The value of 'x' satisfying the equation x^2+p^2=(q-x)^2 is:

Question

anonymous · Answer

$$x^{2}+p^{2} = (q-x)^{2}$$

First I'd expand out the right side of the equation.

anonymous · Answer

Simply divide both sides by 2q and you will get your answer...

jiteshmeghwal9 · Answer

Plz give me full solution:)

anonymous · Answer

Above I have solved it @jiteshmeghwal9 ..

anonymous · Answer

$$p ^{2}-q ^{2}=-2xp$$

anonymous · Answer

the way the equation is opened on the right should be -2px,
but i am not sure what to do from there

anonymous · Answer

You might be overthinking it @Jonask :) Just solve for X, then slap it back into the original equation as a test. Ultimately though it's up to @jiteshmeghwal9

anonymous · Answer

or i see its value not value(s)
but iam still concerned abut the expansion of
$$(x-q)^{2}=x ^{2}-2qx+q ^{2}$$

anonymous · Answer

@jonask What's your concern exactly?

anonymous · Answer

waterineyes's first expantion gives $$x ^{2}+2px+q ^{2}$$
isn't that supposed to be the one i just posted above
-2px

anonymous · Answer

-2QX actually.

anonymous · Answer

Looks better, but you're still not supposed to do the work for them...

anonymous · Answer

@Jonask, you are right I have committed just a little minus mistake there...

anonymous · Answer

Sorry @Wired ..

anonymous · Answer

I have deleted the post...

anonymous · Answer

TYVM :)

anonymous · Answer

its okay
so
$$x=(q ^{2}-p ^{2})/2q$$