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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

Nice!

Log in to reply

Thanks. Are you preparing for olympiad?

Log in to reply

Log in to reply

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

Log in to reply

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

Log in to reply

Simple standard solution.

Log in to reply

Change your Profile pic :3 xD

A good troll anyway :P

I got trolled xD

Log in to reply

Log in to reply

Hey Nihar and Mehul, try this problem.

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

Log in to reply

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 \(a=2,b=2\)

Log in to reply

Log 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 \[\large a=b=c\]

Log in to reply

i did same.....

Log in to reply

How do u know that the roots are equal?

Log in to reply

@Dev Sharma @Swapnil Das @Nihar Mahajan @Mehul Arora @Shivam Jadhav

Log in to reply

Thanks for mentioning😛

Log in to reply

Thanks for mentioning me! @Svatejas Shivakumar

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

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

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 \(a^2 + 3b^2 = 4\)

Log in to reply

@Dev Sharma Can you post the solution?

Log in to reply

I still did not understand how to solve the question

Log in to reply

Bt why

Log in to reply

Thnx

Log in to reply