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# 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 $$a=b=c$$.

Please post your answers after at least one week (although this one is quite simple).

Note by Svatejas Shivakumar
1 year, 8 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 · 1 year, 8 months ago

Simple standard solution. · 1 year, 8 months ago

Hey Nihar and Mehul, try this problem.

If $$x^3 = x + 1$$ then determine integer a, b and c such that $$x^7 = ax^2 + bx + c$$ · 1 year, 8 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 $$a=2,b=2$$ · 1 year, 7 months ago

nice · 1 year, 7 months ago

Change your Profile pic :3 xD

A good troll anyway :P

I got trolled xD · 1 year, 8 months ago

i got confused when i was reading comment. Lol. · 1 year, 8 months ago

How do you simply "remove" $$y^2$$ from these equations? · 1 year, 8 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. · 1 year, 8 months ago

Yeah, I got it :) · 1 year, 8 months ago

Nice! · 1 year, 8 months ago

Thanks. Are you preparing for olympiad? · 1 year, 8 months ago

Yes · 1 year, 8 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 $\large a=b=c$ · 1 year, 8 months ago

How do u know that the roots are equal? · 1 year, 7 months ago

i did same..... · 1 year, 8 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 $$a^2 + 3b^2 = 4$$ · 1 year, 8 months ago

@Dev Sharma Can you post the solution? · 1 year, 7 months ago

Thanks for mentioning me! @Svatejas Shivakumar

@Silas Hundt There is some problem. I am not receiving notifications when someone @mentions me · 1 year, 8 months ago

Which kind of notifications? The ones in the corner on Brilliant or emails or...? Staff · 1 year, 8 months ago

The ones in the corner, sir :) · 1 year, 8 months ago

Is this still happening? I'm still seeing notifications... Staff · 1 year, 8 months ago

It happened with me twice. Hasn't happened very recently. I'll be sure to inform you in any such event :) · 1 year, 8 months ago