So, I found these statements and I need your assistance to prove them since my body condition is not fit enough to think that much.
1. The quadratic equation whose roots are k less than the roots of $\displaystyle ax^2+bx+c=0$ is $\displaystyle a(x+k)^2+b(x+k)+c=0$.
2. The quadratic equation whose roots are k more than the roots of $\displaystyle ax^2+bx+c=0$ is $\displaystyle a(x-k)^2+b(x-k)+c=0$.
3. The quadratic equation whose roots are n times the roots of $\displaystyle ax^2+bx+c=0$ is $\displaystyle ax+bnx+cn^2=0$.
4. The quadratic equation whose roots are negations of the roots of $\displaystyle ax^2+bx+c=0$ is $\displaystyle ax^2-bx+c=0$.
5. The quadratic equation whose roots are inverses of the roots of $\displaystyle ax^2+bx+c=0$ is $\displaystyle cx^2+bx+a=0$.
6. The quadratic equation whose roots are squareroots of the roots of $\displaystyle ax^2+bx+c=0$ is $\displaystyle a^2x^2-(b^2-2ac)+c^2=0$.

1) and 2), set (for example) $y=x+k$, the roots of the quadratic in $y$ are those of the roots of the first quadratic in the question. Replace $y$ with the expression for $x$ and you have your result.
3) is similar: $y = nx$. 4) is a special case of 3).
In 5) "inverse" presumably means "reciprocal". Set $y = \frac1x$ and form a quadratic from the resultant equation.
6) requires something mildly different, but it ought to be pretty clear how to approach it from what I've already written.

Mr.Fly, one is innocent until proven guilty.
Soooo....those 6 statements are correct until proven incorrect...
Soooo....relax

(6)
if we take $a=1,b=-5,c=4$ then the roots of

$a^2 x^2-(b^2-2ac)x +c^2=x^2-17x+16=0$

are the squares (not the square roots) of the roots of

$ax^2+bx+c=x^2-5x+4=0$

Originally Posted by Archie
6) requires something mildly different, but it ought to be pretty clear how to approach it from what I've already written.
Actually, I think it's a lot different. More along the lines of identifying that for roots $r_1,r_2$ of the first equation we have $\frac{c}{a} = r_1r_2$ and $-\frac{b}{a}=r_1+r_2$. For the equation $Ax^2+Bx +C=0$ with roots $r_1^2, r_2^2$ we therefore have $\frac{C}{A} = r_1^2 r_2^2 = \frac{c^2}{a^2}$ and $-\frac{B}{A}=r_1^2+r_2^2 = (r_1+r_2)^2 - 2r_1r_2$.

Originally Posted by Archie
1) and 2), set (for example) $y=x+k$, the roots of the quadratic in $y$ are those of the roots of the first quadratic in the question. Replace $y$ with the expression for $x$ and you have your result.
3) is similar: $y = nx$. 4) is a special case of 3).
In 5) "inverse" presumably means "reciprocal". Set $y = \frac1x$ and form a quadratic from the resultant equation.
6) requires something mildly different, but it ought to be pretty clear how to approach it from what I've already written.
Yes. "Inverse" does mean "reciprocal".
For number 6, I mistook "whose roots are squares" as "squareroots".

Originally Posted by DenisB
Mr.Fly, one is innocent until proven guilty.
Soooo....those 6 statements are correct until proven incorrect...
Soooo....relax
Don't worry, Denis McField, someone at MIF has already proven their innocance. Now they can Rest in Peace (Wait... Doesn't that mean they're executed?).

Originally Posted by Idea
(6)
if we take $a=1,b=-5,c=4$ then the roots of

$a^2 x^2-(b^2-2ac)x +c^2=x^2-17x+16=0$

are the squares (not the square roots) of the roots of

$ax^2+bx+c=x^2-5x+4=0$
Thank you. I did mistook "whose roots are squares" as "squareroots".

Originally Posted by Archie
Actually, I think it's a lot different. More along the lines of identifying that for roots $r_1,r_2$ of the first equation we have $\frac{c}{a} = r_1r_2$ and $-\frac{b}{a}=r_1+r_2$. For the equation $Ax^2+Bx +C=0$ with roots $r_1^2, r_2^2$ we therefore have $\frac{C}{A} = r_1^2 r_2^2 = \frac{c^2}{a^2}$ and $-\frac{B}{A}=r_1^2+r_2^2 = (r_1+r_2)^2 - 2r_1r_2$.
I mistook "whose roots are squares" as "squareroots".