Thread: Finding the values of p and q

1. Finding the values of p and q

$x^2-8x+23-(x-p)^2+q$ for all values of x

find the value of p and q

I did

$(x^2-8x+16)+7=...\rightarrow (x+4)^2+7=(x+p)^2+q$

I'm stuck.. what should I do now, square route the whole thing?

edit:

whittled it down to
$2x*2+23=p^2+q$

what now?

2. The first expression you typed, is it supposed to be an equation?

$x^2-8x+23=(x-p)^2+q$

If so, you can expand on the right hand side.

Giving $x^2-8x+23=x^2 -2px + p^2 + q$

Then, equating the coefficients, you see that:

$-8 = -2p$

and

$23 = p^2+q$

Or, you can take directly from where you reached;

$(x+4)^2+7=(x+p)^2+q$

and equate the coefficients.

3. Thanks, so many people have told me about the coefficient and I've looked on sites but I still don't understand how to calculate it.

All I've learned is that comparing the coefficient is along the lines of ax^2+bx+c=a+b+c

4. Okay, if you have

$ax^2 + bx + c = px^2 + qx + r$

Then, a = p because both a and p are terms in x^2
b = q since both b and q are terms in x,
and c = r since both are independent terms.

5. That's what I was thinking about when I was doing the equation but I was unsure because it seems too perfect. Can you show me a proof?

6. I'm not good at proving things

Have you done vector coordinates yet?

If so, you should know, for example that if you have a vector (ai + bj) and another vector (xi + yj)

Given both are the same vectors, mean that ai + bj = xi + yj

Automatically, a will be equal to x and b will be equal to y.

I know, it's not a very convincing way, but I've learned it through practice

7. Originally Posted by Unknown008
I'm not good at proving things

Have you done vector coordinates yet?

If so, you should know, for example that if you have a vector (ai + bj) and another vector (xi + yj)

Given both are the same vectors, mean that ai + bj = xi + yj

Automatically, a will be equal to x and b will be equal to y.

I know, it's not a very convincing way, but I've learned it through practice

I hate having to merely accept things >.>

yes I have studied them

8. Ok, I thought more about it during the day. I'll use the method of completing the squares to prove it.

Let's have a look at:

$ax^2 + bx + c= px^2 + qx + r$

I have to prove that a = p, b = q and c = r.

$(a-p)x^2 + (b-q)x + (c-r)= 0$

$(a-p)[x^2 + \frac{(b-q)}{(a-p)}x + \frac{(c-r)}{(a-p)}]= 0$

Now, completing the square;

$(a-p)[(x + \frac{(b-q)}{2(a-p)}x)^2 - (\frac{(b-q)}{2(a-p)})^2 + \frac{(c-r)}{a-p}]= 0$

So far so good.

For the equation to be true, either (a-p) = 0, or $(x + \frac{(b-q)}{2(a-p)}x)^2 - (\frac{(b-q)}{2(a-p)})^2 + \frac{(c-r)}{a-p} = 0$

I solve the first one.

a - p = 0

Therefore, a = p (1st shown)

Using that in the initial equation given,

$ax^2 + bx + c= px^2 + qx + r$

Since a = p,

$bx + c= qx + r$

I put all the terms on the same side again.

$(b - q)x + (c-r)= 0$

$(b - q)[x + \frac{(c-r)}{(b-q)}]= 0$

Hence, b - q = 0, and b = q. (2nd shown)

Again, using this new result in the previous equation,

$bx + c= qx + r$

We get:

c = r

I hope this one satisfies you

9. Wow that really is something. Thanks for the effort you put in! It really settles my mind :P