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Mathematics

Given a and b are rational numbers. Find a and b if :

(i)3−53+25=−1911+a5(ii)2+332−23=a−b6(iii)7+57−5−7−57+5=a+711b5\begin{matrix} \text{(i)} & \dfrac{3 - \sqrt{5}}{3 + 2\sqrt{5}} = -\dfrac{19}{11} + a\sqrt{5} \\[1.5em] \text{(ii)} & \dfrac{\sqrt{2} + \sqrt{3}}{3\sqrt{2}-2\sqrt{3}} = a - b\sqrt{6} \\[1.5em] \text{(iii)} & \dfrac{7 + \sqrt{5}}{7 - \sqrt{5}} - \dfrac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + \dfrac{7}{11}b\sqrt{5} \\[1.5em] \end{matrix}

Rational Irrational Nos

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Answer

(i) Since, it is given that

3−53+25\dfrac{3 - \sqrt{5}}{3 + 2\sqrt{5}} = −1911+a5-\dfrac{19}{11} + a\sqrt{5}

On solving,

(3−5)(3+25)×(3−25)(3−25)⇒9−65−35+1032−(25)2⇒19−959−20⇒−95+19−11⇒−1911+9511∴−1911+9511=−1911+a5\dfrac{(3 - \sqrt{5})}{(3 + 2\sqrt{5})} × \dfrac{(3-2\sqrt{5})}{(3-2\sqrt{5})} \\[1.5em] \Rightarrow \dfrac{9 - 6\sqrt{5} - 3\sqrt{5} + 10 }{3^2 - (2\sqrt{5})^2} \\[1.5em] \Rightarrow \dfrac{19 - 9{\sqrt{5}}}{9 - 20} \\[1.5em] \Rightarrow \dfrac{-9{\sqrt{5}} + 19}{-11} \\[1.5em] \Rightarrow -\dfrac{19}{11} + \dfrac{9\sqrt{5}}{11} \\[1.5em] \therefore -\dfrac{19}{11} + \dfrac{9\sqrt5}{11} = -\dfrac{19}{11} + a\sqrt{5}

Hence, a = 911\dfrac{9}{11}.

(ii) Since, it is given that

2+332−23=a−b6\dfrac{\sqrt{2} + \sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} = a - b\sqrt6

On solving,

2+332−23×(32+23)(32+23)⇒6+26+36+6(32)2−(23)2⇒12+5618−12⇒12+566⇒126+566⇒2+566∴2−(−56)6=a−b6\dfrac{\sqrt{2} + \sqrt3}{3\sqrt{2} - 2\sqrt{3}} × \dfrac{(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2} + 2\sqrt{3})} \\[1.5em] \Rightarrow \dfrac{6 + 2\sqrt{6} + 3\sqrt6 + 6 }{(3\sqrt{2})^2 -(2\sqrt{3})^2} \\[1.5em] \Rightarrow \dfrac{12 + 5{\sqrt6}}{18-12} \\[1.5em] \Rightarrow \dfrac{12 + 5{\sqrt{6}}}{6} \\[1.5em] \Rightarrow \dfrac{12}{6} + \dfrac{5\sqrt{6}}{6} \\[1.5em] \Rightarrow 2 + \dfrac{5\sqrt{6}}{6} \\[1.5em] \therefore 2 - \dfrac{(-5\sqrt{6})}{6} = a - b\sqrt{6}

Hence, a = 2 and b = −56\dfrac{-5}{6}.

(iii) Since, it is given that

(7+5)(7−5)−(7−5)(7+5)=a+711b5\dfrac{(7 + \sqrt5)}{(7 - \sqrt{5})} - \dfrac{(7 - \sqrt5)}{(7 + \sqrt{5})} = a+\dfrac{7}{11}b\sqrt{5}

On solving,

(7+5)(7+5)−(7−5)(7−5)(7−5)(7+5)(7+5)2−(7−5)2(7−5)(7+5)⇒(72+2×7×5+(5)2)−(72−2×7×5+(5)2)72−(5)2⇒(49+2×75+5)−(49−2×75+5)72−(5)2⇒49+145+5−49+145−572−(5)2⇒28549−5⇒28544⇒7511∴a+711b5=7511⇒a+711b5=0+711×1×5\dfrac{(7 + \sqrt{5})(7 + \sqrt{5}) - (7-\sqrt{5})(7 - \sqrt{5})}{(7 - \sqrt{5})(7 + \sqrt{5})} \\[1.5em] \dfrac{(7 + \sqrt{5})^2 - (7 - \sqrt{5})^2}{(7 - \sqrt{5})(7+\sqrt{5})} \\[1.5em] \Rightarrow \dfrac{(7^2 + 2 × 7 × \sqrt5 + (\sqrt{5})^2)-(7^2 - 2 × 7 × \sqrt5 + (\sqrt{5})^2)}{7^2-(\sqrt5)^2} \\[1.5em] \Rightarrow \dfrac{(49 + 2 × 7\sqrt{5} + 5)-(49 - 2 × 7\sqrt{5}+ 5)}{7^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow \dfrac{49+14\sqrt{5} + 5 - 49 + 14\sqrt{5} - 5}{7^2 - (\sqrt5)^2} \\[1.5em] \Rightarrow \dfrac{28{\sqrt{5}}}{49 - 5} \\[1.5em] \Rightarrow \dfrac{28{\sqrt{5}}}{44} \\[1.5em] \Rightarrow \dfrac{7{\sqrt{5}}}{11} \\[1.5em] \therefore a+\dfrac{7}{11}b\sqrt{5} = \dfrac{7{\sqrt5}}{11} \\[1.5em] \Rightarrow a + \dfrac{7}{11}b\sqrt{5} = 0 + \dfrac{7}{11} × 1 × \sqrt{5}

Hence,value of a = 0 and b = 1.

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