Why is $p^2-q^2 = p-q$ when $p+q = 1$?

Question

I'm having trouble to see why it is that:

$p^2-q^2 = p-q$ when $p+q = 1$

If I take a number for which $p+q \not= 1$, this doesn't hold. What's so special about the former case? Can someone take me through this, algebraically?

score 13 · Accepted Answer · answered Feb 09 '14 at 18:44

13

$$p^2 - q^2 = (p-q)(p+q){}{}{}{}$$

answered Feb 09 '14 at 18:44

amWhy

209,954

Ah, thanks. :) Now I see why it simplifies to $(p-q)$ when $(p+q)=1$. I never thought of writing out the difference of the squares in that way. I should probably practice factoring the difference of squares. Incidentally, I only just learned the name of this algebraic technique (from @ir7). – BigSmoke Feb 09 '14 at 19:02

ir7 · Answer 2 · 2014-02-09T19:13:10.003

2

Complementing the other answer, the formula has a name: Difference of two squares formula.

Moreover, it also goes the other way, if $p\not = q$. That is: if $p^2-q^2=p-q$ and $p\not=q$, then $p+q=1$.

edited Feb 09 '14 at 19:13

answered Feb 09 '14 at 18:45

ir7

6,249

Why is $p^2-q^2 = p-q$ when $p+q = 1$?

2 Answers2