Quadratic equation -- question about the roots

[Waffle24]

Homework Statement

Consider the most general quadratic equation ax^2 +bx + c = 0 and suppose that the two solutions are x1 and x2
Now if x1 and x2are the roots of the equation then you can ‘work
backwards’ to generate the original equation.
A quadratic with the two solutions x1 and x2 is :
(x -x1)(x - x2) = 0

Question :
My question here is, how and why is the quadratic of the two solutions x1 and x2 : (x - x1)(x - x2) = 0, and not for example : (x + x1)(x + x2) = 0 or (x + x1)(x - x2) = 0 or (x - x1)(x + x2) = 0

Relevant Equations

ax^2 +bx + c = 0
(x -x1)(x - x2) = 0

Maybe there's a formula or rule that I'm not aware of...

 

[fresh_42]

First of all, $ax^2+bx+c \neq (x-x_1)(x-x_2)$. It is $ax^2+bx+c = a (x-x_1)(x-x_2)$.
As to your question: Why do you automatically assume $x_i > 0$? They can be everything, positive, negative or even zero. One writes them with a minus sign, because then they are the so called zeros of the polynomial. That means if $p(x)=ax^2+bx+c=a(x-x_1)(x-x_2)$ then $p(x_1)=0=p(x_2)$. If you wrote them as $x+x_1$ then $-x_1$ would be the zero. Since this would result in a mess, and because the zeros are important, the convention is to write the factors of $p(x)$ with a minus sign, such that the $x_i$ are the zeros of $p(x)$ without any confusing signs.

 

[fresh_42]

There is indeed a formula. Do you know the binomial formulas?
Try to write $p(x)=ax^2+bx+c=a(x-r)^2+s$ and then solve for $x$ in $a(x-r)^2+s=0$.

 

[PeroK]

Question :
My question here is, how and why is the quadratic of the two solutions x1 and x2 : (x - x1)(x - x2) = 0, and not for example : (x + x1)(x + x2) = 0 or (x + x1)(x - x2) = 0 or (x - x1)(x + x2) = 0
Relevant Equations: ax^2 +bx + c = 0
(x -x1)(x - x2) = 0

Maybe there's a formula or rule that I'm not aware of...

You might be able to answer your own question by considering why you wrote a general quadratic as:

$ax^2 + bx + c$

And not:

$ax^2 - bx - c$

Or some other variation of plusses and minuses.

 

[mathwonk]

the reason the equation with solutions x1 and x2 is not (X+x1)(X-x2) = 0, is that when you substitute X=x1 into the left side of this equation, you get 2x1.(x1-x2) which in general is not zero.

I.e. x1 and x2 are solutions of (X-x1)(X-x2) = 0, but they are in general not both solutions of any of your other equations.

("in general" above means as long as 0 ≠ x1 ≠ x2 ≠ 0).

 

[RPinPA]

Maybe there's a formula or rule that I'm not aware of...

There is indeed. It's given the fancy name of the "Zero Factor Theorem" but the idea is actually pretty obvious. If I'm multiplying two numbers and their product is 0, then at least one of them must be 0. Anything times 0 is 0. But the product of two nonzero numbers can't be 0.

and not for example : (x + x1)(x + x2) = 0

At $x = x_1$ the first factor is $2x_1$. If $x_1$ is nonzero, that factor is not zero.

The second factor at $x = x_1$ is $x_1 + x_2$. In general that isn't zero either except in the particular case $x_1 = -x_2$.

If $x_1$ is a root, there must be a factor that is 0 when $x = x_1$. The factor $x - x_1$ has that property.

 

[Merlin3189]

If there is a solution, $x = x_1$ then $(x -x_1) =0$
So $(x -x_1) =0$ is an equation with the root $x = x_1$ (there are others of course)
The negative sign must arise, because you change the solution into the equation by subtracting $x_1$ from both sides.

And if there is a solution $x=x_2$ then $(x -x_2) =0$ and this is an equation with root $x_2$

Now, because of the idea mentioned br RPinPA, we can multiply these expressions to produce an equation with two possible roots - if you multiply two numbers and get zero, then one of the two numbers must be zero.
$(x-x_1)(x-x_2) =0$

Or an equation with 3 possible roots $(x-x_1)(x-x_2)(x-x_3)=0$
etc

The expressions that you multiply to form these equations always have the form (x -root1)(x -root2)...(x-rootn) because they all come from the solutions x=root
and when you subtract root from both sides, that becomes x-root =0