Tessellate S.T.E.M.S - Mathematics - School - Set 1 - Problem 2

Algebra Level 3

You are allowed to do two operations to a quadratic polynomial with integer coefficients. You can either replace f ( x ) f(x) by f ( x + a ) f(x+a) for any integer a a or you can switch the positions of the leading coefficient and the constant term (i.e. you can replace a x 2 + b x + c ax^2+bx+c by c x 2 + b x + a cx^2+bx+a ). You start with the polynomial f = x 2 + x + 1 f=x^2+x+1 . Using the given operations, which of the following polynomials can you arrive at starting from f f ?

This problem is a part of Tessellate S.T.E.M.S.

6 x 2 + 7 x + 9 6x^2 + 7x + 9 none of the these 5 x 2 + 7 x + 3 5x^2 + 7x + 3 6 x 2 + 7 x + 8 6x^2 + 7x + 8

Writabrata Bhattacharya by Writabrata Bhattacharya , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Raushan Sharma
Dec 26, 2017

Note that, on performing the two operations, the discriminant of the quadratic expression remains invariant. So the discriminant of the quadratic expression we started with, was -3, and none of the quadratics in the options has D=-3, so ans is None of These.

11 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice invariant found!

Conversely, if the polynomial has integer coefficients and has a discriminant of -3, does that mean that we can use these operations to reach it?

Calvin Lin Staff - 3 years, 4 months ago
Aniket Messi
Jan 13, 2018

First replacing f(x) by f(x+a) f(x+a) = (x+a)^2 + (x+a) + 1 = x^2 + (2ax + x) + (a^2 + a + 1) Hence, whatever may be the integer 'a' , the co-efficient of the leading term x^2 will be 1 or, we can also say by the 2nd operation, the constant can be 1.

So, no polynomial have a single co-efficient of 1
and thus, the answer is B(None Of the these).

1 Helpful 0 Interesting 0 Brilliant 0 Confused

You have not considered the other operation: switching the constant term and the coefficient of x 2 x^2 is also a valid move.

Agnishom Chattopadhyay - 3 years, 4 months ago

Trying out different combis you can find 5 and 6 are no possible a's

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you explain your idea clearly? What combinations should you try to make this work?

Agnishom Chattopadhyay - 3 years, 4 months ago

1 pending report

Vote up reports you agree with

×

Problem Loading...

Note Loading...

Set Loading...