Relation Between, Quadratic Functions in the Form of Completing the Square And the Product of Binomials

[Ju.]

New?

 

[HallsofIvy]

Looks like basic information you could find in any secondary school algebra text. What was your purpose in posting this?

 

[Ju.]

Looks like basic information you could find in any secondary school algebra text. What was your purpose in posting this?

I just wondered if it was new. I could't fimd nothing anywhere describing this EXACT equation. Cuz if u have an equation that cannot be factored in the form of two binomials, then the equation ain't that obvious, or you just have some pretty complex solutions. My math eacher and a friend who's studying astrophysics didn't see it anywhere, so u know.
But does it only look like somethin, that can be found in high school algebra book? Or have you actually seen it in one?

 

[Dr.Peterson]

The only thing that might not be commonly found in textbooks is actually writing out the factored form containing the quadratic formula. We don't usually do that only because it makes things look more complicated than they are. It's much easier just to explain that if you find the zeros of [MATH]ax^2+bx+c[/MATH], say [MATH]p[/MATH] and [MATH]q[/MATH] (by any method, including the quadratic formula), then you can factor it as [MATH]a(x-p)(x-q)[/MATH].

But did you notice the error in one of your first lines?

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When you soon assume [MATH]a=1[/MATH] this is no longer a problem, but it's a major error in the general case.

 

[Ju.]

The only thing that might not be commonly found in textbooks is actually writing out the factored form containing the quadratic formula. We don't usually do that only because it makes things look more complicated than they are. It's much easier just to explain that if you find the zeros of [MATH]ax^2+bx+c[/MATH], say [MATH]p[/MATH] and [MATH]q[/MATH] (by any method, including the quadratic formula), then you can factor it as [MATH]a(x-p)(x-q)[/MATH].

But did you notice the error in one of your first lines?

View attachment 23229​

When you soon assume [MATH]a=1[/MATH] this is no longer a problem, but it's a major error in the general case.

Thank you for information. Still don't see error lol.

 

[Dr.Peterson]

Thank you for information. Still don't see error lol.

Do you not see that where the left side has ax^2, when you expand the right side it has just x^2? You need to multiply the right by a.

It was only later that you assumed a to be 1.