What is a simple explanation for the general equation of the second degree?
- Can you solve x^2=9?
- Yes: x=3.
- There's another solution.
- Oh yeah, x=-3 works as well.
- Good. How about (x+1)^2=9?
- Well, I know that (x+1)^2=x^2+2x+...
- Stop! Resist the reflex to expand this. It's easier if you don't.
- But how can I solve (x+1)^2=9 without expanding the square?
- You've already solved a very similar equation just a moment ago.
- But turning x into x+1 changes everything.
- There's another way to look at it: give x+1 a new name, like u.
- Ok, how does that help?
- What is the equation now?
- It's u^2=9. But that's cheating!
- No, you've done nothing wrong. What is u?
- It is either 3 or -3.
- Right. So what is x?
- Well u=x+1, so x is either 2 or -4.
- Brilliant. You just solved x^2+2x+1=9 by looking at it in a different way!
- Yeah, well, but you made it up so it works nicely. What if it didn't?
- What general family of equations do you think you can solve now?
- Um, stuff like (x+\mbox{blah})^2=\mbox{whatever}. I just give x+\mbox{blah} a new name, solve, and recover x.
- Exactly. And guess what? This is enough to solve any second-degree equation.
- Huh?
- Suppose I give you x^2+10x = 11. What would you do?
- I've no idea. It doesn't look at all like (x+\mbox{blah})^2=\mbox{anything}.
- Don't be so pessimistic. Do you see that \mbox{blah}=5 almost works?
- How "almost"? Why 5?
- Try it.
- Well (x+5)^2 = x^2+10x+25. But that's not what's in the equation.
- What is?
- Just x^2+10x.
- ...which you now know is (x+5)^2-25.
- Oh. Right. So the equation is (x+5)^2-25=11.
- ...which is...
- (x+5)^2 = 36.
- ...so...
- I rename x+5 to u (it still feels like cheating), get u^2=36, so u is 6 or -6 and x is 1 or -11. That's it?
- Yes. What sort of equations can you solve now?
- Well, it looks like I can handle x^2+\mbox{something}\,\, x = \mbox{whatever}, provided \mbox{something} is even.
- Why do you need \mbox{something} to be even?
- Because what you did there was take that \mbox{something}, which was 10, and take half of it to be the \mbox{blah}.
- That's right. So you can't take half of 7?
- I can, but it's a fraction.
- So?
- I hate fractions.
- LOL, ok. But they work just the same.
- Fine. So I can handle x^2+\mbox{something}\,\, x = \mbox{whatever}. I take half of the \mbox{something}, call it \mbox{blah}, find the correction so I can write (x+\mbox{blah})^2-\mbox{correction} = \mbox{whatever}, move the numbers together, rename x+\mbox{blah} to u, and solve.
- Excellent. Can you do what you just said with letters instead of blahs and whatevers?
- I think so:
x^2+2bx=c
(x+b)^2-b^2=c
(x+b)^2 = c+b^2
x+b = \pm \sqrt{c+b^2}
x = -b \pm \sqrt{c+b^2}.
- Excellent. We usually write a quadratic as ax^2+bx+c=0, but you can just divide through by a and move the constant to the other side to get exactly what you just solved. You're done!