Question

if a and b are two rational numbers then find the value of a and b in following equation.
(í) 6+√3/6-√3 = a + b√3​

Answer 1

Question

If a and b are two rational numbers then find the value of a and b in following equation.

(í) (6+√3 ) / (6-√3) = a + b√3

Answer

$\sf \: \pink a = \red{ \dfrac{13}{11} } \quad \: \red b = \pink{ \dfrac{4}{11}}$

Solution

$\mathcal{ \dfrac{ (6+√3 )}{ (6-√3) }= a + b√3}$

Now , on LHS we have a complex fraction with its denominator containing a surd form , we know that a surd cannot be in denominator if so then we have to rationalise the rational number by multiplying and dividing the number with the conjugates of denominator.

$\dfrac{ (6+√3 )}{ (6-√3) } \times \dfrac{ (6 +√3 )}{ (6 + √3) } \\ = \dfrac{ (6+√3 ) {}^{2} }{(6 + \sqrt{3}) (6-√3) } \quad = \frac{36 + 3 + 12 \sqrt{3} }{36 - 3} \\ = \frac{39 + 12 \sqrt{3} }{33} = \quad \frac{39}{33} + \frac{12 \sqrt{3} }{33} \\ = \red{ \frac{13}{11} + \frac{4 \sqrt{3} }{11} }$

On solving LHS we get the highlighted value

(13 / 11 )+ (4√3 / 11) = a + b√ 3

On comparing both side ;

↦ a = 13 / 11

4√3 / 11 = b√ 3

(cancel √3 on both sides )

↦ b = 4 / 11

$\rule{180pts}{2pts}$

Thankyou