Question

if both a and b are rational numbers,find a and b from the following 3+2 root 3 upon 3-2 root 3 =a+b root 3​

Answer 1

Answer :-

$\boxed{ \sf a = - \dfrac{21}{3} , \: b= - \dfrac{12}{3} }$

Explanation :-

Given :-

$\tt \dfrac{3 + 2 \sqrt{3} }{3 - 2 \sqrt{3} } = a + b \sqrt{3}$

And a, b are rational numbers

To find :-

Values of a and b

Solution :-

$\tt \dfrac{3 + 2 \sqrt{3} }{3 - 2 \sqrt{3} } = a + b \sqrt{3}$

Consider RHS

$\tt \dfrac{3 + 2 \sqrt{3} }{3 - 2 \sqrt{3} }$

First rationalise the denominator

The rationalising factor of 3 - 2√3 is 3 + 2√3. So multiply both numerator and denominator with rationalising factor

$\tt = \dfrac{3 + 2 \sqrt{3} }{3 - 2 \sqrt{3} } \times \dfrac{3 + 2 \sqrt{3} }{3 + 2 \sqrt{3} }$

$\tt = \dfrac{(3 + 2 \sqrt{3})^{2} }{(3)^{2} - (2 \sqrt{3})^{2} }$

[ Since (x + y)(x - y) = x² - y² and above x = 3, y = 2√3]

$\tt = \dfrac{(3 + 2 \sqrt{3})^{2} }{9 - {2}^{2}(( \sqrt{3})^{2})}$

$\tt = \dfrac{(3 + 2 \sqrt{3})^{2} }{9 - 4(3)}$

$\tt = \dfrac{(3 + 2 \sqrt{3})^{2} }{9 - 12}$

$\tt = \dfrac{(3 + 2 \sqrt{3})^{2} }{ - 3}$

$\tt = \dfrac{(3)^{2} + 2(3)(2 \sqrt{3}) + {(2 \sqrt{3})}^{2} }{ - 3}$

[Since (x + y)² = x² + 2xy + y² and above x = 3, b = 2√3 ]

$\tt = \dfrac{9 + 12\sqrt{3} + {2}^{2}(( \sqrt{3}) ^{2})}{ - 3}$

$\tt = \dfrac{9 + 12\sqrt{3} + 4(3)}{ - 3}$

$\tt = \dfrac{9 + 12\sqrt{3} + 12}{ - 3}$

$\tt = \dfrac{21 + 12\sqrt{3}}{ - 3}$

$\tt = \dfrac{21}{ - 3} + \dfrac{12 \sqrt{3} }{ - 3}$

$\tt = - \dfrac{21}{3} + ( - \dfrac{12 \sqrt{3} }{3} )$

$\tt = - \dfrac{21}{3} - \dfrac{12 \sqrt{3} }{3}$

Now consider

$\bf \dfrac{3 + 2 \sqrt{3} }{3 - 2 \sqrt{3} } = a + b \sqrt{3}$

$\tt \implies -\dfrac{21}{3} - \dfrac{12 \sqrt{3} }{3} = a + b \sqrt{3}$

Comparing on both sides

$\tt a = - \dfrac{21}{3}$

$\tt b \sqrt{3} = - \dfrac{12 \sqrt{3} }{3}$

Cancelling √3 on both sides

$\tt b= - \dfrac{12}{3}$

$\Huge{\boxed{ \sf a = - \dfrac{21}{3} , \: b= - \dfrac{12}{3} }}$