# RMO practice problems from past year papers (polynomials)

$\begin{cases} P_{1}(x) = ax^{2}-bx-c \\ P_{2}(x)=bx^{2}-cx-a \\ P_{3}(x)=cx^2-ax-b\ \end{cases}$ Let the above be three quadratic polynomials where a,b,c are non-zero real numbers. Suppose there exists a real number $\alpha$ such that $P_{1}(\alpha)=P_{2}(\alpha)=P_{3}(\alpha)$, then

• Prove that $a=b=c$. Note by A Former Brilliant Member
5 years, 10 months ago

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just put alpha at x. I am taking alpha as y.

$ay^2 - by - c = k$

$by^2 - cy - a = k$

$cy^2 - ay - b = k$

removing $y^2$ from these equations, we get three other equations and add them all the we get

$ab+bc+ca-a^2-b^2-c^2 = 0$

so a = b = c

- 5 years, 10 months ago

Nice!

- 5 years, 10 months ago

Thanks. Are you preparing for olympiad?

- 5 years, 10 months ago

Yes

- 5 years, 10 months ago

How do you simply "remove" $y^2$ from these equations?

- 5 years, 10 months ago

Make the coefficient of $y^2$ same and subtract to eliminate it. For eg: multiply first equation by $b$ and second by $a$ and observe the coefficient of $y^2$ and thereby subtract them.

- 5 years, 10 months ago

Yeah, I got it :)

- 5 years, 10 months ago

Simple standard solution.

- 5 years, 10 months ago

Change your Profile pic :3 xD

A good troll anyway :P

I got trolled xD

- 5 years, 10 months ago

i got confused when i was reading comment. Lol.

- 5 years, 10 months ago

Hey Nihar and Mehul, try this problem.

If $x^3 = x + 1$ then determine integer a, b and c such that $x^7 = ax^2 + bx + c$

- 5 years, 10 months ago

$x^3=x+1$

Squaring both sides yields

$x^6=(x+1)^2=x^2 + 2x+1$

Since $x$ is non zero we can multiply both sides by $x$ and get

$x^7=x^3+2x^2+x$

$x^7=x+1+2x^2+x$

$x^7=2x^2+2x+1$

So we get $a=2,b=2$

- 5 years, 9 months ago

nice

- 5 years, 9 months ago

Since, the three polynomials have same root. So, one of them can be expressed as linear combination of other two. So, $P_{1} (x) = P_{2} (x) + \lambda P_{3} (x)$, where $\lambda$ is a real number. Comparing the coefficients of $x^{2},x$ and constant, we get different values of $\lambda$. They are

$\lambda = \dfrac{c-a}{b} = \dfrac{a-b}{c} = \dfrac{b-c}{a} = \dfrac{a-b+b-c+c-a}{a+b+c} = 0$ which implies that $\large a=b=c$

- 5 years, 10 months ago

i did same.....

- 5 years, 10 months ago

How do u know that the roots are equal?

- 5 years, 9 months ago

- 5 years, 10 months ago

Thanks for mentioning😛

- 5 years, 10 months ago

Thanks for mentioning me! @Svatejas Shivakumar

@Silas Hundt There is some problem. I am not receiving notifications when someone @mentions me

- 5 years, 10 months ago

Which kind of notifications? The ones in the corner on Brilliant or emails or...?

Staff - 5 years, 10 months ago

The ones in the corner, sir :)

- 5 years, 10 months ago

Is this still happening? I'm still seeing notifications...

Staff - 5 years, 10 months ago

It happened with me twice. Hasn't happened very recently. I'll be sure to inform you in any such event :)

- 5 years, 10 months ago

try this

$P(x) = 4x^3 - 2x^2 - 15x + 9$ and $Q(x) = 12x^3 + 6x^2 - 7x + 1$ then prove that each polynomial has 3 distinct real roots and let a and b be greatest roots of p(x) and q(x), respectively then prove that $a^2 + 3b^2 = 4$

- 5 years, 10 months ago

@Dev Sharma Can you post the solution?

- 5 years, 9 months ago

I still did not understand how to solve the question

- 3 years, 10 months ago

Bt why

- 3 years, 9 months ago

Thnx

- 3 years, 9 months ago