Is there a reason Lang uses (x-s)^2 instead of (x+s)^2?

[r0bHadz]

Homework Statement

When learning about quadratic equations in Langs book he gave the example:

let ax^2 + bx+ c = 0 be an equation

he subtracts the c, and says we want to get the left hand side (ax^2 + bx) = (x-s)^2

Homework Equations

The Attempt at a Solution

I'm just interested, either my conclusion that it doesn't matter which one you use, you will get the same solution, is wrong, or is this just a matter of convention for mathematicians to use negative numbers?

Sorry if this questions seems a bit useless but I'm curious about something..

 

[SammyS]

Homework Statement

When learning about quadratic equations in Langs book he gave the example:

let ax^2 + bx+ c = 0 be an equation

he subtracts the c, and says we want to get the left hand side (ax^2 + bx) = (x-s)^2

Homework Equations

The Attempt at a Solution

I'm just interested, either my conclusion that it doesn't matter which one you use, you will get the same solution, is wrong, or is this just a matter of convention for mathematicians to use negative numbers?

Sorry if this questions seems a bit useless but I'm curious about something..

Solving something like $\ (x-s)^2 = D \$ then gives $\ x=s\pm\sqrt{D} \$ rather than $\ x=-s\pm\sqrt{D} \$.

No biggy, but that's likely the reason.

Added in Edit:
I beat @fresh_42 to it, but he gives a better and more complete answer.

Added in Edit #2:
Also:
$\ (x-s)^2 = D \$ Is the equation of a parabola with a symmetry axis of $\ x=s\,.$

 

[fresh_42]

Homework Statement

When learning about quadratic equations in Langs book he gave the example:

let ax^2 + bx+ c = 0 be an equation

he subtracts the c, and says we want to get the left hand side (ax^2 + bx) = (x-s)^2

Homework Equations

The Attempt at a Solution

I'm just interested, either my conclusion that it doesn't matter which one you use, you will get the same solution, is wrong, or is this just a matter of convention for mathematicians to use negative numbers?

Sorry if this questions seems a bit useless but I'm curious about something..

Of course you can use both signs. The choice of a negative $s$ comes from the fact that the equation can be written as $ax^2+bx+c=a(x-s)(x-t)$. The numbers $s,t$ are called zeros of the equation. If you write $ax^2+bx+c=a(x+s)(x+t)$ then the zeros are instead of $x=s$ and $x=t$ at $x=-s$ and $x=-t$, which is confusing.

You can test it. Take e.g. $p(x)=4x^2 + 8x - 12$ then $p(1)=0$. Now divide $p(x)\, : \,(x-1)$. You won't have a remainder. So the zero is at $x=1$, the location at which $p(1)=0$. And $(x+1) \nmid p(x)\,.$ So whenever a polynomial $p(x)$ has a zero at $x=s \Longleftrightarrow x-s=0$, i.e. $p(s)=0$, then we have $(x-s) \,|\,p(x)$ without remainder. That's where the sign convention comes from.

 

[r0bHadz]

Added in Edit #2:
Also:
$\ (x-s)^2 = D \$ Is the equation of a parabola with a symmetry axis of $\ x=s\,.$

this is very good to know! cheers!

 

[r0bHadz]

Of course you can use both signs. The choice of a negative $s$ comes from the fact that the equation can be written as $ax^2+bx+c=a(x-s)(x-t)$. The numbers $s,t$ are called zeros of the equation. If you write $ax^2+bx+c=a(x+s)(x+t)$ then the zeros are instead of $x=s$ and $x=t$ at $x=-s$ and $x=-t$, which is confusing.

You can test it. Take e.g. $p(x)=4x^2 + 8x - 12$ then $p(1)=0$. Now divide $p(x)\, : \,(x-1)$. You won't have a remainder. So the zero is at $x=1$, the location at which $p(1)=0$. And $(x+1) \nmid p(x)\,.$ So whenever a polynomial $p(x)$ has a zero at $x=s \Longleftrightarrow x-s=0$, i.e. $p(s)=0$, then we have $(x-s) \,|\,p(x)$ without remainder. That's where the sign convention comes from.

cheers! very well explained mate.