1
$\begingroup$

I have the equation

$t - (m-q)^2 = v - (m-p)^2$

which I would like to rearrange to be able to apply the quadratic formula, and solve in terms of $q$. Accordingly, it needs to be in the form:

$ax^2 + bx +c = 0$

I have got as far as:

$(t-v)-(m-q)^2 + (m-p)^2 = 0$

$(t-v)-m^2 - q^2 + 2mq + m^2 +p^2 - 2mp = 0$

$(t-v) - q^2 + 2mq +p^2 - 2mp = 0$

But am now stuck.

Can anyone suggest a way forward? Or, as is likely, point out where I've gone wrong?

Thanks very much!

$\endgroup$
4
  • $\begingroup$ What would play the role of $x?$ $\endgroup$ – mfl Jul 23 '14 at 11:49
  • $\begingroup$ Apologies! I'm trying to solve in terms of $q$. I'll edit the question to reflect this. $\endgroup$ – user2728808 Jul 23 '14 at 11:51
  • $\begingroup$ I think you are on the right track, for $x=q$, you have $a=-1$, $b=2m$, and $c=p^2-2mp+t-v$ $\endgroup$ – Tymric Jul 23 '14 at 11:59
  • $\begingroup$ Thanks Timmy - your answer is just the same as kleineg. Thanks very much! $\endgroup$ – user2728808 Jul 23 '14 at 12:15
5
$\begingroup$

You have $(t-v) - q^2 + 2mq +p^2 - 2mp = 0$ and want to get the equation into the form $aq^2+bq+c$

I would combine the terms to get $- q^2 + 2mq + (p^2 - 2mp+t-v) = 0 $

This gives you $a=-1$, $b=2m$, and $c =p^2 - 2mp+t-v$

$\endgroup$
2
  • 1
    $\begingroup$ Many thanks, this is great! Sometimes it's hard to see what's right in front of you... $\endgroup$ – user2728808 Jul 23 '14 at 12:14
  • $\begingroup$ Glad to be of help. $\endgroup$ – kleineg Jul 23 '14 at 12:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.