\[\begin{cases} P_{1}(x) = ax^{2}-bx-c \\ P_{2}(x)=bx^{2}-cx-a \\ P_{3}(x)=cx^2-ax-b\ \end{cases} \] 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).*

## Comments

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TopNewestjust 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 – Dev Sharma · 1 year, 3 months ago

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– Nihar Mahajan · 1 year, 3 months ago

Simple standard solution.Log in to reply

If \(x^3 = x + 1\) then determine integer a, b and c such that \(x^7 = ax^2 + bx + c\) – Dev Sharma · 1 year, 3 months ago

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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 \(a=2,b=2\) – Ravi Dwivedi · 1 year, 3 months ago

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– Dev Sharma · 1 year, 3 months ago

niceLog in to reply

A good troll anyway :P

I got trolled xD – Mehul Arora · 1 year, 3 months ago

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– Dev Sharma · 1 year, 3 months ago

i got confused when i was reading comment. Lol.Log in to reply

– Mehul Arora · 1 year, 3 months ago

How do you simply "remove" \(y^2\) from these equations?Log in to reply

– Nihar Mahajan · 1 year, 3 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.Log in to reply

– Mehul Arora · 1 year, 3 months ago

Yeah, I got it :)Log in to reply

– Svatejas Shivakumar · 1 year, 3 months ago

Nice!Log in to reply

– Dev Sharma · 1 year, 3 months ago

Thanks. Are you preparing for olympiad?Log in to reply

– Svatejas Shivakumar · 1 year, 3 months ago

YesLog in to reply

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 \[\large a=b=c\] – Surya Prakash · 1 year, 3 months ago

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– Sneha Iyer · 1 year, 3 months ago

How do u know that the roots are equal?Log in to reply

– Sourav Kumar Surya · 1 year, 3 months ago

i did same.....Log in to reply

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 \(a^2 + 3b^2 = 4\) – Dev Sharma · 1 year, 3 months ago

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@Dev Sharma Can you post the solution? – Svatejas Shivakumar · 1 year, 2 months ago

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@Dev Sharma @Swapnil Das @Nihar Mahajan @Mehul Arora @Shivam Jadhav – Svatejas Shivakumar · 1 year, 3 months ago

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@Svatejas Shivakumar

Thanks for mentioning me!@Silas Hundt There is some problem. I am not receiving notifications when someone @mentions me – Mehul Arora · 1 year, 3 months ago

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– Silas Hundt Staff · 1 year, 3 months ago

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

– Mehul Arora · 1 year, 3 months ago

The ones in the corner, sir :)Log in to reply

– Silas Hundt Staff · 1 year, 3 months ago

Is this still happening? I'm still seeing notifications...Log in to reply

– Mehul Arora · 1 year, 3 months ago

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

– Swapnil Das · 1 year, 3 months ago

Thanks for mentioning😛Log in to reply