Question

do you see it?

Answer 1

The following operations are permitted with the quadratic polynomial: \(ax^2 + bx + c\):

(a) switch \(a\) and \(c\),
(b) replace \(x\) by \(x+t\) where \(t\) is any real number.

Can we transform \(x^2 - x - 2\) into \(x^2 - x - 1\)?

Answer 2

Another one. I'm finally able to solve problems like these.

Each of the numbers \(1\) to \(10^6\) is repeatedly replaced by its digital sum until we reach
\(10^6\) one-digit numbers. Will these have more 1’s or 2’s?

Answer 3

the first one is basically a comparison.. right?

Answer 4

i dont see it :(
there only two ways to do it, switch a, c first or and then shift by 't', or the other way round, either case, i cant see any solution

Answer 5

if we have any equation and we transform it
for example if we have this- ax^2 +bx+c
1st step we get this-
cx^2+bx+a

the coefficient of x^2 can never change even when we proceed for the 2nd step i.e., (x+t)

so
a of the equation x^2-x-1 is positive so it will be upwrd facing

when u transform the equation x^2-x-2
the 1st step u get this-
-2x^2-x+1

the a of this equation will always be -2 i.e., it will always be negative
so it will be downwrd facing
so we can't transform it like that

Answer 6

My proof is this: in either steps, the discriminant remains the same.

Answer 7

omg yes xD

Answer 8

i dont get it...so what if the discriminant remains the same?

Answer 9

Then the initial and final discriminant must be the same.

Answer 10

ah..your saying the the given polynomials have different discriminats, and D before/after transformation is the same

Answer 11

When you say "a and c" does this represent a valid transformation:

\[ax^2+bx+bt+c \to (bt+c)x^2+bx+a\]

Answer 12

erm, ok whatever I left out shifting the x^2 but you get the question