Question

find the value of 'a' and 'b' if
$\frac{ \sqrt{3} - 1}{ \sqrt{3} + 1} = a + b \sqrt{3}$

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Answer 1

Answer :-

$\sf \dfrac{\sqrt 3 - 1}{\sqrt 3 + 1 } = a + b\sqrt3$

Solving the LHS :-

$\implies\sf \dfrac{\sqrt 3 - 1}{\sqrt 3 + 1 }$

Rationalizing the denominator :-

$\implies\sf \left(\dfrac{\sqrt 3 - 1}{\sqrt 3 + 1 }\right)\times \left(\dfrac{\sqrt 3 - 1}{\sqrt 3 - 1 } \right)$

$\implies\sf \dfrac{(\sqrt 3 - 1)^2}{(\sqrt3-1)(\sqrt3+1)}$

Using the identities :-

• $\sf (a - b)^2 = a^2 + b^2 - 2ab$
• $\sf (a+b)(a-b) = a^2 - b^2$

$\implies\sf \dfrac{(\sqrt 3)^2 + 1^2 - 2 \times \sqrt 3 \times 1}{(\sqrt{3})^2 - (1)^2}$

$\implies\sf \dfrac{3 + 1 - 2\sqrt 3}{3 - 1}$

$\implies\sf \dfrac{4 -2 \sqrt 3}{2}$

$\implies\sf 2 - \sqrt 3$

Comparing LHS and RHS :-

$\implies\sf 2 - \sqrt 3 = a + b\sqrt3$

• a = 2
• b = -1

Value of a = 2 & b = -1

Answer 2

Answer:

read the above pic ^_^

hope you are also doing well