What do they mean by 'for all x' in (a)?
It means find $\displaystyle p$ and $\displaystyle q$ such that whatever value $\displaystyle x$ takes:
$\displaystyle
x^2-8x=(x-p)^2+q
$
That is find $\displaystyle p$ and $\displaystyle q$ so that this is an identity.
To do this expand the right hand side, and equate the coefficients of like
powers of $\displaystyle x$ on both sides of the equation.
RonL
a) $\displaystyle x^2-8x=(x-p)^2+q\Leftrightarrow x^2-8x=x^2-2px+p^2+q,\forall x\in\mathbf{R}$. That means all coefficients of both members are equal.
Then $\displaystyle \left\{\begin{array}{c}-2p=-8\\p^2+q=0\end{array}\right.\Rightarrow p=4,q=-16$.
b) The equation can be written
$\displaystyle x^2-8x+16+y^2=16\Leftrightarrow (x-4)^2+y^2=16$.
c) The graph is a circle with center $\displaystyle C(4,0)$ and radius $\displaystyle r=4$.
I think red_dog did an excellent job on this question. However, i shall take baby steps and leave out the formalism, hoefully you can get the concept from this.
We have $\displaystyle x^2 - 8x = (x - p)^2 + q$
we want this to be true no matter what $\displaystyle x$ we choose. we will do the classic move of equating coefficients to accomplish this. Let's expand the right side, we get:
$\displaystyle x^2 - 8x = x^2 - 2xp + p^2 + q$
Now group like powers of $\displaystyle x$
$\displaystyle \Rightarrow x^2 - 8x = x^2 - (2p)x + \left( p^2 + q \right)$
so now we match up the coefficients. since the two sides are equal, we must have the same number of each power of $\displaystyle x$ on both sides.
we have 1 $\displaystyle x^2$ on each side, so we're good there
on the left we have -8x on the right we have -2px, they must be equal, so we must have:
$\displaystyle -8 = -2p \implies \boxed { p = 4 }$
on the left we have 0 as the constant term, on the right we have $\displaystyle p^2 + q$ as the constant term. since we want the constants to be the same as well for the sake of equality, we must have:
$\displaystyle 0 = p^2 + q$
$\displaystyle \Rightarrow 0 = 16 + q$ ..........since $\displaystyle p = 4$
$\displaystyle \Rightarrow \boxed { q = -16 }$
so those are the two values we need. so no matter what $\displaystyle x$ is, those values for $\displaystyle p$ and $\displaystyle q$ will work
did you get it?