Question

If [underroot3+1/underroot3-1=a+b underroot 3 ]find thevalue of a and b

Answer 1

Answer:-

Given:

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

Multiplying numerator and denominator by √3 + 1 in LHS we get,

$\implies \sf \: \frac{( \sqrt{3 } + 1)( \sqrt{3} + 1)}{( \sqrt{3} - 1)( \sqrt{3} + 1)} = a + b \sqrt{3}$

using (a + b)(a - b) = a² - b² & (a + b)(a + b) = (a + b)² = a² + b² + 2ab we get,

$\implies \sf \: \frac{ {( \sqrt{3} )}^{2} + {1}^{2} + 2 \sqrt{3} }{( \sqrt{3)} ^{2} - {1}^{2} } = a + b \sqrt{3} \\ \\ \implies \sf \: \frac{3 + 1 + 2 \sqrt{3} }{3 - 1} = a + b \sqrt{3} \\ \\ \implies \sf \: \frac{4 + 2 \sqrt{3} }{2} = a + b \sqrt{3} \\ \\ \implies \sf \: \frac{4}{2} + \frac{2 \sqrt{3} }{2} = a + b \sqrt{3} \\ \\ \implies \boxed{ \sf \:2 + \sqrt{3} = a + b \sqrt{3} }$

On comparing both sides we get,

• a = 2

• b = 1

Answer 2

(√3 + 1)/(√3 - 1) = a + b√3

→ (√3 + 1)/(√3 - 1) . (√3 + 1)/(√3 + 1) = a + b√3 {This is called rationalising}

→ (3 + 1 + 2√3)/(3 - 1) = a + b√3 {Use suitable identities and simplify}

→ (4 + 2√3)/2 = a + b√3

→ 2 + √3 = a + b√3

Thus:- a = 2, b = 1