Question

Question 11
Find the values of a and b in each of the following
$\left(i\right)\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=a-6\sqrt{3}$
$\left(ii\right)\frac{3-\sqrt{5}}{3+2\sqrt{5}}=a\sqrt{5}-\frac{19}{11}$
$\left(iii\right)\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=2-b\sqrt{6}$
$\left(iv\right)\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11}\sqrt{5}b$

Answer 1

(i) We have $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=a-6\sqrt{3}$
For rationalizing the above equation, we multiply numerator and denominator of LHS by 7 - 4 $\sqrt{3}$, we get,
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}}=a-6\sqrt{3}\frac{5(7-4\sqrt{3})+2\sqrt{3}(7-4\sqrt{3})}{7^2-(4\sqrt{3}{)}^2}=a-6\sqrt{3}$
[Using identity,$(a+b)(a-b)=a^2-b^2]$
$\Rightarrow \frac{35-20\sqrt{3}+14\sqrt{3}-24}{49-48}=a-6\sqrt{3}\Rightarrow 11-6\sqrt{3}=a-6\sqrt{3}\Rightarrow a=11$

(ii) We have, $\frac{3-\sqrt{5}}{3+2\sqrt{5}}=a\sqrt{5}-\frac{19}{11}$
For rationalizing the above equation, we multiply numerator and denominator of LHS by $3-2\sqrt{5}$, we get
$\Rightarrow \frac{(3-\sqrt{5})}{3+2\sqrt{5}}\times \frac{3-2\sqrt{5}}{3-2\sqrt{5}}=a\sqrt{5}-\frac{19}{11}\Rightarrow \frac{3(3-2\sqrt{5})-\sqrt{5}(3-2\sqrt{5})}{(3{)}^2-2\sqrt{5}{)}^2}=a\sqrt{5}-\frac{19}{11}$
[Using identity, $(a-b)(a+b)=a^2-b^2]$
$\Rightarrow \frac{9-6\sqrt{5}-3\sqrt{5}+10}{9-4\times 5}=a\sqrt{5}-\frac{19}{11}\Rightarrow \frac{19-9\sqrt{5}}{9-20}=a\sqrt{5}-\frac{19}{11}\Rightarrow \frac{19-9\sqrt{5}}{-11}=a\sqrt{5}-\frac{19}{11}\Rightarrow \frac{9\sqrt{5}}{11}-\frac{19}{11}=a\sqrt{5}-\frac{19}{11}\Rightarrow \frac{9\sqrt{5}}{11}=a\sqrt{5}\therefore a=\frac{9}{11}$

(iii) We have, $\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=2-b\sqrt{6}$
For rationalising the above equation, we multiply numerator and denominator of LHS by $3\sqrt{2}+2\sqrt{3}$, we get
$\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}\times \frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}=2-b\sqrt{6}\Rightarrow \frac{\sqrt{2}(3\sqrt{2}+2\sqrt{3})+\sqrt{3}(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2}{)}^2-(2\sqrt{3}{)}^2}=2-b\sqrt{6}$
[Using identity,$(a-b)(a+b)=a^2-b^2]$
$\Rightarrow \frac{6+2\sqrt{6}+3\sqrt{6}+6}{18-12}=2-b\sqrt{6}\Rightarrow \frac{12+5\sqrt{6}}{6}=2-b\sqrt{6}\Rightarrow 2+\frac{5\sqrt{6}}{6}=2-b\sqrt{6}\Rightarrow -b\sqrt{6}=\frac{5\sqrt{6}}{6}\therefore b=-\frac{5}{6}$

(iv) We have, $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11}\sqrt{5}b$
$\Rightarrow \frac{(7+\sqrt{5}{)}^2-(7-\sqrt{5}{)}^2}{(7-\sqrt{5})(7+\sqrt{5})}=a+\frac{7}{11}\sqrt{5}b\Rightarrow \frac{[7^2+(\sqrt{5}{)}^2+2\times 7\times \sqrt{5}]-(7^2+(\sqrt{5}{)}^2-2\times 7\times \sqrt{5}]}{7^2-(\sqrt{5}{)}^2}=a+\frac{7}{11}\sqrt{5}b$
$\left\{\begin{array}{cc}Usingidentity,& (a+b{)}^2=a^2+2ab+b^2\\ & (a-b{)}^2=a^2-2ab+b^2\\ and& (a-b)(a+b)=a^2-b^2\end{array}\right\}$
$\Rightarrow \frac{49+5+14\sqrt{5}-49-5+14\sqrt{5}}{49-5}=a+\frac{7}{11}\sqrt{5}b\Rightarrow \frac{28\sqrt{5}}{44}=a+\frac{7}{11}\sqrt{5}b\Rightarrow 0+\frac{7}{11}\sqrt{5}=a+\frac{7}{11}\sqrt{5}b$
On comparing both sides, we get
a = 0 and b = 1