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.

# Problem

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.

## Key ideas you will need to solve this problem

- Theory of Equations:
- Discriminant of a quadratic is non -negative iff roots are real.
- If a, b, c are in Geometric Progression then \( ac = b^2 \)

Also see

## Hint 1: Compute P(0), P(1), P(2)

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.

## Hint 2: Express b in terms of a

( 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 )

## Hint 3: Use the discriminant of the quadratic

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.

## Hint 4: Final Lap

$$ -4a^2 + 8a – 2 \geq 0 \Rightarrow 2a^2 – 4a + 1 \leq 0 $$

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] )

# Reference:

- These ideas are usually discussed in the
**Polynomials I**module of Cheenta Math Olympiad Program. **Challenges and Thrills of Pre -College Mathematics by Venkatchala**is a good reference for some these ideas.

Also see

Google