In each of the following determine rational numbers a and b:
Question:

In each of the following determine rational numbers a and b:

(i) $\frac{\sqrt{3}-1}{\sqrt{3}+1}=a-b \sqrt{3}$

(ii) $\frac{4+\sqrt{2}}{2+\sqrt{2}}=a-\sqrt{b}$

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

(iv) $\frac{5+3 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3}$

(v) $\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=a-b \sqrt{77}$

(vi) $\frac{4+3 \sqrt{5}}{4-3 \sqrt{5}}=a+b \sqrt{5}$

 

Solution:

Given,

(i) $\frac{\sqrt{3}-1}{\sqrt{3}+1}=a-b \sqrt{3}$

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$\sqrt{3}-1$

$=\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}$

As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\frac{3-2 \sqrt{3}+1}{3-1}=\frac{4-2 \sqrt{3}}{2}=2-\sqrt{3}$

$2-\sqrt{3}=a-b \sqrt{3}$

On comparing the rational and irrational parts of the above equation, we get, a = 2 and b = 1

(ii) $\frac{4+\sqrt{2}}{2+\sqrt{2}}=\mathrm{a}-\sqrt{\mathrm{b}}$

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$2-\sqrt{2}$

$=\frac{(4+\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}$

As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\frac{(8-4 \sqrt{2}+2 \sqrt{2}-2)}{4-2}$

$=\frac{(6-2 \sqrt{2})}{2}$

$=3-\sqrt{2}$

$3-\sqrt{2}=a-\sqrt{b}$

On comparing the rational and irrational parts of the above equation, we get, $a=3 a n d b=2$

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

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$3+\sqrt{2}$

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

As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\frac{(9+6 \sqrt{2}+2)}{9-2}$

$=\frac{(11+6 \sqrt{2})}{7}$

$=\frac{11}{7}+\frac{6 \sqrt{2}}{7}$

$\frac{11}{7}+\frac{6 \sqrt{2}}{7}=a+b \sqrt{2}$

On comparing the rational and irrational parts of the above equation, we get,

$7-4 \sqrt{3}$

$=\frac{(5+3 \sqrt{3})(7-4 \sqrt{3})}{(7+4 \sqrt{3})(7-4 \sqrt{3})}$

As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\frac{(35-20 \sqrt{3}+21 \sqrt{3}-36)}{49-48}=-1+\sqrt{3}$

$-1+\sqrt{3}=a+b \sqrt{3}$

On comparing the rational and irrational parts of the above equation, we get, $a=-1$ and $b=1$

(v) $\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=a-b \sqrt{77}$

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$\sqrt{11}-\sqrt{7}$

$=\frac{(\sqrt{11}-\sqrt{7})(\sqrt{11}-\sqrt{7})}{(\sqrt{11}+\sqrt{7})(\sqrt{11}-\sqrt{7})}$

As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\frac{(11-\sqrt{77}-\sqrt{77}+7)}{11-7}$

$=\frac{(18-2 \sqrt{77})}{4}$

$=\frac{9}{2}-\frac{\sqrt{77}}{2}$

$\frac{9}{2}-\frac{\sqrt{77}}{2}=a-b \sqrt{77}$

On comparing the rational and irrational parts of the above equation, we get, a = 92 and b = 12

(vi) $\frac{4+3 \sqrt{5}}{4-3 \sqrt{5}}=\mathrm{a}+\mathrm{b} \sqrt{5}$

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$4+3 \sqrt{5}$

$=\frac{(4+3 \sqrt{5})(4+3 \sqrt{5})}{(4-3 \sqrt{5})(4+3 \sqrt{5})}$

As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\frac{(16+24 \sqrt{5}+45)}{-29}$

$=\frac{(61+24 \sqrt{5})}{-29}$

$=\frac{-61}{29}-\frac{(24 \sqrt{5})}{29}$

$\frac{-61}{29}-\frac{(24 \sqrt{5})}{29}=a+b \sqrt{5}$

On comparing the rational and irrational parts of the above equation, we get,

$a=\frac{-61}{29}$, and $b=\frac{-24}{29}$

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