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.