# Find the values of a and b in each of the following

Question:

Find the values of $a$ and $b$ in each of the following

(i) $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a-6 \sqrt{3}$

(ii) $\frac{3-\sqrt{5}}{3+2 \sqrt{5}}=a \sqrt{5}-\frac{19}{11}$

(iii) $\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6}$

(iv) $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11} \sqrt{5} b$

Solution:

(i) We have, $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a-6 \sqrt{3}$

For rationalising 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, $(a+b)(a-b)=a^{2}-b^{2}$ ]

$\Rightarrow$        $\frac{35-20 \sqrt{3}+14 \sqrt{3}-24}{49-48}=a-6 \sqrt{3}$

$\Rightarrow$        $11-6 \sqrt{3}=a-6 \sqrt{3}=a=11$

(ii) We have, $\frac{3-\sqrt{5}}{3+2 \sqrt{5}}=a \sqrt{5}-\frac{19}{11}$

For rationalising 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, $(a-b)(a+b)=a^{2}-b^{2}$ ]

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

$\therefore$                   $b=-\frac{5}{6}$

(iv) We have, $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11} \sqrt{5} b$

$\Rightarrow$            $\frac{(7+\sqrt{5})^{2}-(7-\sqrt{5})^{2}}{(7-\sqrt{5})(7+\sqrt{5})}=a+\frac{7}{11} \sqrt{5} b$

$\Rightarrow \frac{\left.\left[7^{2}+(\sqrt{5})^{2}+2 \times 7 \times \sqrt{5}\right]-7^{2}+(\sqrt{5})^{2}-2 \times 7 \times \sqrt{5}\right]}{7^{2}-(\sqrt{5})^{2}}=a+\frac{7}{11} \sqrt{5} b$

\left\{\begin{aligned} \text { using identity, }(a+b)^{2} &=a^{2}+2 a b+b^{2}, \\(a-b)^{2} &=a^{2}-2 a b+b^{2} \\ \text { and } \quad(a-b)(a+b) &=a^{2}-b^{2} \end{aligned}\right\}

$\Rightarrow \quad \frac{49+5+14 \sqrt{5}-49-5+14 \sqrt{5}}{49-5}=a+\frac{7}{11} \sqrt{5} b$

$\Rightarrow$           $\frac{28 \sqrt{5}}{44}=a+\frac{7}{11} \sqrt{5} b \Rightarrow \frac{7}{11} \sqrt{5}=a+\frac{7}{11} \sqrt{5} b$

$\Rightarrow$          $0+\frac{7}{11} \sqrt{5}=a+\frac{7}{11} \sqrt{5} b$

On comparing both sides, we get

$a=0$ and $b=1$