# Weird question

1. Jul 29, 2012

### phospho

$6pq+r=\sqrt{r^2-4ps}, \ \ p\neq 0 \ \ (1)$

Without squaring (1) or any of it's rearrangements, express p in terms of only q, r and s.

I realised this is some sort of quadratic formula so I got $$p = \dfrac{-r+\sqrt{r^2-4ps}}{6q}$$

so for the quadratic ap^2 + bp + c,
a = 3q, b = r, c = s,
3qp^2 + rp + s

i'm stuck from here.

2. Jul 29, 2012

### SammyS

Staff Emeritus
Does the quadratic formula count as squaring equation (1) ? Just asking ...

If it's OK to use the quadratic formula, then notice that in relation to the standard quadratic equation,
$ax^2+bx+c=0$​
the quantities under the radical are the coefficients, a, b, and c.
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$​

Puttying this in a form similar to eq. (1) gives
$\displaystyle 2ax+b=\pm\sqrt{b^2-4ac}$​

3. Jul 29, 2012

### gabbagabbahey

errm... $4ac \neq 4ps$ so this isn't quite so nice

4. Jul 30, 2012

### phospho

It doesn't no.

I somehow managin to get $$p = \dfrac{-(rq+s)}{3q^2}$$ which is incorrect

could you explain?

Last edited: Jul 30, 2012
5. Jul 30, 2012

### phospho

never mind I rearranged into $3q = \dfrac{-r+\sqrt{r^2-4ps}}{2p}$ and got the right answer.

thanks!

Last edited: Jul 30, 2012
6. Jul 30, 2012

### eumyang

Not sure how you got the right answer, because what you wrote doesn't follow from the original. Also, I thought the question was to...
Color me confused.

7. Jul 30, 2012

### SammyS

Staff Emeritus
So, what did you get for p ?

8. Jul 30, 2012

### phospho

well the original equation is $6pq + r = \sqrt{r^2-4ps}$ I rearranged to get $3q = \dfrac{-r+\sqrt{r^2-4ps}}{2p}$ I then compared it with $x = \dfrac{-b+\sqrt{b^2-4ac}}{2a}$ giving x = 3q, b = r, c = s so: $p(3q)^2 +r3q +s = 0$ rearranging to get $p = \dfrac{-(3qr + s)}{9q^2}$ which is correct (or so it says.)

9. Jul 30, 2012

### SammyS

Staff Emeritus
I agree.