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Explore the Back-Story 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

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

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