60} \end{array}\]
Comparing the squared values: \( 8 + \sqrt{28} \) and \( 8 + \sqrt{60} \).
Since \( 60 > 28 \), it follows that \( \sqrt{60} > \sqrt{28} \).
Therefore, \( 8 + \sqrt{60} > 8 + \sqrt{28} \).
Hence, \( (\sqrt{5}+\sqrt{3})^2 > (\sqrt{7}+1)^2 \).
Taking the positive square root of both sides (since both original expressions are positive):
\( \sqrt{5}+\sqrt{3} > \sqrt{7}+1 \)
Answer: \( \sqrt{5}+\sqrt{3} \) is greater.
📘 10. Very short answer type question (V.S.A.)
(A) M.C.Q.Explanation:
Given \( x = 2+\sqrt{3} \)
\[\begin{array}{l} \frac{1}{x} = \frac{1}{2+\sqrt{3}} \\ = \frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})} \\ = \frac{2-\sqrt{3}}{2^2 - (\sqrt{3})^2} \\ = \frac{2-\sqrt{3}}{4 - 3} \\ = 2-\sqrt{3} \end{array}\]Now, substitute the values:
\[\begin{array}{l} x + \frac{1}{x} = (2+\sqrt{3}) + (2-\sqrt{3}) \\ = 4 \end{array}\]Correct Option: (c) 4
Explanation:
We know the algebraic formula:
\[\begin{array}{l} 4pq = (p+q)^2 - (p-q)^2 \end{array}\]Substituting the given values:
\[\begin{array}{l} 4pq = (\sqrt{13})^2 - (\sqrt{5})^2 \\ 4pq = 13 - 5 \\ 4pq = 8 \\ pq = \frac{8}{4} \\ pq = 2 \end{array}\]Correct Option: (a) 2
Explanation:
We know the algebraic formula:
\[\begin{array}{l} 2(a^2+b^2) = (a+b)^2 + (a-b)^2 \end{array}\]Substituting the given values:
\[\begin{array}{l} 2(a^2+b^2) = (\sqrt{5})^2 + (\sqrt{3})^2 \\ 2(a^2+b^2) = 5 + 3 \\ 2(a^2+b^2) = 8 \\ (a^2+b^2) = \frac{8}{2} \\ (a^2+b^2) = 4 \end{array}\]Correct Option: (b) 4
Explanation:
Given expression:
\[\begin{array}{l} \sqrt{125} - \sqrt{5} \\ = \sqrt{25 \times 5} - \sqrt{5} \\ = 5\sqrt{5} - 1\sqrt{5} \\ = 4\sqrt{5} \end{array}\]Now, expressing \( 4\sqrt{5} \) purely inside the square root to match the options:
\[\begin{array}{l} 4\sqrt{5} = \sqrt{4^2 \times 5} \\ = \sqrt{16 \times 5} \\ = \sqrt{80} \end{array}\]Correct Option: (a) \( \sqrt{80} \)
Explanation:
Given expression:
\[\begin{array}{l} (5-\sqrt{3})(\sqrt{3}-1)(5+\sqrt{3})(\sqrt{3}+1) \end{array}\]Rearranging the terms to group the conjugate pairs:
\[\begin{array}{l} = [(5-\sqrt{3})(5+\sqrt{3})] \times [(\sqrt{3}-1)(\sqrt{3}+1)] \end{array}\]Applying the formula \( (a-b)(a+b) = a^2 - b^2 \):
\[\begin{array}{l} = [5^2 - (\sqrt{3})^2] \times [(\sqrt{3})^2 - 1^2] \\ = [25 - 3] \times [3 - 1] \\ = 22 \times 2 \\ = 44 \end{array}\]Correct Option: (b) 44
📘 10. Very short answer type question (V.S.A.)
(B) True or False\( \sqrt{75} \) and \( \sqrt{147} \) are similar surds.
Explanation:
Let's simplify both surds:
\[\begin{array}{l} \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \\ \sqrt{147} = \sqrt{49 \times 3} = 7\sqrt{3} \end{array}\]Since both have the same irrational factor \( \sqrt{3} \), they are similar surds.
Answer: True
\( \sqrt{\pi} \) is a quadratic surd.
Explanation:
A quadratic surd is the square root of a positive rational number which cannot be exactly determined (i.e., it is not a perfect square). Since \( \pi \) is an irrational number, \( \sqrt{\pi} \) does not satisfy the definition of a quadratic surd.
Answer: False
📘 10. Very short answer type question (V.S.A.)
(C) Fill up the blankExplanation:
The product of a non-zero rational number (like 5) and an irrational number (like \( \sqrt{11} \)) is always an irrational number.
Answer: irrational
Explanation:
The conjugate surd is found by changing the sign of the irrational (surd) part of the mixed quadratic surd. The given surd is \( -5 + \sqrt{3} \). Changing the sign of the surd part gives \( -5 - \sqrt{3} \), which can be written as \( -\sqrt{3} - 5 \).
Answer: \( -\sqrt{3}-5 \)
Explanation:
By definition, if both the sum and the product of two mixed quadratic surds are rational numbers, then they are called conjugate surds of each other (e.g., \( a+\sqrt{b} \) and \( a-\sqrt{b} \)).
Answer: conjugate
📘 11. Short answer type (S.A.)
Quadratic SurdGiven: \( x = 3 + 2\sqrt{2} \)
First, we find the value of \( \frac{1}{x} \) by rationalising the denominator:
\[\begin{array}{l} \frac{1}{x} = \frac{1}{3 + 2\sqrt{2}} \\ = \frac{1 \times (3 - 2\sqrt{2})}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} \\ = \frac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} \\ = \frac{3 - 2\sqrt{2}}{9 - (4 \times 2)} \\ = \frac{3 - 2\sqrt{2}}{9 - 8} \\ = 3 - 2\sqrt{2} \end{array}\]Now, substitute the values into \( x + \frac{1}{x} \):
\[\begin{array}{l} x + \frac{1}{x} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) \\ = 3 + 3 \\ = 6 \end{array}\]Answer: 6
Let's compare the squares of the two expressions to determine which is greater.
\[\begin{array}{l} (\sqrt{15} + \sqrt{3})^2 = (\sqrt{15})^2 + 2(\sqrt{15})(\sqrt{3}) + (\sqrt{3})^2 \\ = 15 + 2\sqrt{45} + 3 \\ = 18 + 2\sqrt{45} \end{array}\]\[\begin{array}{l} (\sqrt{10} + \sqrt{8})^2 = (\sqrt{10})^2 + 2(\sqrt{10})(\sqrt{8}) + (\sqrt{8})^2 \\ = 10 + 2\sqrt{80} + 8 \\ = 18 + 2\sqrt{80} \end{array}\]
Now, comparing \( 18 + 2\sqrt{45} \) and \( 18 + 2\sqrt{80} \):
Since \( 80 > 45 \), it follows that \( \sqrt{80} > \sqrt{45} \).
Therefore, \( 18 + 2\sqrt{80} > 18 + 2\sqrt{45} \).
This implies \( (\sqrt{10} + \sqrt{8})^2 > (\sqrt{15} + \sqrt{3})^2 \).
Taking the positive square root of both sides gives \( (\sqrt{10} + \sqrt{8}) > (\sqrt{15} + \sqrt{3}) \).
Answer: \( (\sqrt{10} + \sqrt{8}) \) is greater.
Explanation:
Two mixed quadratic surds whose product is a rational number are generally conjugate surds of each other.
For example, let us take \( (2 + \sqrt{3}) \) and \( (2 - \sqrt{3}) \).
Let's check their product:
\[\begin{array}{l} (2 + \sqrt{3})(2 - \sqrt{3}) \\ = 2^2 - (\sqrt{3})^2 \\ = 4 - 3 \\ = 1 \end{array}\]Since 1 is a rational number, the selected surds satisfy the condition.
Answer: \( (2 + \sqrt{3}) \) and \( (2 - \sqrt{3}) \) (Other conjugate pairs are also correct valid answers).
Let the number to be subtracted be \( y \).
\[\begin{array}{l} \sqrt{72} - y = \sqrt{32} \\ \Rightarrow y = \sqrt{72} - \sqrt{32} \end{array}\]Now, simplify the surds:
\[\begin{array}{l} \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2} \\ \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \end{array}\]Substitute these values back:
\[\begin{array}{l} y = 6\sqrt{2} - 4\sqrt{2} \\ y = 2\sqrt{2} \end{array}\]We can also write \( 2\sqrt{2} \) as \( \sqrt{4 \times 2} = \sqrt{8} \).
Answer: \( 2\sqrt{2} \) (or \( \sqrt{8} \))
We will rationalise the denominator of each term separately.
1st term:
\[\begin{array}{l} \frac{1}{\sqrt{2} + 1} = \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)} \\ = \frac{\sqrt{2} - 1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1 \end{array}\]2nd term:
\[\begin{array}{l} \frac{1}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} \\ = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2} \end{array}\]3rd term:
\[\begin{array}{l} \frac{1}{\sqrt{4} + \sqrt{3}} = \frac{\sqrt{4} - \sqrt{3}}{(\sqrt{4} + \sqrt{3})(\sqrt{4} - \sqrt{3})} \\ = \frac{\sqrt{4} - \sqrt{3}}{(\sqrt{4})^2 - (\sqrt{3})^2} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3} \end{array}\]Now, substituting the rationalised terms back into the original expression:
\[\begin{array}{l} = (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3}) \\ = \sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + 2 - \sqrt{3} \\ = -1 + 2 \\ = 1 \end{array}\]Answer: 1
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