Question
If $$a = 1 + \left( {\sqrt 3 - 1} \right) + {\left( {\sqrt 3 - 1} \right)^2} + {\left( {\sqrt 3 - 1} \right)^3} + ....$$ and $$ab = 1,$$ then $$a$$ and $$b$$ are the roots of the equation
A.
$${x^2} + 4x - 1 = 0$$
B.
$${x^2} - 4x - 1 = 0$$
C.
$${x^2} + 4x + 1 = 0$$
D.
$${x^2} - 4x + 1 = 0$$
Answer :
$${x^2} - 4x + 1 = 0$$
Solution :
$$a = \frac{1}{{1 - \left( {\sqrt 3 - 1} \right)}} = \frac{1}{{2 - \sqrt 3 }} = 2 + \sqrt 3 $$
and $$ab = 1$$
$$ \Rightarrow b = 2 - \sqrt 3 $$
so, $$a$$ and $$b$$ are roots of $${x^2} - 4x + 1 = 0$$