Question

find the value of a and b ; 5+2√3/5-2√3=a+b√3​

Answer 1

Answer :-

$\tt a = \dfrac{37}{13} , \: b = \dfrac{20}{13}$

Explanation :-

Given :-

$\tt \dfrac{5 + 2 \sqrt{3} }{5 - 2 \sqrt{3} } = a + b \sqrt{3}$

To find :-

Values of a and b

Solution :-

$\tt \dfrac{5 + 2 \sqrt{3} }{5 - 2 \sqrt{3} } = a + b \sqrt{3}$

Consider LHS

$\tt \dfrac{5 + 2 \sqrt{3} }{5 - 2 \sqrt{3} }$

Rationalise the denominator

The rationalising factor of 5 - 2√3 is 5 + 2√3. So multiply both numerator and denominator with rationalising factor.

$\tt = \dfrac{5 + 2 \sqrt{3} }{5 - 2 \sqrt{3} } \times \dfrac{5 + 2 \sqrt{3} }{5 + 2 \sqrt{3} }$

$\tt = \dfrac{(5 + 2 \sqrt{3})^{2} }{5^{2} - (2 \sqrt{3})^{2} }$

[Since (x - y)(x + y) = x² - y² and above x = 5, y = 2√3 ]

$\tt = \dfrac{(5 + 2 \sqrt{3})^{2} }{25 - {2}^{2}( { (\sqrt{3})}^{2})}$

$\tt = \dfrac{(5 + 2 \sqrt{3})^{2} }{25 - 4(3)}$

$\tt = \dfrac{(5 + 2 \sqrt{3})^{2} }{25 -12}$

$\tt = \dfrac{(5 + 2 \sqrt{3})^{2} }{13}$

$\tt = \dfrac{5^{2} + {(2 \sqrt{3})}^{2} + 2(5)(2 \sqrt{3}) }{13}$

[Since (x + y)² = x² + y² + 2xy and above x = 5, y = 2√3 ]

$\tt = \dfrac{25 + {2}^{2}(( \sqrt{3})^{2}) + 2(5)(2 \sqrt{3}) }{13}$

$\tt = \dfrac{25 +4(3) + 2(5)(2 \sqrt{3}) }{13}$

$\tt = \dfrac{25 +12+ 2(5)(2 \sqrt{3}) }{13}$

$\tt = \dfrac{25 +12+ 20 \sqrt{3}}{13}$

$\tt = \dfrac{37+ 20 \sqrt{3}}{13}$

$\tt = \dfrac{37}{13} + \dfrac{20 \sqrt{3} }{13}$

$\tt \dfrac{5 + 2 \sqrt{3} }{5 - 2 \sqrt{3} } = \dfrac{37}{13} + \dfrac{20 \sqrt{3} }{13}$

Now consider

$\tt \dfrac{5 + 2 \sqrt{3} }{5 - 2 \sqrt{3} } = a + b \sqrt{3}$

$\tt \implies\dfrac{37}{13} + \dfrac{20 \sqrt{3} }{13} = a + b \sqrt{3}$

Comparing on both sides

$\tt a = \dfrac{37}{13}$

$\tt b \sqrt{3} = \dfrac{20 \sqrt{3} }{13}$

By cancelling √3 on both sides

$\tt b \cancel{\sqrt{3}} = \dfrac{20 \cancel{\sqrt{3}}}{13}$

$\tt b = \dfrac{20}{13}$

$\Huge{\boxed{ \tt a = \dfrac{37}{13} , \: b = \dfrac{20}{13} }}$

Answer 2

$\dfrac{5 \: + \: 2 \sqrt{3} }{5 \: - \: 2 \sqrt{3}}$ = $a \: + \: b \sqrt{3}$

_____________ [GIVEN]

• We have to find the value of a and b.

______________________________

$\implies$ $\dfrac{5 \: + \: 2 \sqrt{3} }{5 \: - \: 2 \sqrt{3}}$

• Rationalize it

$\implies$ $\dfrac{5 \: + \: 2 \sqrt{3} }{5 \: - \: 2 \sqrt{3}}$ × $\dfrac{5 \: - \: 2 \sqrt{3} }{5 \: - \: 2 \sqrt{3}}$

Now..

(a + b) (a + b) = (a + b)² = a² + b² + 2ab

(a + b) (a - b) = a² - b²

$\implies$ $\dfrac{ {(5 \: + \: 2 \sqrt{3} )}^{2} }{( {5)}^{2} \: - \: {(2 \sqrt{3}) }^{2} }$

$\implies$ $\dfrac{ {(5)}^{2} \: + \: {(2 \sqrt{3}) }^{2} \: + \: 2(5)(2 \sqrt{3} ) }{25\: - \: (2)(2)(3) }$

$\implies$ $\dfrac{ 25 \: + \: 4(3) \: + \: 20 \sqrt{3}}{13 }$

$\implies$ $\dfrac{ 25 \: + \: 12 \: + \: 20 \sqrt{3}}{13 }$

$\implies$ $\dfrac{ 37\: + \: 20 \sqrt{3}}{13 }$

Now write it in terms of $a \: + \: b \sqrt{3}$

$\implies$ $\dfrac{37}{13}$ + $\dfrac{20 \sqrt{3} }{13}$

So..

$a \: + \: b \sqrt{3}$ = $\dfrac{37}{13}$ + $\dfrac{20 \sqrt{3} }{13}$

______________________________

$\huge{\bold{a\: =}}$ $\huge{\dfrac{37}{13}}$

and

$\huge{\bold{b\:=}}$ $\huge{\dfrac{20 }{13}}$

____________ $\bold{[ANSWER]}$