RMO 2018 Tamil Nadu Problem 2 is from Theory of Equations. We present sequential hints for this problem. Do not read all hints at one go. Try it yourself.
Find the set of all real values of a for which the real polynomial equation $ P(x) = x^2 - 2ax + b = 0 $ has real roots, given that $ P(0) \cdot P(1) \cdot P(2) \neq 0$ and $( P(0), P(1), P(2) )$ form a geometric progression.
Also see
P(0) = b (plug in x = 0 in the given quadratic). P(1) = 1 - 2a + b and P(2) = 4 - 4a + b.
Since P(0), P(1) and P(2) are in Geometric Progression, hence $P(0) \times P(2) = P(1)^2$. This balls down to $$ b \cdot (4 - 4a + b) = (1 - 2a + b)^2 $$. Simplify this expression.
$ b \cdot (4 - 4a + b) = (1 - 2a + b)^2 \ \Rightarrow 4b - 4ab + b^2 = 1 + 4a^2 + b^2 - 4a - 4ab + 2b \ \Rightarrow 2b = 4a^2 - 4a + 1 $
Discriminant of a quadratic equation $ ax^2 + bx + c = 0$ is $ b^2 - 4ac $. The quadratic equation has real roots iff $ b^2 - 4ac \geq 0 $.
The discriminant of the given quadratic is $$ 4a^2 - 4b $$ From Hint 2, we know $$ 4b = 8a^2 - 8a + 2 $$ Hence the discriminant is $$ 4a^2 - (8a^2 - 8a + 2) = -4a^2 + 8a - 2 $$
We want this discriminant to be non negative.
$$ -4a^2 + 8a - 2 \geq 0 \Rightarrow 2a^2 - 4a + 1 \leq 0 $$
Use the quadratic formula to find the roots of the quadratic $ 2a^2 - 4a + 1 $(this will allow us to factorize the expression).
The roots are $$ \frac { 4 \pm \sqrt {16 - 8}}{2} = 2 \pm \sqrt {2} $$
Hence $a \in [2 - \sqrt 2, 2 + \sqrt 2] $
Also see
