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  • (i)Find the values of a and b in each of the following : 5+2sqroot 3 by 7+4sqroot 3 = a-6 sqroot 3

(i) Find the values of a in each of the following : \frac{5+2\sqrt{3}}{7+4\sqrt{3}}= a-6\sqrt{3}
(ii) Find the values of a in the following : \frac{3-\sqrt{5}}{3+2\sqrt{5}}= a\sqrt{5}-\frac{19}{11}

(iii) Find the values of b in the following : \frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6}
(iv) Find the values of a and b in the following : \frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}= a+\frac{7}{11}\sqrt{5b}

Answers (1)

(i) Answer.   a = 11
Solution.   We have, \frac{5+2\sqrt{3}}{7+4\sqrt{3}}= a-6\sqrt{3}
LHS = \frac{5+2\sqrt{3}}{7+4\sqrt{3}}
Rationalising the denominator, we get:
= \frac{5+2\sqrt{3}}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}}
= \frac{\left ( 5+2\sqrt{3} \right )\left ( 7-4\sqrt{3} \right )}{\left ( 7+4\sqrt{3} \right )\left ( 7-4\sqrt{3} \right )}
{Using (a – b) (a + b) = a2 – b2}

= \frac{35+14\sqrt{3}-20\sqrt{3}-24}{7^{2}-\left ( 4\sqrt{3} \right )^{2}}
= \frac{11-6\sqrt{3}}{49-48}
= 11-6\sqrt{3}
Now RHS = a-6\sqrt{3}
\Rightarrow 11-6\sqrt{3}= a-6\sqrt{3}
\Rightarrow 11-6\sqrt{3}= a-6\sqrt{3}
\Rightarrow a= 11
Hence a = 11 is the required answer

(ii)Answer.   a= \frac{9}{11}
Solution.  Given that, \frac{3-\sqrt{5}}{3+2\sqrt{5}}= a\sqrt{5}-\frac{19}{11}

LHS = \frac{3-\sqrt{5}}{3+2\sqrt{5}}
Rationalising the denominator, we get:
LHS = \frac{3-\sqrt{5}}{3+2\sqrt{5}}\times \frac{3-2\sqrt{5}}{3-2\sqrt{5}}
{Using (a – b) (a + b) = a2 – b2}
= \frac{9-3\sqrt{5}-6\sqrt{5}+10}{3^{2}-\left ( 2\sqrt{5} \right )^{2}}
= \frac{19-9\sqrt{5}}{9-20}
Now RHS = a\sqrt{5}-\frac{19}{11}
\Rightarrow \frac{9\sqrt{5}}{11}-\frac{19}{11}= a\sqrt{5}-\frac{19}{11}
Comparing both , we get
\Rightarrow a= \frac{9}{11}
Hence a= \frac{9}{11}is the correct answer

(iii) Answer:b = -\frac{5 }{6}

Solution:

Given:

\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6}

LHS = \frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}

Rationalize

= \frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}} \times \frac{3 \sqrt{2}+2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}

=\frac{\sqrt{2}(3 \sqrt{2}+2 \sqrt{3})+\sqrt{3}(3 \sqrt{2}+2 \sqrt{3})}{(3 \sqrt{2})^{2}-(2 \sqrt{3})^{2}}

= \frac{6+2 \sqrt{6}+3 \sqrt{6}+6}{18-12}

= 2+\frac{5 \sqrt{6}}{6}

2+\frac{5 \sqrt{6}}{6}=2-b \sqrt{6}

b = -\frac{5 }{6}

 

(iv) Answer. a = 0, b = 1
Solution.         Given,
\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}= a+\frac{7}{11}\sqrt{5b}
LHS = \frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}
=\frac{\left ( 7+\sqrt{5} \right )\times\left ( 7+\sqrt{5} \right )-\left ( 7-\sqrt{5} \right ) \times \left ( 7-\sqrt{5} \right )}{\left (7 -\sqrt{5} \right )\left (7 +\sqrt{5} \right )}
Using (a – b) (a + b) = a2 – b2
(a + b)2 = a2 + b2 + 2ab
(a - b)2 = a2 + b2 - 2ab
= \frac{\left ( 7^{2} +\sqrt{5}^{2}+2\cdot 7\cdot \sqrt{5}\right )-\left ( 7^{2} +\sqrt{5}^{2}-2\cdot 7\cdot \sqrt{5} \right )}{7^{2}-\sqrt{5}^{2}}
= \frac{\left ( 49+5+14\sqrt{5} \right )-\left ( 49+5-14\sqrt{5} \right )}{49-5}
= \frac{54+14\sqrt{5}-54+14\sqrt{5}}{44}
= \frac{28\sqrt{5}}{44}
RHS = a+\frac{7}{11}\sqrt{5b}
Now LHS = RHS
\Rightarrow \frac{28\sqrt{5}}{44}= a+\frac{7}{11}\sqrt{5b}
\Rightarrow 0+\left ( \frac{4}{4} \right )\frac{7}{11}\sqrt{5}= a+\frac{7}{11}\sqrt{5b}

\Rightarrow a = 0, b = 1
Hence the answer is a = 0, b = 1

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