1. Find integers

What do they mean by 'for all x' in (a)?

2. Originally Posted by GAdams
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

3. 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 $\displaystyle r=4$.

4. Originally Posted by red_dog
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 $\displaystyle r=4$.
What does the upside down A and the R mean?
I don't understand how you got part 'b'.
Thanks.

5. Originally Posted by GAdams
What does the upside down A and the R mean?
I don't understand how you got part 'b'.
Thanks.
the upside down A means "for all" so he's saying "for all x"

the "R" means the set of real numbers, so he is really saying "for all real x" or "for all x in the set of real numbers"

6. Thanks, but how can you write the equation as it's written in part (b)?

7. Originally Posted by GAdams
Thanks, but how can you write the equation as it's written in part (b)?
complete the square for the x-term, you know how to complete the square right?

8. Ok, I got part (b) thus:

x^2 + y^2 -8x =0

by completing the square (which I just learnt how to do)

x^2 + y^2 -8x + 16 = 16

(x-4)^2 + y^2 = 16

Part (c) is clear

Part (a) is still confusing me. I don't understand it all.

9. Originally Posted by GAdams
Part (a) is still confusing me. I don't understand it all.
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?

10. Got it!! Thanks. Great explanation!