October 29, 2018

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