I have an equation,

$-(x_m - x_q)^2 = -(x_m - x_p)^2$

which I want to solve in terms of $x_q$. I can see (by using a number line) that $q$ can have two solutions:

$x_q = x_p$


$x_q = 2x_m-x_p$

But I'd really like to be able to understand the algebra that could lead me to this. Could anyone walk me through it? Or, alternatively, direct me to the right resources to be able to work out the solution? Afraid I'm very new to all this and so realise this question is totally naive!

Many thanks!


The equation you want to solve is a quadratic equation which has usually 2 solutions (I already eliminated the minus): $$\begin{align} (-[x_m-x_p])^2 &= \\ (x_m-x_q)^2 &= (x_m-x_p)^2 \\ &= (-[x_m-x_p])^2 \\ \end{align}$$

So for solving we have to consider both possibilities

1.) $x_m-x_q = x_m-x_p$ (which is aequivalent to $ -[x_m-x_q] = -[x_m-x_p] $) and

2.) $x_m-x_q = -[x_m-x_p]$ (which is aequivalent to $ -[x_m-x_q] = x_m-x_p $)

  • $\begingroup$ This is great, very helpful for me, thanks so much. As an extension - if you're interested - can we extend this to the case where there is an additional additive term on either side of the equation? I.e. $a-(x_m - x_q)^2 = b - (x_m - x_p)^2$ $\endgroup$ Jul 23 '14 at 9:15
  • $\begingroup$ Not directly that way. Do you know how to solve general quadratic equations? You'd have to expand the quadratic terms and then rearrange them to the form $ax^2+bx+c = 0$ And then you could solve the equation with this formula. Because the equation is quadratic you will get again two solutions (or one or zero in special cases). I hope that helps! $\endgroup$
    – flawr
    Jul 23 '14 at 10:08

You should draw the parabola $y=(x-a)^2.$

The symmetry is clear. Just as $z^2= (-z)^2$ and there's no other number whose square equals $z^2$, you'll see the shape of your parabola.

The value $A^2$ is attained by $A$ and by $2a-A$, because the symmetry axis of the parabola is $x=A$. A picture is worth 1000 words, so do it yourself and check it online.

This happens also for all even powers.


when you take square root of a number which itself is squared , you'll get both negative and positive of that number $\sqrt{x^2}=\pm x$

  • 1
    $\begingroup$ There are a couple of problems with what you wrote. First, when applied to positive real numbers, the $\sqrt{{\cdot}}$ symbol denotes the positive square root; second, it's not the case that square rooting gives you both, it's that if you want to solve the equation $x^2=a$ (where $a > 0$) that there are two possible solutions, either $\sqrt{a}$ or $-\sqrt{a}$. (Note 'or' not 'and'.) I'm just being pedantic, but I think it's important to be pedantic with a question like this one where the details are being asked for. FWIW, $\sqrt{x^2} = |x|$, where $|x|$ is the absolute value of $x$. $\endgroup$ Jul 23 '14 at 9:03

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