- Yes: [math]x=3[/math].
- There's another solution.
- Oh yeah, [math]x=-3[/math] works as well.
- Good. How about [math](x+1)^2=9[/math]?
- Well, I know that [math](x+1)^2=x^2+2x+[/math]...
- Stop! Resist the reflex to expand this. It's easier if you don't.
- But how can I solve [math](x+1)^2=9[/math] without expanding the square?
- You've already solved a very similar equation just a moment ago.
- But turning [math]x[/math] into [math]x+1[/math] changes everything.
- There's another way to look at it: give [math]x+1[/math] a new name, like [math]u[/math].
- Ok, how does that help?
- What is the equation now?
- It's [math]u^2=9[/math]. But that's cheating!
- No, you've done nothing wrong. What is [math]u[/math]?
- It is either 3 or -3.
- Right. So what is [math]x[/math]?
- Well [math]u=x+1[/math], so [math]x[/math] is either 2 or -4.
- Brilliant. You just solved [math]x^2+2x+1=9[/math] 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 [math](x+\mbox{blah})^2=\mbox{whatever}[/math]. I just give [math]x+\mbox{blah}[/math] a new name, solve, and recover [math]x[/math].
- Exactly. And guess what? This is enough to solve any second-degree equation.
- Huh?
- Suppose I give you [math]x^2+10x = 11[/math]. What would you do?
- I've no idea. It doesn't look at all like [math](x+\mbox{blah})^2=\mbox{anything}[/math].
- Don't be so pessimistic. Do you see that [math]\mbox{blah}=5[/math] almost works?
- How "almost"? Why 5?
- Try it.
- Well [math](x+5)^2 = x^2+10x+25[/math]. But that's not what's in the equation.
- What is?
- Just [math]x^2+10x[/math].
- ...which you now know is [math](x+5)^2-25[/math].
- Oh. Right. So the equation is [math](x+5)^2-25=11[/math].
- ...which is...
- [math](x+5)^2 = 36[/math].
- ...so...
- I rename [math]x+5[/math] to [math]u[/math] (it still feels like cheating), get [math]u^2=36[/math], so [math]u[/math] is 6 or -6 and [math]x[/math] is 1 or -11. That's it?
- Yes. What sort of equations can you solve now?
- Well, it looks like I can handle [math]x^2+\mbox{something}\,\, x = \mbox{whatever}[/math], provided [math]\mbox{something}[/math] is even.
- Why do you need [math]\mbox{something}[/math] to be even?
- Because what you did there was take that [math]\mbox{something}[/math], which was 10, and take half of it to be the [math]\mbox{blah}[/math].
- 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 [math]x^2+\mbox{something}\,\, x = \mbox{whatever}[/math]. I take half of the [math]\mbox{something}[/math], call it [math]\mbox{blah}[/math], find the correction so I can write [math](x+\mbox{blah})^2-\mbox{correction} = \mbox{whatever}[/math], move the numbers together, rename [math]x+\mbox{blah}[/math] to [math]u[/math], and solve.
- Excellent. Can you do what you just said with letters instead of blahs and whatevers?
- I think so:
[math]x^2+2bx=c[/math]
[math](x+b)^2-b^2=c[/math]
[math](x+b)^2 = c+b^2[/math]
[math]x+b = \pm \sqrt{c+b^2}[/math]
[math]x = -b \pm \sqrt{c+b^2}[/math].
- Excellent. We usually write a quadratic as [math]ax^2+bx+c=0[/math], but you can just divide through by [math]a[/math] and move the constant to the other side to get exactly what you just solved. You're done!
Read other answers by Alon Amit on Quora:
- Which grows faster: [math]e^{e^x}[/math] or [math]x![/math] ?
- How can I solve an equation of the third degree?
- What's missing in Maxwell's Equations? What physics doesn't fit?
from Quora http://ift.tt/2cijmNl
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